Answer:
The volume of the sample given is 850 ml, the density given is 0.79 gram per cm. Now the weight of the sample will be,
Weight = volume × density = 850 × 0.79
= 671.5 grams
Weight of the suspended solids given is 0.001 gram
The concentration of the sample can be determined by using the formula,
Concentration = wt. of sample/volume
= [671.5 - 0.001) 10³ mg / 0.85 L
= 789998.82 mg/L or 789998.82 ppm
Now the concentration of suspended solids is.
Css = 0.001 × 10³ mg / 0.85 L = 1.1764 mg per L or 1.1764 ppm
Next, a chemical reaction of interest was conducted in the same constant volume calorimeter. The neutralization reaction of HCl(aq) with NaOH(aq) caused the temperature of the calorimeter to rise by 2.46 °C. What is the change in internal energy ΔU of the neutralization reaction in kJ?
Answer:
A constant volume calorimeter (bomb calorimeter) was calibrated by performing in it a reaction in which 5.23 kJ of heat energy was released, causing the calorimeter to rise by 7.33 °C. What is the heat capacity Cy of the calorimeter
Explanation:
the heat capacity of the calorimeter is
[tex]C_v = \frac{q}{\Delta T} \\\\=\frac{5.23}{7.33} \\\\=0.714kJ /^\circ C[/tex]
Now that our heat capacity of the calorimeter is 0.714kJ/°C
we can easily calculate the change in internal energy ΔU of the neutralization reaction
[tex]\Delta U = C_v \ dt[/tex]
or
[tex]\Delta U = C_V \Delta T[/tex]
Δ T = 2.46 °C
[tex]C_v = 0.714 kJ/ ^\circ C[/tex]
[tex]\Delta U = 0.714*2.46\\\\=1.7564kJ[/tex]
Which is the best method for collection of co2
Answer:
The best way is to burn fuel with oxygen instead of airNeed help with these 7 problems
Answer:
1. 2.3
2. 4.5
3. 10.5
4. 5.6
5. 3.5
6. 3.9
7. 1.6
Explanation:
pH = - log [H+]
1. Lemon juice
[H+] = 0.005
pH = - log [H+]
pH = - log 0.005
pH = 2.3
2. Beer
[H+] = 3.16x10^-5
pH = - log [H+]
pH = - log 3.16x10^-5
pH = 4.5
3. Milk of Magnesia
[H+] = 3.16x10^-11
pH = - log [H+]
pH = - log 3.16x10^-11
pH = 10.5
4. Rain water
[H+] = 2.51x10^-6
pH = - log [H+]
pH = - log 2.51x10^-6
pH = 5.6
5. Soda
[H+] = 3.16x10^-4
pH = - log [H+]
pH = - log 3.16x10^-4
pH = 3.5
6. Tomatoes
[H+] = 1.23x10^-4M
pH = - log [H+]
pH = - log 1.23x10^-4
pH = 3.9
7. Bleach
[OH-] = 2.3x10^-2M
pOH = - log [OH-]
pOH = - log 2.3x10^-2
pOH = 1.6
How many moles of NaOH will be required to produce 375.4 g of any Na2S04
Answer:
5.28 moles of NaOH.
Explanation:
Step 1:
Determination of the number of mole in
375.4 g of any Na2S04.
Molar mass of Na2SO4 = (2x23) + 32 + (16x4) = 142g
Mass of Na2SO4 = 375.4g
Number of mole of Na2SO4 =..?
Mole = Mass /Molar Mass
Number of mole of Na2SO4 = 375.4/142 = 2.64 moles
Step 2:
The balanced equation for the reaction.
H2SO4 + 2NaOH —> Na2SO4 + 2H2O
Step 3:
Determination of the number of mole of NaOH needed for the reaction.
From the balanced equation above,
2 moles of NaOH reacted to produce 1 mole of Na2SO4.
