Answer:
pH =1 2.84
Explanation:
First we have to start with the reaction between HCl and KOH:
[tex]HCl~+~KOH->~H_2O~+~KCl[/tex]
Now for example, we can use a volume of 10 mL of HCl. So, we can calculate the moles using the molarity equation:
[tex]M=\frac{mol}{L}[/tex]
We know that [tex]10mL=0.01L [/tex] and we have the concentration of the HCl [tex]0.282M[/tex], when we plug the values into the equation we got:
[tex]0.282M=\frac{mol}{0.01L}[/tex]
[tex]mol=0.282*0.01 [/tex]
[tex]mol=0.00282[/tex]
We can do the same for the KOH values ([tex]50mL=0.05L[/tex] and [tex]0.141M[/tex]).
[tex]0.141M=\frac{mol}{0.05L}[/tex]
[tex]mol=0.141*0.05 [/tex]
[tex]mol=0.00705[/tex]
So, we have so far 0.00282 mol of HCl and 0.00705 mol of KOH. If we check the reaction we have a molar ratio 1:1, therefore if we have 0.00282 mol of HCl we will need 0.00282 mol of KOH, so we will have an excess of KOH. This excess can be calculated if we substract the amount of moles:
[tex]0.00705-0.00282=0.00423mol~of~KOH[/tex]
Now, if we want to calculate the pH value we will need a concentration, in this case KOH is in excess, so we have to calculate the concentration of KOH. For this, we already have the moles of KOH that remains left, now we need the total volume:
[tex]Total~volume=50mL+10mL=60mL[/tex]
[tex]60mL=0.06L[/tex]
Now we can calculate the concentration:
[tex]M=\frac{0.00423mol}{0.06L}[/tex]
[tex]M=0.0705[/tex]
Now, we can calculate the pOH (to calculate the pH), so:
[tex]pOH=-Log(0.0705)[/tex]
[tex]pOH=1.15[/tex]
Now we can calculate the pH value:
[tex]14=~pH~+~pOH[/tex]
[tex]pH=14-1.15=12.84[/tex]
If 14 moles of oxygen react with 14 moles of hydrogen to produce water, what is the
limiting reactant?
A. oxygen
B. there is no limiting reactant
C. hydrogen
D. water
How many grams of o2 gas are contained in 890.0 ml at 21.0 degrees Celsius and 750.0 psi
Answer:
60.288 grams of O₂ gas are contained in 890.0 mL at 21.0 degrees Celsius and 750.0 psi.
Explanation:
Ideal gases are a simplification of real gases that is done to study them more easily. It is considered to be formed by point particles, do not interact with each other and move randomly. It is also considered that the molecules of an ideal gas, in themselves, do not occupy any volume.
The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:
P*V = n*R*T
where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas.
In this case:
P= 750 psi= 51.0345 atm (being 1 psi = 0.068046 atm)V= 890 mL= 0.890 L (being 1 L= 1,000 mL)n= ?R= 0.082 [tex]\frac{atm*L}{mol*K}[/tex]T= 21 °C= 294 °KReplacing:
51.0345 atm*0.890 L= n*0.082 [tex]\frac{atm*L}{mol*K}[/tex] *294 °K
Solving:
[tex]n=\frac{51.0345 atm*0.890 L}{0.082\frac{atm*L}{mol*K} *294 K}[/tex]
n=1.884 moles
Being 32 g / mole the molar mass of oxygen O₂, the following rule of three applies: if 1 mole contains 32 grams, 1,884 moles, how much mass does it have?
[tex]mass=\frac{1.884 moles*32 grams}{1mole}[/tex]
mass= 60.288 grams
60.288 grams of O₂ gas are contained in 890.0 mL at 21.0 degrees Celsius and 750.0 psi.
Answer:
The mass of the O2 gas is 60.16 grams
Explanation:
Step 1: Data given
Volume of O2 gas = 890.0 mL = 0.890 L
Temperature of the O2 gas = 21.0 °C = 294 K
Pressure = 750.0 psi = 51.03 atm
Step 2: Calculate moles of O2 gas
p*V = n*R*T
⇒with p = the pressure of the O2 gas = 51.03 atm
⇒with V = the volume = 0.890 L
⇒with R = the gas constant == 0.08206 L*atm/mol*K
⇒with T = the temperature = 294 K
⇒with n = the number of moles of O2 gas
n = (p*V) / (R*T)
n = (51.03 atm * 0.890 L) / (0.08206L*atm/mol*K * 294K)
n = 1.88 moles of O2
Step 3: Calculate the mass of O2 gas
Mass = moles * molar mass O2
Mass O2 gas = 1.88 moles * 32.0 g/mol
Mass O2 gas = 60.16 grams
The mass of the O2 gas is 60.16 grams
Suppose you are titrating an acid of unknown concentration with a standardized base. At the beginning of the titration, you read the base titrant volume as 1.12 mL. After running the titration and reaching the endpoint, you read the base titrant volume as 26.78 mL. What volume of base was required for the titration
Answer:
The volume of base required is = 25.66 mL
Explanation:
Volume of titrant required is given by:
Final titre value - Initial titre value
=26.78 - 1.12
= 25.66 mL
Note: repeat and average volumes for accuracy of result.
