Answer:
Option C
Explanation:
Consider the ionic equation of this chemical equation. We are given barium chloride and silver nitrate as the reactants, and silver chloride and barium nitrate as the products. We can thus conclude that the ionic equation ( not balanced yet ) should be as follows -
Ba( 2 + ) + Cl ( - ) + Ag ( + ) + NO3 ( - ) ------> AgCl + Ba( 2 + ) + NO3( - )
As you can see these compounds are present in aqueous solutions, and are thus dissociated.
______________________________________________________
Now let us take a look at the number of elements on the reactant and product sides, and balance this chemical equation out -
Ba( 2 + ) + 2Cl ( - ) + 2Ag ( + ) + 2NO3 ( - ) ------> 2AgCl + Ba( 2 + ) + 2NO3( - )
Solution = Option C!
what is a molecule containing only carbon and hydrogen called?
Answer:
d
Explanation:
it's called the hydrocarbon
Answer:
A hydrocarbon
Explanation:
Calculate the free energy change for the reaction:
2 NO(g) + O2(g) → 2 NO2(g)
(A) - 9.3 kcal
(B) + 24.9 kcal
(C) + 9.3 kcal
(D) - 16.6 kcal
(E) + 16.6 kcal
Answer:
(D) - 16.6 kcal
Explanation:
Hello,
In this case, the Gibbs free energy for the given reaction is computed in terms of the Gibbs free energy of formation of each species involved in the chemical reaction:
[tex]\Delta _rG=2\Delta _fG_{NO_2}-2\Delta _fG_{NO}-\Delta _fG_{O_2}[/tex]
Thus, it is found for nitrogen monoxide, oxygen and nitrogen dioxide the following Gibbs free energies of formation: 87.6, 0 and 51.3 kJ/mol respectively, therefore we compute:
[tex]\Delta _rG=2(51.3kJ/mol)-2(87.6kJ/mol)=-72.6kJ*\frac{1kcal}{4.184kJ} \\\\\Delta _rG=-17.35kcal[/tex]
The closest result is (D) - 16.6 kcal, as such difference is noticed when different sources for thermochemical data are used, in this case, the NIST data were used.
Best regards.
. In which reaction is nitric acid acting as an oxidising agent?
A. CuO + 2HNO3 → Cu(NO3)2 + H2O
B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2
C. Na2CO3 + 2HNO3 → 2NaNO3 + H2O + CO2
D. NaOH + HNO3 → NaNO3 + H2O. In which reaction is nitric acid acting as an oxidising agent?
A. CuO + 2HNO3 → Cu(NO3)2 + H2O
B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2
C. Na2CO3 + 2HNO3 → 2NaNO3 + H2O + CO2
D. NaOH + HNO3 → NaNO3 + H2O
Answer:
B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2
Explanation:
Hello,
In this case, we should understand oxidizing agents as those substances able to increase the oxidation state of another substance, therefore, in B. reaction we notice that copper oxidation state at the beginning is zero (no bonds are formed) and once it reacts with nitric acid, its oxidation states raises to +2 in copper (II) nitrate, thus, in B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2 nitritc acid is acting as the oxidizing agent.
Moreover, in the other reactions, copper (A.), sodium (C. and D.) remain with the same initial oxidation state, +2 and +1 respectively.
Regards.
Answer:
B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2
Explanation:
A substance is acting as oxidising agent when is reducing its chemical state in the reaction.
In HNO3, Nitrogen has an oxidation state of +5 (Each oxygen is -2, 3 oxygens, -6, 1 hydrogen, +1). Thus, if the oxidation state of the nitrogen in the products is < +5, the nitrogen is reducing its oxidation state acting as oxidising agent.
In the reactions:
A. CuO + 2HNO3 → Cu(NO3)2 + H2O . In Cu(NO₃)₂, Nitrogen has an oxidation state of +5. Thus, nitric acid is not acting as oxidising agent.