Therefore, Xmol of NaOH will react to produce 2.64 moles of Na2SO4 i.e
Xmol of NaOH = 2 x 2.64
Xmol of NaOH = 5.28 moles
Therefore, 5.28 moles of NaOH is needed for the reaction.
Draw a mechanism for the reaction of methylamine with 2-methylpropanoic acid. Draw any necessary curved arrows. Show the products of the reaction. Include any nonzero formal charges and all lone pairs of electrons. Indicate which side of the reaction is favored at equilibrium.
Answer:
See figure 1
Explanation:
On this case we have a base (methylamine) and an acid (2-methyl propanoic acid). When we have an acid and a base an acid-base reaction will take place, on this specific case we will produce an ammonium carboxylate salt.
Now the question is: ¿These compounds can react by a nucleophile acyl substitution reaction? in other words ¿These compounds can produce an amide?
Due to the nature of the compounds (base and acid), the nucleophile (methylamine) doesn't have the ability to attack the carbon of the carbonyl group due to his basicity. The methylamine will react with the acid-producing a positive charge on the nitrogen and with this charge, the methylamine loses all his nucleophilicity.
I hope it helps!
Steam reforming of methane (CH4 ) produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas, which is the starting point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills a tank with of methane gas and of water vapor, and when the mixture has come to equilibrium measures the amount of carbon monoxide gas to be 18 mol . Calculate the concentration equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to significant digits.
The question is incomplete, here is the complete question:
Steam reforming of methane [tex](CH_4)[/tex] produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas, which is the starting point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills a 125 L tank with 20 mol of methane gas and 10 mol of water vapor at 38 degrees celsius. He then raises the temperature, and when the mixture has come to equilibrium measures the amount of hydrogen gas to be 18 mol . Calculate the concentration equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to significant digits.
Answer: The equilibrium constant for the reaction is [tex]3.99\times 10^{-2}[/tex]
Explanation:
We are given:
Initial moles of methane gas = 20 moles
Initial moles of water vapor = 10 moles
Equilibrium moles of carbon monoxide = 18 moles
Volume of the tank = 125 L
The chemical equation for the reaction of methane and water vapor follows:
[tex]CH_4(g)+H_2O(g)\rightleftharpoons CO(g)+3H_2(g)[/tex]
Initial: 20 10
At eqllm: 20-x 10-x x 3x
Evaluating the value of 'x':
[tex]\Rightarrow 3x=18\\x=6[/tex]
So, equilibrium moles of methane gas = (20 - x) = [20 - 6] = 14 mol
Equilibrium moles of water vapor = (10 - x) = [10 - 6] = 4 mol
Equilibrium moles of carbon monoxide gas = x = 6 mol
The expression of equilibrium constant for the above reaction follows:
[tex]K_{eq}=\frac{[H_2]^3[CO]}{[CH_4][H_2O]}[/tex]
We are given:
[tex][H_2]=\frac{18}{125}=0.144M[/tex]
[tex][CO]=\frac{6}{125}=0.048M[/tex]
[tex][CH_4]=\frac{14}{125}=0.112M[/tex]
[tex][H_2O]=\frac{4}{125}=0.032M[/tex]
Putting values in above expression, we get:
[tex]K_{eq}=\frac{(0.144)^3\times 0.048}{0.112\times 0.032}\\\\K_{eq}=3.99\times 10^{-2}[/tex]
Hence, the equilibrium constant for the reaction is [tex]3.99\times 10^{-2}[/tex]
Draw the molecule by placing atoms on the grid and connecting them with bonds. Include all lone pairs of electrons. Show the formal charges of all atoms in the correct structure. To change the symbol of an atom, double-click on the atom and enter the letter of the new atom for:SeO2 and CO2?3
Answer:
See explanation below
Explanation:
In this case, let's see both molecules per separate:
In the case of SeO₂ the central atom would be the Se. The Se has oxidation states of 2+, and 4+. In this molecule it's working with the 4+, while oxygen is working with the 2- state. Now, how do we know that Se is working with that state?, simply, let's do an equation for it. We know that this molecule has a formal charge of 0, so:
Se = x
O = -2
x + (-2)*2 = 0
x - 4 = 0
x = +4.