The formation of the silver(I) ammine complex ion is a reversible reaction that is allowed to reach equilibrium. For each subsequent change to the system, indicate how the concentration of each species in the chemical equation will change to reestablish equilibrium. An up arrow indicates an increase in concentration, a down arrow indicates a decrease in concentration and leaving t blank means there is no change in the concentration Ag+(aq) + 2NH3 (aq)-----------------> Ag(NH3 (aq) <----------------- increasing the concentration of Ag+ decreasing the concentration of NH increasing the concentration of Ag(NH)
Explanation:
Ag+(aq) + 2NH3 (aq) ⇄ Ag(NH3)2 (aq)
When it comes to question of this sort, the LeChatelier principle should come to mind. The LeChatelier principle states that changes in the temperature, pressure, volume, or concentration of a system will result in predictable and opposing changes in the system in order to achieve a new equilibrium state.
increasing the concentration of Ag+
This would lead to an increase in concentration of reactants, the system would annul or oppose this change by moving towrds the right, favouring the forward reaction.
decreasing the concentration of NH3
This would lead to a decrease in concentration of reactants, the system would annul this change by moving to the left, favourin the backward reaction.
increasing the concentration of Ag(NH3)2
This would lead to an increase in concnetration of products, the system would annul or oppose this change by moving towards the left, favouring the backward reaction.
which would you think would be a stronger interaction and why: an interaction between a sodium ion and the prtial negative charge on the oxygen in ethanol, or the interaction between two ethanol molecules
Answer:
The interaction between sodium ion and the partial negative charge on the oxygen is stronger.
Explanation:
The predominant interaction that exists between sodium ion and the partial negative charge on the oxygen is ion-dipole interaction.
The predominant interaction that exists between two ethanol molecules is hydrogen bonding interaction.
The order of strength of the intermolecular interactions in decreasing order:
Ionic bond > ion-dipole interaction > hydrogen bonding > dipole-dipole interaction > ion-induced dipole interaction > induced dipole- dipole interaction > london force
So, ion-dipole interaction is stronger than hydrogen bonding.
Hence, the interaction between sodium ion and the partial negative charge on the oxygen is stronger.
Matter does NOT exist in which form?
Answer:
vacuum
Explanation:
vacuum is the form by which matter cant exist
what is a molecule containing only carbon and hydrogen called?
Answer:
d
Explanation:
it's called the hydrocarbon
Answer:
A hydrocarbon
Explanation:
The following initial rate data are for the reaction of mercury(II) chloride with oxalate ion: 2 HgCl2 + C2O42- 2 Cl- + Hg2Cl2 + 2 CO2
Experiment [HgCl2] [C2O42-]o, M Initial Rate
1 0.124 0.115 1.61E-5
2 0.248 0.015 3.23E-5
3 0.124 0.229 6.40E-5
4 0.248 0.229 1.28E-4
1) Complete the rate law for this reaction in the box below. Use the form k[A]m[B]n , where '1' is understood for m or n and concentrations taken to the zero power do not appear. Don't enter 1 for m or n.
2) From these data, the rate constant is ____ M-2s-1?
Answer:
Explanation:
2 HgCl₂ + C₂O₄²⁻ = 2 Cl⁻ + Hg₂Cl₂ + 2CO₂
1 )
Rate of reaction [tex]= k[HgCl_2]^m[C_2O_4^{-2}]^n[/tex]
[HgCl₂] [C₂O₄²⁻ ] Rate
1 . .124 .115 1.61 x 10⁻⁵
2 . .248 .115 3.23 x 10⁻⁵
3 . .124 .229 6.4 x 10⁻⁵
4 . .248 .229 1.28 x 10⁻⁴
comparing 1 and 3 , when concentration of HgCl₂ remains constant and concentration of C₂O₄²⁻ becomes twice , rate becomes 4 times so rate is proportional to square of concentration of C₂O₄²⁻ .
Hence n = 2
comparing 1 and 2 , when concentration of HgCl₂ becomes twice and concentration of C₂O₄²⁻ remains constant , rate becomes 2 times so rate is proportional to simply concentration of C₂O₄²⁻ .
Hence m = 1
Putting the data of 1 in the rate equation found
1.61 x 10⁻⁵ = k x .124 x .115²
k = 11.3 x 10⁻⁴ M⁻² s⁻¹
Lithium reacts more slowly with cold water than sodium.
State two ways the reaction can be made to go faster.
Answer:
1. Increase in the temperature of the water
2. Increasing the surface area of the lithium
Explanation:
1. Increase in the temperature of the water
The activation energy for the lithium water reaction is +161 kJ/mol while the activation energy for the sodium is +109 kJ/mol, hence for increased reaction rate, the water temperature will be raised to enable more lithium atoms enter into reaction with the water molecules as their energy is increased lowering the activation energy required for the reaction.