B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2 . Also, oxidation state in Cu(NO₃)₂ does not have any change but there is another product, NO₂, where nitrogen has an oxidation state of +4. That means nitric acid is acting as oxidising agent.
C. Na2CO3 + 2HNO3 → 2NaNO3 + H2O + CO2 . Here, in NaNO₃, oxidation state of nitrogen is +5 (Na: +1, 3 O: -6). Thus, nitrogen is not changing is oxidation state and nitric acid is not acting as oxidising agent.
D. NaOH + HNO3 → NaNO3 + H2O. Here, also, the product is NaNO₃. That means nitric acid is not acting as oxidising agent.
QUICK!!!
The specific heat of mercury is 0.138 J/g Celsius. If 452 g of mercury at 85.0 Celsius are placed in 145 g of water at 23.0 Celsius, what will be the final temperature for both the mercury and the water?
Answer: Thus the final temperature for both the mercury and the water is [tex]28.8^0C[/tex]
Explanation:
The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.
[tex]heat_{released}=heat_{absorbed}[/tex]
[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]
[tex]-[m_1\times c_1\times (T_{final}-T_1)]=[m_2\times c_2\times (T_{final}-T_2)][/tex]
Q = heat absorbed or released
[tex]m_1[/tex] = mass of mercury= 452 g
[tex]m_2[/tex]= mass of water = 145 g
[tex]T_{final}[/tex] = final temperature = ?
[tex]T_1[/tex] = temperature of mercury = [tex]85.0^0C[/tex]
[tex]T_2[/tex] = temperature of water = [tex]23.0^0C[/tex]
[tex]c_1[/tex] = specific heat of mercury = [tex]0.138J/g^0C[/tex]
[tex]c_2[/tex] = specific heat of water= [tex]4.184J/g^0C[/tex]
Now put all the given values in equation (1), we get
[tex]-[452\times 0.138\times (T_f-85.0)^0C]=145\times 4.184\times (T_f-23.0)^0C[/tex]
[tex]T_f=28.8^0C[/tex]
Thus the final temperature for both the mercury and the water is [tex]28.8^0C[/tex]
How many moles of HCl can be produced from 0.226 g of SOCl2? SOCl2 + H2O ----> SO2 + 2HCl
Explanation would be helpful c:
Answer:
3.80*10⁻³ moles of HCl can be produced from 0.226 g of SOCl₂
Explanation:
The balanced reaction is:
SOCl₂ + H₂O ----> SO₂ + 2 HCl
By stoichiometry of the reaction they react and produce:
SOCl₂: 1 moleH₂O: 1 moleSO₂: 1 moleHCl: 2 moleBeing:
S: 32 g/moleO: 16 g/moleCl: 35.45 g/moleH: 1 g/molethe molar mass of the compounds participating in the reaction is:
SOCl₂: 32 g/mole + 16 g/mole + 2*35.45 g/mole= 118.9 g/moleH₂O: 2*1 g/mole + 16 g/mole= 18 g/moleSO₂: 32 g/mole + 2*16 g/mole= 64 g/moleHCl: 1 g/mole + 35.45 g/mole= 36.45 g/moleThen, by stoichiometry of the reaction, the following amounts of mass react and are produced:
SOCl₂: 1 mole* 118.9 g/mole= 118.9 gH₂O: 1 mole* 18 g/mole= 18 gSO₂: 1 mole* 64 g/mole= 64 gHCl: 2 mole* 36.45 g/mole= 72.9 gThen the following rule of three can be applied: if by stoichiometry of the reaction 118.9 grams of SOCl₂ produce 2 moles of HCl, 0.226 grams of SOCl₂ how many moles of HCl do they produce?
[tex]moles of HCl=\frac{0.226 grams of SOCl_{2}*2 mole of HCl }{118.9 grams of SOCl_{2} }[/tex]
moles of HCl= 3.80*10⁻³
3.80*10⁻³ moles of HCl can be produced from 0.226 g of SOCl₂
How many moles of oxygen in 3.5 Mols of CH2O
The correct name for CaH2 is
Answer:
Calcium hydride
Explanation:
A 0.04380 g sample of gas occupies 10.0 mL at 290.5 K and 1.10 atm. Upon further analysis, the compound is found to be 25.305% C and 74.695% Cl. What is the molecular formula of the compound?