Therefore, Selenium is working with +4 state, the only way to bond this molecule is with a covalent bond, and in the case of the oxygen will be with double bond. See picture below.
In the case of CO₂ happens something similar. Carbon is working with +4 state, so in order to stabilize the charges, it has to be bonded with double bonds with both oxygens. The picture below shows.
Analyses of enzymes found in the blood are used as indicators that tissue damage (heart, liver, muscle) has occurred and resulted in the leakage of cellular enzymes into the bloodstream. One of the common enzymes measured for this purpose is the one that interconverts lactate and pyruvate. Identify the name of this enzyme.
Answer:
Lactate dehydrogenase (LDH)
Explanation:
LDH is tetrameric enzyme found in the muscles (M-type) and the heart (H-type) of living cells, responsible for the conversion of pyruvate to lactate and back, to promote generation of nicotinamide adenine dinucleotide (NAD).
For the general reaction aA + bB cC + DD, has the general rate law, rate k[A] [B].". What is the
correct algebraic equation for determining the units of the rate constant, k, when concentration is in M
and time is in seconds.
O
M
M's
k=
k=
MX+y.s
M*+y
O
Ma+b
O
M:s
k=
k
s
MX+y
O
M
o
S
k=
k=
Ma+b.s
M
O
M*+y
k=
S
Answer:
K=CHANGE IN CONCENTRATION/TIME TAKEN
Explanation:
Write condensed and bond-line structural formulas for all of the constitutional isomers Practice problem 4.1 with the molecular formula C7H16. (There are a total of nine constitutional isomers.)
Answer:
See figure 1.
Explanation:
In this case have to take into account that all structures must have the formula: [tex]C_7H_16[/tex]. If we remember the general formula for alkanes: [tex]C_nH_2_n_+_2[/tex] if we have "7" carbons (n=7) we will have 16 hydrogens. Therefore all the structures that fit with this formula are alkanes.
Using the relationship given in part 1, calculate Ecell if [Cu2+] = 1.453 M and [Cr3+] = 0.00176 M, assuming that the temperature remains at 25°C. Give your answer to two places after the decimal.
Answer:
1.05 V
Explanation:
Now recall Nernst equation;
Ecell= E°cell - 0.0592/n log [Red]/[Ox]
Now we look at the values of E°cathode and E°anode
E°cathode= 0.34 V
E°anode = -0.74 V
E°cell= 0.34-(-0.74)
E°cell= 1.08 V
[Red] = 1.453 M
[Ox] = 0.00176 M
n= 6
Ecell= E°cell - 0.0592/n log [Red]/[Ox]
Ecell= 1.08 - 0.0592/6 × log[1.453]/[0.00176]
Ecell= 1.08 - 0.0592/6 × 2.917
Ecell= 1.08 - 0.02878
Ecell= 1.05 V
Draw the structure of the bromohydrin formed when (Z)-3-hexene reacts with Br2/H2O. Use the wedge/hash bond tools to indicate stereochemistry where it exists. If the reaction produces a racemic mixture, draw both stereoisomers. Separate multiple products using the sign from the drop-down menu.
Answer:
(3R,4R)-4-bromohexan-3-ol
Explanation:
In this case, we have reaction called halohydrin formation. This is a markovnikov reaction with anti configuration. Therefore the halogen in this case "Br" and the "OH" must have different configurations. Additionally, in this molecule both carbons have the same substitution, so the "OH" can go in any carbon.
Finally, in the product we will have chiral carbons, so we have to find the absolute configuration for each carbon. On carbon 3 we will have an "R" configuration on carbon 4 we will have also an "R" configuration. (See figure 1)
I hope it helps!