2. Increasing the surface area of the lithium
As the lithium floats on the water, due to its low temperature and the heat evolved from the reaction of lithium with the cold water is below the melting point of lithium, the reaction rate can be increased by increasing the surface area of lithium sample by grinding so as to increase the number of lithium water reaction sites.
The tabulated data were collected for this reaction:
CH3Cl(g) + 3Cl2(g) → CCl4(g) + 3HCl(g)
[CH3Cl] (M) [Cl2] (M) Initial Rate (M/s)
0.050 0.050 0.014
0.100 0.050 0.029
0.100 0.100 0.041
0.200 0.200 0.115
(a) Write an expression for the reaction rate law and calculate the value of the rate constant, k.
(b) What is the overall order of the reaction?
Answer:
ai) Rate law, [tex]Rate = k [CH_3 Cl] [Cl_2]^{0.5}[/tex]
aii) Rate constant, k = 1.25
b) Overall order of reaction = 1.5
Explanation:
Equation of Reaction:
[tex]CH_{3} Cl (g) + 3 Cl_2 (g) \rightarrow CCl_4 (g) + 3 HCl (g)[/tex]
If [tex]A + B \rightarrow C + D[/tex], the rate of backward reaction is given by:
[tex]Rate = k [A]^{a} [B]^{b}\\k = \frac{Rate}{ [A]^{a} [B]^{b}}\\k = \frac{Rate}{ [CH_3 Cl]^{a} [Cl_2]^{b}}[/tex]
k is constant for all the stages
Using the information provided in lines 1 and 2 of the table:
[tex]0.014 / [0.05]^a [0.05]^b = 00.029/ [0.100]^a [0.05]^b\\0.014 / [0.05]^a [0.05]^b = 00.029/ [2*0.05]^a [0.05]^b\\0.014 / = 0.029/ 2^a\\2^a = 2.07\\a = 1[/tex]
Using the information provided in lines 3 and 4 of the table and insering the value of a:
[tex]0.041 / [0.100]^a [0.100]^b = 0.115 / [0.200]^a [0.200]^b\\0.041 / [0.100]^a [0.100]^b = 0.115 / [2 * 0.100]^a [2 * 0.100]^b\\[/tex]
[tex]0.041 = 0.115 / [2 ]^a [2]^b\\ \[[2 ]^a [2]^b = 0.115/0.041\\ \[[2 ]^a [2]^b = 2.80\\\[[2 ]^1 [2]^b = 2.80\\\[[2]^b = 1.40\\b = \frac{ln 1.4}{ln 2} \\b = 0.5[/tex]
The rate law is: [tex]Rate = k [CH_3 Cl] [Cl_2]^{0.5}[/tex]
The rate constant [tex]k = \frac{Rate}{ [CH_3 Cl]^{a} [Cl_2]^{b}}[/tex] then becomes:
[tex]k = 0.014 / ( [0.050] [0.050]^(0.5) )\\k = 1.25[/tex]
b) Overall order of reaction = a + b
Overall order of reaction = 1 + 0.5
Overall order of reaction = 1.5
The rate law is for this reaction is [tex]r = 1.25 M^{-0.5} s^{-1} [CH_3Cl] [Cl_2]^{0.5}[/tex]
where the rate constant is k = [tex]1.25 M^{-0.5} s^{-1}[/tex] and the overall order of the reaction is 1.5.
Let's consider the following reaction.
CH₃Cl(g) + 3 Cl₂(g) → CCl₄(g) + 3 HCl(g)
What is the rate law?The rate law for a chemical reaction is an equation that relates the reaction rate with the concentrations of the reactants.
The rate law for this reaction is:
r = k [CH₃Cl]ᵃ [Cl₂]ᵇ
where,
r is the initial rate.k is the rate constant.a is the reaction order for CH₃Cl.b is the reaction order for Cl₂.If we write the ratio r₂/r₁, we get:
r₂/r₁ = k [CH₃Cl]₂ᵃ [Cl₂]₂ᵇ / k [CH₃Cl]₁ᵃ [Cl₂]₁ᵇ
r₂/r₁ = {[CH₃Cl]₂/ [CH₃Cl]₁}ᵃ
(0.029 M/s)/(0.014 M/s) = {0.100M/0.050 M}ᵃ
a ≈ 1
If we write the ratio r₃/r₂, we get:
r₃/r₂ = k [CH₃Cl]₃ᵃ [Cl₂]₃ᵇ / k [CH₃Cl]₂ᵃ [Cl₂]₂ᵇ
r₃/r₂ = {[Cl₂]₃/[Cl₂]₂}ᵇ
(0.041 M/s)/(0.029 M/s) = {0.100 M/0.050 M}ᵇ
b ≈ 0.5
So far, the rate law is:
[tex]r = k [CH_3Cl] [Cl_2]^{0.5}[/tex]
Let's use the values of the first experiment to find the value of the rate constant.