Answer:
THE MOLECULAR FORMULA FOR THE COMPOUND IS C2 Cl2
Explanation:
mass = 0.04380 g
Volume = 10 mL = 10 / 1000 L = 0.010 L
Pressure = 1.10 atm
R = 0.082 L atm mol^-1 K^-1
Temperature = 290.5 K
Carbon = 25.305 %
Chlorine = 74.695 %
Atomic mass of carbon = 12
Atomic mass of chlorine = 35.5
First calculate the empirical formula by following these steps:
1. Write the percentage composition of the elements involved and divide by its atomic mass
Carbon = 25.305 / 12 = 2.10875
Chlorine = 74.695 /35.5 = 2.1040
2. Divide each by the smaller value
Carbon = 2.10875 / 2.1040 = 1.002
Chlorine = 2.1040 / 2.1040 = 1
3. Round up to the whole number
The empirical formula is C Cl
Next is to calculate the molar mass of the compound using ideal gas equation
PV = mRT/ MM
Mm = mRT / PV
Mm = 0.04380 * 0.082 * 290.5 / 1.10 * 0.010
Mm = 1.0433 / 0.011
Mm = 94.845 g/mol
Mm = 94.85 g/mol
Now that we know the molar mass, we can go on to calculate the molecular formula:
(C Cl) n = Molar mass
( 12 + 35.5) n = 94.85
(47.5)n = 94.85
n = 94.85 / 47.5
n = 1.9968
n ~ 2
The molecular formula can then be written as C2Cl2.
Which equation represents the reaction of a weak acid with water
Answer:
Which equation represents the reaction of a weak acid with water? the equation is : HCl + H2O H3O+ + Cl- HCO3– + H2O H2CO3 + OH– H2O H + + OH- HCOOH + H2O H3O+ + HCOO
Explanation:
hope it helps : )
Answer:
Hey mate, here is your answer. Hope it helps you.
HCOOH + H2O ↔ H3O+ + HCOO-
Explanation:-
A strong acid is one which completely dissociates into its corresponding ions in aqueous medium.
In contrast, a weak will only partially dissociate such that there is an equilibrium between the dissociated ions and the undissociated acid.
In the given examples:
HCl, HCO3- and H2CO3 are all strong acids. However, HCOOH i.e. formic acid is a weak acid which dissociates in water to form H3O+ and formate ion, HCOO-
HCOOH + H2O ↔ H3O+ + HCOO-
How many moles are in 50 grams of cobalt?
help asapp
hey there!
atomic mass cobalt = 58.9332 amu
therefore:
1 mol Co ------------ 58.9332 g
moles ------------ 50 g
moles = 50 x 1 / 58.9332
moles = 50 / 58.9332
= 0.8484 moles of Co
Hope this helps!
0.8484 moles are in 50 grams of cobalt.
What are moles?A mole is defined as 6.02214076 × [tex]10^{23}[/tex] of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.
Atomic mass cobalt = 58.9332 amu
Therefore:
1 mol Co ------------ 58.9332 g
moles ------------ 50 g
moles = 50 x 1 ÷ 58.9332
moles = 50 ÷ 58.9332
= 0.8484 moles of Co
Learn more about moles here:
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What functional groups are in ch2=chch2oh?
Explanation:
H
|
H-C=C-C-OH
| | |
H H H
in this organic compound the functional groups are OH and alkene
prop-2-enol
since it has a double bond, alkene is the functional group. and also it has OH group so hydroxyl group also the other functional group
How many grams of CO2 are used when 6.0 g of O2 are produced? Express your answer with the appropriate units.
Answer:
that is why co2 is in the power of 2ik
What is economic racism?