A 1424 gram sample of a liquid at an initial temperature of 30.0 degrees C absorbs 1560 J of heat. Given the specific heat of 2.44 J/g degree C, what is the final temperature of the liquid?
____________________ degree C
Answer:
THE FINAL TEMPERATURE OF THE LIQUID SAMPLE IS 30.45 DEGREE CELSIUS
Explanation:
Mass of the liquid sample = 1424 g
Initail temperature = 30 degree C
Heat evolved = 1560 J
Specific heat of the liquid = 2.44 J/g degree C
Final temperature = unknown
Since the heat evolved by a substance is the product of the mass, specific heat capacity and the change in temperature of the sample
Heat = Mass * Specific heat * change in temp.
H = m c (T2-T1)
Re-arranging the formula by making T2 (final temperature) the subject of the equation, we have:
T2= H/ m c + T1
So therefore, introducing the value of the variables and solving for T2, we have:
T2 = 1560 / 1424 * 2.44 + 30
T2 = 1560 / 3474.56 + 30
T2 = 0.4487 + 30
T2 = 30.4487 degree C
The final temperature of the liquid sample is approximately 30.45 degree C
Consider the reaction below. 2C6H14 + 19O2 Right arrow. 12CO2 + 14H2O How many moles of hexane (C6H14) must burn to form 18.4 mol of carbon dioxide?
Answer:
[tex]3.06mol~C_6H_14[/tex]
Explanation:
We have to start with the reaction:
[tex]2C_6H_1_4 + 19O_2->12CO_2 + 14H_2O[/tex]
We have 2 carbons, 6 hydrogens and 38 oxygens on both sides.
Now the molar ratio between [tex]C_6H_14[/tex] and [tex]CO_2[/tex] is 2:12 or [tex]2~mol~C_6H_14=~12~mol~CO_2[/tex]. With this in mind we can calculate the moles of [tex]C_6H_14[/tex], so:
[tex]18.4~mol~CO_2\frac{2~mol~C_6H_14}{12~mol~CO_2}=3.06mol~C_6H_14[/tex]
I hope it helps!
Answer:
B) 3.07 mol
Explanation:
Aqueous solutions of barium chloride and silver nitrate are mixed to form solid silver chloride and aqueous barium nitrate. The complete ionic equation contains which of the following species (when balanced in standard form)?
A. NO (aq)
B. 2Ba (aq)
C. 2Ag (aq)
D. CI(aq)
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!
50 mL of CH3CH2Br (bromoethane) and 50 mL of water are poured into a separatory funnel. Bromoethane is a water‑insoluble compound with a density of 1.460gmL . The funnel is stoppered and the mixture is shaken vigorously. After standing, two layers separate. Which substance is in which layer? Explain. Bromoethane is more dense than water. Water is added second. The top layer is bromoethane and the bottom layer is water. The top layer is water and the bottom layer is bromoethane. Bromoethane has a higher molar mass than does water. b) Into the same funnel is poured carefully 50 mL of hexane (density = 0.660 g/mL) so that the other two layers are not disturbed. The hexane forms a third layer. The funnel is stoppered, and the mixture is shaken vigorously. After standing, two layers separate. Which compound(s) are in which layer? The top layer contains: The bottom layer contains:
Answer:
See explanation below
Explanation:
In this case, we can explain this in a very basic way.
We know that heavier objects will go always at the bottom when we are carrying two objects, one heavier than the other right?
In the same manner works the density of two liquids. In this case, we have a mix of water and bromoethane. Bromoethane is an organic compound and it's less polar than water which is extremely polar. When we mix these two liquids, we can see that both of them are insoluble, so no matter how much we shake the funnel, the liquids will not mix to form a solution.
Instead of that, both of them will be in the funnel, and they'll be gradually separating into two layers. The bromoethane has a higher density than water, this means that in the bottom layer we will have the bromoethane and in the top layer we will have the water.