[tex]k = \frac{r}{[CH_3Cl][Cl_2]^{0.5} } = \frac{0.014 M/s}{(0.050 M)(0.050 M)^{0.5} } = 1.25 M^{-0.5} s^{-1}[/tex]
The final rate law is:
[tex]r = 1.25 M^{-0.5} s^{-1} [CH_3Cl] [Cl_2]^{0.5}[/tex]
What is the overall order of the reaction?It is the sum of the individual orders of reaction.
a + b = 1 + 0.5 = 1.5
The rate law is for this reaction is [tex]r = 1.25 M^{-0.5} s^{-1} [CH_3Cl] [Cl_2]^{0.5}[/tex]
where the rate constant is k = [tex]1.25 M^{-0.5} s^{-1}[/tex] and the overall order of the reaction is 1.5.
Learn more about the rate law here: https://brainly.com/question/14945022
Select the statement which explains the meaning of the symbol 6d5.
A. The principal quantum number (n) is 6, the angular momentum quantum number (ell) is 5, and there are 2 electrons in the subshell.
B. The principal quantum number (n) is 6, the angular momentum quantum number (ell) is 2, and there are 5 electrons in the subshell.
C. The principal quantum number (n) is 5, the angular momentum quantum number (ell) is 5, and there are 5 electrons in the subshell.
D. The principal quantum number (n) is 5, the angular momentum quantum number (ell) is 4, and there are 4 electrons in the subshell.
Answer:
B. The principal quantum number (n) is 6, the angular momentum quantum number (ell) is 2, and there are 5 electrons in the subshell.
Explanation:
Given;
electronic configuration of 6d⁵
where;
6 in the configuration shows that, the principal quantum number (n) is 6
d is the subshell
d orbital has angular momentum quantum number (l) of 2
s orbital, l = 0
p orbital, l = 1
d orbital, l = 2
f orbital, l = 3
5 in the configuration means that there are 5 electrons in the subshell
Therefore, 6d⁵ symbol means that "The principal quantum number (n) is 6, the angular momentum quantum number (ell) is 2, and there are 5 electrons in the subshell".
A 2.300×10^−2 M solution of glycerol (C3H8O3) in water is at 20.0°C. The sample was created by dissolving a sample of C3H8O3 in water and then bringing the volume up to 1.000 L. It was determined that the volume of water needed to do this was 999.0 mL . The density of water at 20.0°C is 0.9982 g/mL.
Required:
a. Calculate the molality of the glycerol solution.
b. Calculate the mole fraction of glycerol in this solution.
c. Calculate the concentration of the glycerol solution in percent by mass.
d. Calculate the concentration of the glycerol solution in parts per million.
Answer:
a) Based on the given information, the volume of the solvent given is 999 ml and the density of water given is 0.9982 gram per ml.
The mass of solvent can be calculated by the formula mass = density * volume
mass = 0.9982 * 999 = 0.997 Kg
The molarity of the solution given is 2.300 * 10^-2 M
Molality of the glycerol solution can be calculated by using the formula,
Molality = molarity/solvent (kg) = 2.300 * 10^-2 / 0.997 = 0.023 m
b) Molarity or the moles of the solute given is 2.300 * 10^-2 moles
The moles of solvent can be determined by using the formula, n = mass of solvent/mol.wt = 997/18 (mol.wt of solvent is 18 g/mol), now putting the values we get,
n = 997/18 = 55.4
The mole fraction of the glycerol will be = 2.300 * 10^-2 M/(2.300 * 10^-2)+55.40
= 4.15 * 10^-4
c) The mass percent of glycerol can be determined by using the formula,
mass of solute/mass of solution * 100% ----- (i)
The mass of solute can be determined by using the formula,
n = mass of solute/mol.wt of solute
The n or the no. of moles is 2.300 * 10^-2 and the molecular weight of glycerol is 92.09 g/mol. Now putting the values we get,
mass = 2.300 * 10^-2 * 92.09 = 2.12 grams
Now putting the values in equation (i) we get,
mass percent = 2.12 / 997 * 100% = 0.21%
d) Based on the above calculation, the mass of solute (glycerol) is 2.12 g or 2.12 * 1000 mg
The volume of water is 999 ml or 999 * 10^-3 L
The concentration of the glycerol solution will be,
Concentration = 2.12 * 10^3 mg/999 * 10^-3 L
= 2.12/999 * 10^6 mg/L
= 2122.1 ppm
Considering the solution of molality, mole fraction, mass percentage and ppm, you obtain that:
a) The molality of the glycerol solution is 0.02306 molal.
b) The mole fraction of glycerol in the solution is 4.15×10⁻⁴.
c) The percent by mass is 0.212%.
d) The concentration of the glycerol solution is 2118.07 ppm.
a. MolalityMolality is the ratio of the number of moles of any dissolved solute to kilograms of solvent.