OA. Something that causes financial inequality for certain ethnic groups
OB. A stock market crash
OC. Laws and policies that go against a particular race
OD. All of the above
Please
Answer:
A
Explanation:
Glyceraldehyde is an aldose monosaccharide. The Fischer projection of D-glyceraldehyde is given below. Draw D-glyceraldehyde using wedge and dash bonds around the chirality center and including ALL hydrogen atoms.
Image is not given in the question, so the image for the question is given below.
Answer:
Fischer projection is a two-dimensional representation of a three-dimensional organic molecule by projection.
Wedge and dash bonds are used to represent the three-dimensional structure of a molecule, in which wedges indicates bonds towards the viewer, solid lines indicates bonds in the plane of the image and dashed lines indicates bonds away from the viewer.
Wedge and dash bonds structure of D-glyceraldehyde is attached below.
chemical reactions occur in predictable ways
Answer:
Yes that is why you can even see how some chemicals might react but that is if you are not considering temperature, pressure, or if catalyst are involved.
In a photoelectric experiment a student uses a light source whose frequency is greater than that needed to eject electrons from a certain metal. However, after continuously shining the light on the same area of the metal for a long period of time the student notices that the maximum kinetic energy of ejected electrons begins to decrease, even though the frequency of the light is held constant. How would you account for this behaviour?
Answer:
The metal surface becomes more positive as electrons are lost from it.
Explanation:
Let us note that photoelectric effect refers to a phenomenon in which electrons are ejected from a clean metal surface irradiated with light of appropriate frequency. This photon must possess a frequency above the threshold frequency of the metal and its energy must exceed the work function of the metal. When these conditions are met, electrons are emitted from a clean metal surface, having a constant kinetic energy as long as the frequency of the incident photon remains constant.
However, as photoelectric effect progresses and electrons are lost from the metal surface, the metal surface becomes more positive. The more positive the surface, the greater the attraction of the positive surface for the emitted electrons. This reduces the kinetic energy of the emitted photons even though the frequency of incident photons is held constant.
4. If the DNA nitrogen bases were TACCGGAT, how would the other half of
the attached DNA strand read?
Answer:
ATGGCCTA
Explanation:
For this we have to keep in mind that we have a specific relationship between the nitrogen bases:
-) When we have a T (thymine) we will have a bond with A (adenine) and viceversa.
-) When we have C (Cytosine) we will have a bond with G (Guanine) and viceversa.
Therefore if we have: TACCGGAT. We have to put the corresponding nitrogen base, so:
TACCGGAT
ATGGCCTA
I hope it helps!
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
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
What product(s) are formed during the complete combustion reaction that occurs when methane (CH4) and molecular oxygen (O2) react? CO2 and H4 C and H2O CO2 and H2O C(OH)4
Answer: CO2 and H2O
Explanation: I already took the test it's right
Consider the following reaction. 2C2H2(g)+5O2(g)⟶4CO2(g)+2H2O(g) 2C2H2(g)+5O2(g)⟶4CO2(g)+2H2O(g) Compound ΔH∘f (kJ/mol)ΔHf∘ (kJ/mol) C2H2(g)C2H2(g) 227.4227.4 O2(g)O2(g) 00 CO2(g)CO2(g) −393.5−393.5 H2O(g)H2O(g) −241.8−241.8 What is the ΔH∘ΔH∘ of the reaction?
Answer:
ΔH° = -1255.8 kJ
Explanation:
Step 1: Data given
ΔHf∘ C2H2 = 227.4 kJ/mol
ΔHf∘ O2 = 0 kJ/mol
ΔHf∘ CO2 = -393.5 kJ/mol
ΔHf∘ H2O = -241.8 kJ/mol
Step 2: The balanced equation
2C2H2(g)+5O2(g) ⟶ 4CO2(g)+2H2O(g)
C2H2(g)+5/2O2(g) ⟶ 2CO2(g)+H2O(g)
Step 3: Calculate ΔH° of the reaction
ΔH° = (2*ΔHf∘ CO2 + ΔHf∘ H2O) - (ΔHf∘ C2H2)
ΔH° = (2* -393.5 kJ/mol + (-241.8) kJ/mol) - 227.4 kJ/mol
ΔH° = -787 - 241.8 - 227. kJ/mol
ΔH° = -1255.8 kJ
According to the Clausius theorem, the cyclic integral of
for a reversible cycle is zero.