In case you are wondering what happens if we added water first and then, the bromoethane?, it will happen the same, it does not matter the order you add the liquid, because density here is a very important factor, so when the water is added no matter which position, it will go to the top layer after the bromoethane is added.
Now when the hexane is added, it will form now three layers, and again, density plays an important factor. The higher density will go to the bottom, and the lowest to the top.
In this case, the order of layer will be:
Top layer: hexane (d = 0.66 g/mL)
Middle layer: water (d = 1 g/mL)
Bottom layer: bromoethane (d = 1.46 g/mL)
Hope this helps
A 1.34 mole sample of LiCl dissolves in water, The volume of the final
solution is 0.86L. Find the Molarity of the Solution.
Answer:
[tex]1.56 \,\,mol/L[/tex]
Explanation:
Molarity of the solution measures number of moles of a solute per litre of solution.
Molarity = volume of solution in litres/number of moles of solute dissolved in solution
Volume of solution in litres = 0.86 L
Also, 1.34 mole sample of LiCl dissolves in water
So,
Molarity of the Solution = [tex]\frac{1.34}{0.86}=1.56 \,\,mol/L[/tex]
A reaction is at equilibrium when...
A. the amounts of the reactants and products are not longer changing.
B. the reactants have all turned into products.
C. the products have all turned into reactants.
D. the amounts of the reactants and products are equal.
Answer:
Option A. is correct
Explanation:
A reaction is at equilibrium when the amounts of the reactants and products are no longer changing.
Chemical equilibrium means that the rate of formation of products is equal to the rate at which the products re-form reactants.
Products are formed by the forward reaction and by the reverse reaction, the products re-form reactants.
Consider the experimental data for multiple experiments with different pressures and volumes of gases.
Experiment PV Value (Pa-mL)
1 1.10 x 106
2 1.14 x 106
3 1.18 x 106
4 1.16 x 106
5 1.19 x 106
Boyle's law that pressure and volume are inversely related because___________.
Answer: Boyle's law states that pressure and volume are inversely related because the product of pressure and volume is approximately constant.
Explanation:
In each of the given values product of P and V, that is, PV remains constant. As according to Boyle's law, at constant temperature the pressure will be inversely proportional to volume of the gas.
Mathematically, [tex]P \propto \frac{1}{V}[/tex]
or, PV = k
where, k = Proportionality constant
P = pressure of gas
V = volume of gas
Product of values will only remain constant when pressure is inversely proportional to volume.
Therefore, we can conclude that Boyle's law states that pressure and volume are inversely related because the product of pressure and volume is approximately constant.
Boyle's law mentioned that pressure and volume should be inversely related since the product of pressure and volume is approximately constant.
Boyle's lawAccording to this law, at a constant temperature, the pressure should be inversely proportional to the volume of the gas.
PV = k
Here
k = Proportionality constant
P = Pressure
V = volume of gas
So based on this we can conclude that Boyle's law mentioned that pressure and volume should be inversely related since the product of pressure and volume is approximately constant.
Learn more about volume here: https://brainly.com/question/21960832
A sample tube consisted of atomic hydrogen in their ground state. A student illuminated the atoms with monochromatic light, that is, light of a single wavelength. If only two separate emission lines in the visible region are observed, what is the wavelength (or wavelengths) of the incident radiation?
Answer:
The wavelength of the monochromatic light is 486.2 nm.
Explanation:
The illumination of the hydrogen atom by the monochromatic light causes an absorption of energy by its electrons which causes an excitation. After a period, the particle de-excites (decays) losing the absorbed energy and falls back to its initial state releasing the energy in the form of a photon. This photon can be observed as a colored light of the Balmer series.