The Molality of a solution is determined by the expression:
[tex]molality=\frac{number of moles of solute}{kilogramsof solvent}[/tex]
In this case, you have a 2.300×10⁻² M solution of glycerol (C₃H₈O₃) in water. The sample was created by dissolving a sample of C₃H₈O₃ in water and then bringing the volume up to 1.000 L.
So, being the molarity the number of moles of solute that are dissolved in a certain volume, the number of moles of glycerol can be calculated as:
number of moles of glycerol= 2.300×10⁻² M× 1 L
number of moles of glycerol= 2.300×10⁻² moles
On the other side, the volume of water needed was 999 mL and the density of water at 20.0∘C is 0.9982 [tex]\frac{g}{mL}[/tex]. So, the mass of water needed can be calculated as:
999 mL×0.9982[tex]\frac{g}{mole}[/tex] = 997.2 grams of water= 0.9972 kg of water
Then, the molality of the solution is:
[tex]molality=\frac{2.300x10^{-2} moles}{0.9972 kg}[/tex]
molality= 0.02306 molal
Finally, the molality of the glycerol solution is 0.02306 molal.
b. Mole fractionThe molar fraction is a way of measuring the concentration that expresses the proportion in which a sustance is found with respect to the total moles of the solution.
In this case, the number of moles of glycerol= 2.300×10⁻² moles
Having 997.2 grams of water, the moles of solvent can be determined knowing that the molar mass of water is 18[tex]\frac{g}{mole}[/tex]:
[tex]number of moles of water=997.2 gramsx\frac{1 mole}{18 grams}[/tex]
number of moles of water= 55.4 moles of water
Being the number of total moles the sum of the moles of grycerol and the numbers of moles of water, the mole fraction can be calculated as:
[tex]mole fraction=\frac{2.300x10^{-2} moles}{2.300x10^{-2}moles+55.4 moles}[/tex]
mole fraction= 4.15×10⁻⁴
In summary, the mole fraction of glycerol in the solution is 4.15×10⁻⁴.
c. Percent by massThe Percentage Composition is a measure of the amount of mass that an element occupies in a compound and indicates the percentage by mass of each element that is part of a compound.
The mass percentage of a component of the solution is defined as the ratio of the mass of the solute to the mass of the solution, expressed as a percentage.
The mass percentage is calculated as the mass of the solute divided by the mass of the solution, the result of which is multiplied by 100 to give a percentage. This is:
[tex]percent by mass=\frac{mass of solute}{mass of solution}x100[/tex]
In this case, remmeber that you have a number of moles of glycerol of 2.300×10⁻² moles .
Being the molar mass glycerol 92.09 [tex]\frac{g}{mole}[/tex], the mass of glycerol can be calculated as:
2.300×10⁻² moles×92.09 [tex]\frac{g}{mole}[/tex]= 2.11807 grams of glycerol
Remembering that you have 997.2 grams of water, the percent by mass is calculated as:
[tex]percent by mass=\frac{mass of glycerol}{mass of glycerol + mass of water}x100[/tex]
Solving:
[tex]percent by mass=\frac{2.11807 grams}{2.11807 grams + 997.2 grams}x100[/tex]
[tex]percent by mass=\frac{2.11807 grams}{999.31807 grams}x100[/tex]
percent by mass= 0.212%
Finally, the percent by mass is 0.212%.
d. Parts per millionParts per million (ppm) is a unit of measurement of concentration that measures the number of units of substance in each million units of the whole. In this case, the concentration measurement refers to mg of glycerol per L of solution.
Being the mass of glycerol 2.11807 grams equal to 2118.07 mg (1 g=1000mg), the concentration is:
[tex]concentration=\frac{2118.07 mg}{1 L solution}[/tex]
Solving:
concentration= 2118.07 ppm
In summary, the concentration of the glycerol solution is 2118.07 ppm.
Learn more about:
molality: brainly.com/question/20366625?referrer=searchResults brainly.com/question/4580605?referrer=searchResultsmole fraction: brainly.com/question/14434096?referrer=searchResults brainly.com/question/10095502?referrer=searchResultsmass percentage: brainly.com/question/19168984?referrer=searchResults brainly.com/question/18646836?referrer=searchResults brainly.com/question/24201923?referrer=searchResults brainly.com/question/9779410?referrer=searchResults brainly.com/question/17030163?referrer=searchResultsppm: brainly.com/question/16727593?referrer=searchResults brainly.com/question/13565240?referrer=searchResultsWhich element is undergoing oxidation?
1) Ag
2) Ag+
3) Pb
4)Pb+2
Answer:
Explanation:
Oxidation In chemistry is the gain in oxygen, loss of hydrogen and loss of electrons. Oxidation occur when an element with gain oxygen, loss hydrogen or loss electrons, which means the element is oxidized. Compared to reduction which is loss ofoxygen, gain of hydrogen and gain of electrons.
From the the question , Ag( silver) is undergoing oxidation.