Answer:
Yes it is true
Explanation:
This is because the Clausius theorem states that The cyclic integral always has two defined results. The results include it being less than or equal to zero under certain conditions.
When the system consists of only reversible processes, the cyclic integral is equal to zero. If it consists of and irreversible processes, the integral is usually less than zero.
2H2+O2->2H20 How many moles of water can be produced if 8 moles H2 are used
Calculate the molar solubility of CaCO3 in 0.250M Na2CO3
Kps CaCO3 is 4.96x10-9
Answer:
solubility is 1.984x10⁻⁹M
Explanation:
When CaCO₃ is in water, the equilibrium that occurs is:
CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)
Kps = [Ca²⁺] [CO₃²⁻] = 4.96x10⁻⁹
If you have a 0.250M solution of Na₂CO₃, [CO₃²⁻] = 0.250M:
[Ca²⁺] [0.250M] = 4.96x10⁻⁹
Assuming you are adding an amount of CaCO₃:
[X] [0.250 + X] = 4.96x10⁻⁹
Where X is the amoun of CaCO₃ you can add, that means, solubility
X² + 0.250X - 4.96x10⁻⁹ = 0
Solving for X:
X = -0.25M → False answer, there is no negative concentrations.
X = 1.984x10⁻⁹M.
That means, solubility is 1.984x10⁻⁹M
Answer:
[tex]1.984x10^{-8}M[/tex]
Explanation:
Hello,
In this case, the equilibrium reaction for the solubility of calcium carbonate is:
[tex]CaCO_3(s) \rightleftharpoons Ca^{2+}(aq)+CO_3^{-2}(aq)[/tex]
In such a way, since 0.250 M sodium carbonate solution is the solvent, we assume an initial concentration of carbonate anion to be also 0.250 M since sodium carbonate is completely dissolved, for that reason the equilibrium equation turns out:
[tex]Ksp=[Ca^{2+}][CO_3^{2-}]\\\\4.96x10^{-9}=x*(0.250+x)[/tex]
Hence, solving for [tex]x[/tex] we have:
[tex]x=1.984x10^{-8}M[/tex]
Which corresponds to the molar solubility if calcium carbonate as well.
Regards.
Under which set of conditions is ΔGrxn for the reaction A(g)→B(g) most likely to be negative? PA=10.0atm ; PB=0.050atm PA=10.0atm; PB=10.0atm PA=0.050atm; PB=0.050atm PA=0.050atm; PB=10.0atm
Answer:
ΔGrxn for the reaction A(g)→B(g) will most likely to be negative under the PA=10.0atm ; PB=0.050atm condition
The FIRST option is correct
Explanation:
From Thermodynamics,
Using the Gibbs Free Energy (G) formula to determine the reaction A(g)→B(g) most likely to be negative.
Gibbs Free Energy (G) which is the energy associated with a chemical reaction that can be used to do work.