From Rydberg's expression,
1/λ=−R([tex]\frac{1}{n_{2} ^{2} }[/tex] − [tex]\frac{1}{n_{1} ^{2} }[/tex])
The transition of the electron is from n = 2 to 4, so that;
1/λ = R ([tex]\frac{1}{2^{2} }[/tex] - [tex]\frac{1}{4^{2} }[/tex])
= 1.097 x [tex]10^{7}[/tex] ([tex]\frac{1}{2^{2} }[/tex] - [tex]\frac{1}{4^{2} }[/tex])
1/λ = 2056875
So that,
λ = [tex]\frac{1}{2056875}[/tex]
= 4.8617 x [tex]10^{-7}[/tex] m
The wavelength of the monochromatic light is 486.2 nm.
How do I calculate the moles consumed in a vinegar titration?
Answer: Use the titration formula
Explanation:
Which type of semiconductor is created by doping with atoms that contain more valence electrons than the semiconductor material ?
Answer:
N-Type Semiconductor
Explanation:
Question 5 Tungsten is a solid phase of tungsten still unknown to science. The only difference between it and ordinary tungsten is that Tungsten forms a crystal with an fcc unit cell and a lattice constant . Calculate the density of Tungsten .
Complete Question
The complete question is shown on the first uploaded image
Answer:
The density is [tex]\rho = 21.1 \ g/cm^3[/tex]
Explanation:
From the question we are told that
The lattice constant is [tex]a = 0.387 nm = 0.387 *10^{-9} \ m[/tex]
Generally the volume of the unit cell is [tex]V = a^3[/tex]
=> [tex]V = [0.387 *10^{-9}]^2[/tex]
[tex]V = 5.796 *10^{-29} \ m^3[/tex]
Converting to [tex]cm^3[/tex] We have [tex]5.796 *10^{-29} * 1000000 = 5.796 *10^{-23} cm^3[/tex]
The molar mass of Tungsten is constant with a value [tex]Z = 184 g/mol[/tex]
One mole of Tungsten contains [tex]6.022*10^{23}[/tex] unit cells
Where [tex]6.022*10^{23}[/tex] is a constant for the number of atom in one mole of a substance(Tungsten) which is known as Avogadro's constant
Now for FCC distance the number of atom per unit cell is n = 4
Mass of Tungsten (M) = [tex]= \frac{Z * n }{1 \ mole \ of \ Tungsten}[/tex]
=> Mass of Tungsten (M) = [tex]= \frac{184 * 4 }{6.023*10^{23}}[/tex]
=> Mass of Tungsten (M) = [tex]= 1.222*10^{-21} \ g[/tex]
Now
The density of Tungsten is
[tex]\rho = \frac{M}{V}[/tex]
substituting values
[tex]\rho = \frac{1.222*10^{-21}}{5.796*10^{-23}}[/tex]
[tex]\rho = 21.1 \ g/cm^3[/tex]
2NO2(g)= 2NO(g) + O2(g)
Nitrogen dioxide is dissociated to the extent of 56.6 percent and 494°C and 99kPa pressure. At what pressure will the dissociation be 80 percent at 494°C?
Answer:
AT THE END OF 80% DISSOLUTION, THE PRESSURE OF NO2 HAS CHANGED FROM 99kPa TO 139.97kPa
Explanation:
P1 = 99 kPa
P2 = unknown
From the reaction,
2 mole of NO2 will produce 2 mole of NO
We can also say that 1 mole of NO2 will produce 1 mole of NO
At 56.6 % of NO2, 0.566 mole of NO2 will be consumed
At STP, 1 mole of a substance will occupy 22.4 dm3 volume
0.566 mole will occupy ( 22.4 * 0.566 / 1) dm3 volume
= 39.58 dm3 volume
V1 = 39.56 dm3
At the new percent of 80%, 0.80 mole of NO2 will be consumed
Since, 1 mole = 22.4 dm3
0.80 mole = (22.4 / 0.80) dm3
= 28 dm3
V2 = 28 dm3
Using the equation of Boyle's law which shows the relationship between pressure and volume of a given mass of gas at constant temperature, we have:
P1 V1 = P2 V2
Re-arranging to make P2 the subject of formula:
P2 = P1V1 / V2
P2 = 99 kPa * 39.56 / 28
P2 = 3916.44 kPa / 28
P2 = 139.87 kPa
So at 80 % dissociation of NO2, the pressure has changed from 99 kPa to 139.97 kPa.
in what type of reaction or process does heat flow into the system
Answer: The correct answer is: " endothermic . "
______________________________________
Note: Heat flows into [heat may be absorbed within] an "endothermic" reaction or system
To the contrary, heat flows out [heat may exit from or may be released from] an "exothermic" reaction or process.