A 3.00g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of 192./gmol, is burned completely in excess oxygen, and the mass of the products carefully measured: product mass carbon dioxide 4.13g water 1.13g Use this information to find the molecular formula of X.
Answer: The molecular formula for the given organic compound X is [tex]C_6H_{8}O_7[/tex]
Explanation:
We are given:
Mass of [tex]CO_2=4.13g[/tex]
Mass of [tex]H_2O=1.13g[/tex]
We know that:
Molar mass of carbon dioxide = 44 g/mol
Molar mass of water = 18 g/mol
For calculating the mass of carbon:
In 44g of carbon dioxide, 12 g of carbon is contained.
So, in 4.13 g of carbon dioxide, =[tex]\frac{12}{44}\times 4.13=1.13g[/tex] of carbon will be contained.
For calculating the mass of hydrogen:
In 18g of water, 2 g of hydrogen is contained.
So, in 1.13 g of water, [tex]\frac{2}{18}\times 1.13=0.125g[/tex] of hydrogen will be contained.
Mass of oxygen in the compound = (3.00) - (1.13+ 0.125) = 1.75 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.
Moles of Carbon =[tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.13g}{12g/mole}=0.094moles[/tex]
Moles of Hydrogen =[tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.125g}{1g/mole}=0.125moles[/tex]
Moles of Oxygen =[tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.75g}{16g/mole}=0.109moles[/tex]
Step 2: Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles
For Carbon = [tex]\frac{0.094}{0.094}=1[/tex]
For Hydrogen = [tex]\frac{0.125}{0.094}=1.33[/tex]
For Oxygen = [tex]\frac{0.109}{0.094}=1.16[/tex]
Step 3: Taking the mole ratio as their subscripts.
The ratio of C : H : O = 1: 1.33: 1.16
Converting them into whole number ratios by multiplying by 6:
The ratio of C : H : O = 6: 8: 7
Hence, the empirical formula for the given compound is [tex]C_6H_8O_7[/tex]
Empirical mass = [tex]6\times 12+8\times 1+7\times 16=192g[/tex]
The equation used to calculate the valency is :
[tex]n=\frac{\text{molecular mass}}{\text{empirical mass}}[/tex]
Putting values in above equation, we get:
[tex]n=\frac{192g/mol}{192g/mol}=1[/tex]
Multiplying this valency by the subscript of every element of empirical formula, we get:
[tex]C_6H_8O_7\times 1=C_6H_{8}O_7[/tex]
Thus molecular formula for the given organic compound X is [tex]C_6H_{8}O_7[/tex]
Three apprentice tailors (X, Y, and Z) are assigned the task of measuring the seam of a pair of trousers. Each one makes three measurements. The results in inches are X (26.5, 26.6, 26.4); Y (27.8, 27.3, 27.7); Z (26.9, 27.2, 27.1). The true length is 27.0 in. Comment on the precision and the accuracy of each tailor's measurements.
Answer:
Explanation:
Precision generally refers to the repeatability of measurements. That is, the closeness of repeated meaurements in values. Accuracy, on the other hand, refers to the closeness of a measured value to the true value.
For apprentice X, the three measurements taken (26.5, 26.6, 26.4) were close in values with an average of 26.5 and a mean deviation of 0.07. Hence, the apprentice is said to be precise in measurement. However, the average of the measurement is quite far from the true length of meaurement. Hence, the apprentice is inaccurate.
Apprentice Y had a mean measurement value of 27.6 and a mean deviation of 0.2. The measurements are far from each other and the apprentice is therefore said to be far from being precise. The mean measurement value is also far from true value. Hence the apprentice is also inaccurate in his/her measurement.
Apprentice Z has a mean measurement of 27.07 and a mean deviation of 0.11. The repeated measurements were close enough to be precise while the average is also close to the true length, and hence, the measurement is said to be accurate.
Find pH for lemon juice pH= -log[0.00500]
Answer:
2.3
Explanation:
Your calculator can evaluate this expression. It tells you the pH of lemon juice is 2.3.
A 25.0g sample of brass, which has a specific heat capacity of 0.375·J·g−1°C−1, is dropped into an insulated container containing 250.0g of water at 25.0°C and a constant pressure of 1atm. The initial temperature of the brass is 96.7°C. Assuming no heat is absorbed from or by the container, or the surroundings, calculate the equilibrium temperature of the water. Be sure your answer has 3 significant digits.