CHECK THE ATTACHMENT FOR THE DETAITALED EXPLANATION
1. Potassium (K) has an atomic mass of 39.0983 amu and only two naturally-occurring isotopes. The K-41 isotope (40.9618 amu) has a natural abundance of 6.7302%. What is the mass (in amu) of the other isotope
Answer:
38.96383282 amu
Explanation:
39.0983 = (40.9618 [tex]\times[/tex] 0.067302) + ( ? [tex]\times[/tex] (1-0.067302)
39.0983 = 2.756811064 + ( ? [tex]\times[/tex] 0.932698)
subtract 2.756811064 from both sides
36.34148894 = ( ? [tex]\times[/tex] 0.932698)
divide both sides by 0.932698
? = 38.96383282 amu
Answer:
38.96383282 amu
Explanation:
39.0983 = (40.9618 0.067302) + ( ? (1-0.067302)
39.0983 = 2.756811064 + ( ? 0.932698)
subtract 2.756811064 from both sides
36.34148894 = ( ? 0.932698)
divide both sides by 0.932698
? = 38.96383282 amu
Helium is a very important element for both industrial and research applications. In its gas form it can be used for welding, and since it has a very low melting point (only 0.95 K under 2.5 MPa) it can be used in liquid form to cool superconducting magnets, such as those found in particle physics experiments. Say we have a cylinder of n = 125 moles of Helium gas at room temperature (T = 20° C). The cylinder has a radius of r = 17 cm and a height h = 1.64 m.What pressure is Helium gas under?
Answer:
P = 20.1697 atm
Explanation:
In this case we need to use the ideal gas equation which is:
PV = nRT (1)
Where:
P: Pressure (atm)
V: Volume (L)
n: moles
R: universal gas constant (=0.082 L atm / K mol)
T: Temperature
From here, we can solve for pressure:
P = nRT/V (2)
According to the given data, we have the temperature (T = 20 °C, transformed in Kelvin is 293 K), the moles (n = 125 moles), and we just need the volume. But the volume can be calculated using the data of the cylinder dimensions.
The volume for any cylinder would be:
V = πr²h (3)
Replacing the data here, we can solve for the volume:
V = π * (17)² * 164
V = 148,898.93 cm³
This volume converted in Liters would be:
V = 148,898.93 mL * 1 L / 1000 mL
V = 148.899 L
Now we can solve for pressure:
P = 125 * 0.082 * 293 / 148.899
P = 20.1697 atmWhat would be the volume of a balloon containing 64g of oxygen gas at STP (standard temperature and pressure)? You should be able to obtain this answer by calculation or by using logical reasoning.
Answer: 44.8 L
Explanation:
To find the volume, we would need to use the ideal gas law.
Ideal Gas Law: PV=nRT
We take the given information and plug it into the equation, but first, we have to manipulate the equation so that we are finding volume.
[tex]V=\frac{nRT}{P}[/tex]
P= 1.00 atm (STP)
T= 273.15 K (STP)
R= 0.08206 Latm/Kmol
n= [tex]64g*\frac{1mol}{31.998 g O_{2} }=2 mol[/tex]
[tex]V=\frac{(2 mol)(0.08206Latm/Kmol)(273.15 K)}{1.00atm} =44.8 L[/tex]
A booth renter is responsible for conducting all necessary business blank
Answer:
building and managing their own clients, purchasing supplies and keeping records
hope this helps you
1. Which of the following choices demonstrates the law of constant composition? (Slides 2 ‒ 3: Laws on Matter) (a) Nitrogen and oxygen are both found in nature as diatomic molecules. (b) When 20.0 g of nitrogen and 32.0 g of oxygen are combined and allowed to react in two separate experiments, both times the product isolated from reaction contains 14.0 g of nitrogen and 32.0 g of oxygen. (c) Nitrogen and oxygen gases are mixed to produce a sample consisting of 20.0 g of nitrogen and 32.0 g of oxygen. (d) When 14.0 g of nitrogen and 32.0 g of oxygen are mixed and react, there is no measurable change in mass during the reaction. (e) Nitrogen and oxygen are gases found in air, nitrogen is approximately 79% of air and oxygen is approximately 21%.
Answer:
(b) When 20.0 g of nitrogen and 32.0 g of oxygen are combined and allowed to react in two separate experiments, both times the product isolated from reaction contains 14.0 g of nitrogen and 32.0 g of oxygen.
Explanation:
The law of definite proportion states that a gen chemical compound always contains its constituent elements in a fixed ratio by mass, independent on the method of preparation.
The molar mass of Nitrogen and Oxygen would always remain the same, allowing for exact reactant masses (or mole ratio) irrespective of the given amount of sample.