Answer:
The equilibrium temperature of water is 25.6 °C
Explanation:
Step 1: Data given
Mass of the sample of brass = 25.0 grams
The specific heat capacity = 0.375 J/g°C
Mass of water = 250.0 grams
Temperature of water = 25.0 °C
The initial temperature of the brass is 96.7°C
Step 2: Calculate the equilibrium temperature
Heat lost = heat gained
Q(sample) = -Q(water)
Q = m*C* ΔT
m(sample)*c(sample)*ΔT(sample) = - m(water)*c(water)*ΔT(water)
⇒m(sample) = the mass of the sample of brass = 25.0 grams
⇒with c(sample) =The specific heat capacity = 0.375 J/g°C
⇒with ΔT = the change of temperature = T2 - T1 =T2 - 96.7 °C
⇒with m(water) = the mass of the water = 250.0 grams
⇒with c(water) = the specific heat capacity = 4.184 J/g°C
⇒with ΔT(water) = the change of temperature of water = T2 - T1 = T2 - 25.0°C
25.0 * 0.375 * (T2 - 96.7) = - 250.0 * 4.184 J/g°C * (T2 - 25.0°C)
9.375T2 - 906.56 = -1046T2 + 26150
9.375T2 + 1046T2 = 26150 + 906.56
1055.375T2 = 27056.26
T2 = 25.6 °C
The equilibrium temperature of water is 25.6 °C
The standard heat of combustion of ethanol, C2H5OH, is 1372 kJ/mol ethanol. How much heat (in kJ) would be liberated by completely burning a 20.0 g sample
Answer:
The correct answer is 596.5 kJ.
Explanation:
The mass of ethanol or C2H5OH mentioned in the question is 20 gm.
The molar mass of ethanol is 46 g/mol.
The moles of the compound can be determined by using the formula,
n = weight of the compound/molar mass
= 20/46 = 0.435 moles
It is mentioned in the question that standard heat of combustion of ethanol is 1372 kJ/mole, that is, one mole of ethanol is producing 1372 kilojoules of energy at the time of combustion.
Therefore, the energy liberated by completely burning the 20 grams of ethanol is 0.435*1372 = 596.5 kJ.
Since the standard heat of combustion of ethanol is -1372 kJ/mol, the heat released by the combustion of a 20.0 g sample is -593 kJ.
Let's consider the thermochemical equation for the combustion of ethanol.
C₂H₅OH(l) + 3 O₂(g) → 2 CO₂(g) + 3 H₂O(l) ΔH°c = -1372 kJ/mol
We can calculate the heat released by burning 20.0 g of ethanol considering the following relationships.
The molar mass of C₂H₅OH is 46.07 g/mol.1372 kJ of heat is released by the combustion of 1 mole of C₂H₅OH.[tex]20.0 g EtOH \times \frac{1molEtOH}{46.07gEtOH} \times \frac{(-1367kJ)}{1molEtOH} = -593 kJ[/tex]
Since the standard heat of combustion of ethanol is -1372 kJ/mol, the heat released by the combustion of a 20.0 g sample is -593 kJ.
Learn more: https://brainly.com/question/2874342
The standard heat of combustion of ethanol, C₂H₅OH, is -1372 kJ/mol ethanol. How much heat (in kJ) would be liberated by completely burning a 20.0 g sample.
is the total disappearance of all members of a species.
Answer: EXTINCTION
Explanation:
Answer:
extinction
Explanation:
A sample of thionylchloride, SOCl2, contains 0.206 mol of the compound. What is the mass of the sample, in grams?
Explanation:
Moles=mass/molar mass
moles × molar mass = mass
0.206 x 119= mass
Mass= 24. 51grams
Answer:
24.51 g to the nearest hundredth
Explanation:
1 mole of SOCl2 = 32 + 16 + 2 *35.5 = 119 g
0.206 mol = 0.206 * 119
= 24.51 g.
If methyl orange is added to a substance and it turns the substance red, what does this tell you about the pH of the substance? Would this substance be an acid or a base?
Explanation:
methyl orange turns pink in an acidic solution
yellow in basic and orange in neutral
Answer: The substance would be a strong acid.
Explanation:
Methyl orange turns orange when it is added to a medium acid and turns red when it is added to a strong acid. The pH of the acid would be above 4 because it is a strong acid. Methyl orange turns yellow when it is added to a medium base.
How many moles of oxygen in 3.5 Mols of CH2O
How many moles of iron(lll) sulfide, Fe2S3, would be produced from the complete reaction of 449 g iron(lll) bromide, FeBr3?
Answer:
.76
Explanation:
f 23.6 mL of 0.200 M NaOH is required to neutralize 10.00 mL of a H3PO4 solution , what is the concentration of the phosphoric acid solution?Start by balancing the equation for the reaction: H3PO4(aq) + NaOH(aq) → Na3PO4(aq) + H2O(l)
Answer:
Explanation:
H3PO4(aq) + 3NaOH(aq) → Na3PO4(aq) + 3H2O(l)
mole of NaOH = 23.6 * 10 ⁻³L * 0.2M
= 0.00472mole
let x be the no of mole of H3PO4 required of 0.00472mole of NaOH
3 mole of NaOH required ------- 1 mole of H3PO4
0.00472mole of NaOH ----------x
cross multiply
3x = 0.0472
x = 0.00157mole
[H3PO4] = mole of H3PO4 / Vol. of H3PO4
= 0.00157mole / (10*10⁻³l)
= 0.157M
The concentration of unknown phosphoric acid is 0.157MAnswer:
[tex]M_{H_3PO_4}=0.157M[/tex]
Explanation:
Hello,
In this case, the balanced chemical reaction is:
[tex]H_3PO_4(aq) + 3NaOH(aq) \rightarrow Na_3PO_4(aq) + 3H_2O(l)[/tex]
Therefore, we compute the moles of used NaOH by the 0.2000-M solution:
[tex]n_{NaOH}=0.200\frac{mol}{L}*0.0236L=4.72x10^{-3}molNaOH[/tex]
Then, we compute the moles of neutralized phosphoric acid by their 1:3 molar ratio:
[tex]n_{H_3PO_4}=4.72x10^{-3}molNaOH*\frac{1molH_3PO_4}{3molNaOH}=1.57x10^{-3}molH_3PO_4[/tex]
Finally, the concentration:
[tex]M_{H_3PO_4}=\frac{1.57x10^{-3}molH_3PO_4}{0.010L}\\ \\M_{H_3PO_4}=0.157M[/tex]
Best regards.
Since some amino acids, such as Glu and His, can be viewed as weak acids or bases, the ratio of their protonated form to their deprotonated form is dependent on pH. Furthermore, you can use this knowledge to predict whether a given amino acid will exist predominantly in its protonated or deprotonated form at a given pH. To do this, it is easier to think about the pK, of the amino acid functional group. Recall from general chemistry that the pK, is equal to -log(K). Using the K, value for the Glu side chain, which is 5.62 x 10-5, calculate the pK, for the Glu side chain. pKa =
Answer:
The pKa of Glu is [tex]pK_a = 4.25[/tex]
Explanation:
From the question we are told that
The [tex]K_a[/tex] value of Glu is [tex]K_a = 5.62*10^{-5}[/tex]
The [tex]pKa[/tex] is mathematically represented as
[tex]pK_a = -log (K_a )[/tex]
substituting values
[tex]pK_a = -log (5.62 *10^{-5} )[/tex]
[tex]pK_a = 4.25[/tex]
what did thomson’s model of the atom called
Answer:
Plum pudding model
Explanation:
Thomson’s model of the atom called: PLUM PUDDING MODEL
How many grams of H2 are needed to produce 12.50g of NH3?
Answer:2.2059
Explanation:find their total ram is(14+(1×3))
Find the portion occupied by
H2 which is three
Divide h2 by total ram then multiply by mass
Gaseous methane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water . If of carbon dioxide is produced from the reaction of of methane and of oxygen gas, calculate the percent yield of carbon dioxide. Round your answer to significant figures.
The question is incomplete, the complete question is;
Gaseous methane (CH4) reacts with gaseous oxygen gas (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). If 28.2 g of carbon dioxide is produced from the reaction of 15.1 g of methane and 81.2 g of oxygen gas, Calculate the percent yield of carbon dioxide. Be sure your answer has the correct number of significant digits in it.
Answer:
71.1%
Explanation:
The balanced reaction equation must first be written;
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g)
Let us obtain the number of moles of carbon dioxide corresponding to 28.2 g
Number of moles = mass/ molar mass= 28.2/44.01 g/mol = 0.64 moles of CO2
Next, we obtain the limiting reactant. This is the reactant that yields the least amount of product.
For methane;
Number of moles in 15.1 g= mass/molar mass= 15.1/16gmol-1 = 0.9 moles
From the reaction equation;
1 mole of methane yields 1 mole of carbon dioxide
Hence 0.9 moles of methane yields 0.9 moles of carbon dioxide.
For oxygen
Number of moles of oxygen corresponding to 81.2 g of oxygen= mass/ molar mass= 81.2g/32gmol-1 = 2.5 moles of oxygen
From the reaction equation;
2 moles of oxygen gas yields 1 mole of carbon dioxide
2.5 moles of oxygen gas yields 2.5 × 1 /2 = 1.25 moles of carbon dioxide.
Methane is the limiting reactant.
Theoretical yield of carbon dioxide= 0.9 moles
Actual yield of carbon dioxide= 0.64 moles
% yield = actual yield/ theoretical yield × 100
% yield= 0.64/0.9 ×100
% yield = 71.1%
Which part of a hypothesis should the student add to this that
could also be tested in other experiments?
an explanation based on prior scientific knowledge or
observations
an explanation of how the hypothesis will inspire new
questions
O an independent variable that the student controls during
the experiment
O an independent variable that is affected by the other
experimental variable
Answer:
I believe the answer would be: D
an independent variable that is affected by the other
experimental variable
Correct me if im wrong
Explanation:
Oxygen is a non-metal true or false
Answer:
This is true because oxygen belongs to the non-metal part of the periodic table.
Answer:hello
Oxygen is a non-metal it's exactly true.
Explanation:
There's a good rule that if the number of electrons in the last layer is 1,2,3,and sometimes 4 it's a metal but if this number increase it's a non-metal like Oxygen or Nitrogen.
Good luck.