Hint: Think of the "prefixes" of: "endothermic" and "exothermic" :
_____________________________________
1) endo- = "within" (as in "endothermic" —heat tends to be absorbed/"within"/"released within"/released within"/into" ;
2) exo- = " outwards"/"exit" (as in "exothermic") —heat tends to '"exit"/leave/escape from/"be released out of/form".
_____________________________________
Hope this is helpful to you!
Best wishes to you in your academic pursuits
—and within the "Brainly" community"!
_____________________________________
What is the concentration of a 34% solution converted to parts per million (ppm).?
34,000,000 ppm
340.000 ppm
34,000 ppm
3,400 ppm
Answer:
340.000 ppm
Explanation:
Parts per million (ppm) is a unit of measure for concentration that measures the number of units of substance per million units of the set. It is a concept homologous to the percentage, only in this case it is not parts per percent but per million. The ppm calculation method is different for solids, liquids and gase. There are many formulas, however in this case, you can use this fact:
[tex]10000ppm=1\%[/tex]
Using Cross-multiplication:
[tex]\frac{10000ppm}{x} =\frac{1\%}{34\%}[/tex]
Solving for [tex]x[/tex] :
[tex]x=34*10000=340000ppm[/tex]
Therefore, the concentration of a 34% solution converted to parts per million is:
[tex]340000ppm[/tex]
Determine the expected diffraction angle for the first-order diffraction from the (111) set of planes for FCC nickel (Ni) when monochromatic radiation of wavelength 0.1937 nm is used. The lattice parameter for Ni is 0.3524 nm
Answer:
56°
Explanation:
First calculate [tex]a:[/tex]
[tex]a=2 R \sqrt{2}=2(0.1246) \sqrt{2}=0.352 \mathrm{nm}[/tex]
The interplanar spacing can be calculated from:
[tex]d_{111}=\frac{a}{\sqrt{1^{2}+1^{2}+1^{2}}}=\frac{0.352}{\sqrt{3}}=0.203 \mathrm{nm}[/tex]
The diffraction angle is determined from:
[tex]\sin \theta=\frac{n \lambda}{2 d_{111}}=\frac{1(0.1927)}{2(0.2035)}=0.476[/tex]
Solve for [tex]\theta[/tex]
[tex]\theta=\sin ^{-1}(0.476)=28^{\circ}[/tex]
The diffraction angle is:
[tex]2 \theta=2\left(28^{\circ}\right)=56^{\circ}[/tex]
express your answer to three significant figures and include appropriate units
Answer:
58.702
Explanation:
Data obtained from the question include:
Mass number of isotope 1 (M1) = 57.93
Abundance of isotope 1 (M1%) = 67.76%
Mass number of isotope 2 (M2) = 59.93
Abundance of isotope 2 (M2%) = 26.16%
Mass number of isotope 3 (M3) = 60.93
Abundance of isotope 3 (M3%) = 1.25%
Mass number of isotope 4 (M4) = 61.93
Abundance of isotope 4 (M4%) = 3.66%
Mass number of isotope 5 (M5) = 63.93
Abundance of isotope 5 (M5%) = 1.16%
Relative atomic mass =...?
The relative atomic mass(RAM) of the element can be obtained by doing the following:
RAM = [(M1×M1%) /100] +[(M2×M2%) /100] + [(M3×M3%) /100] + [(M4×M4%) /100] + [(M5×M5%) /100]
RAM = [(57.93×67.76)/100] + [(59.93×26.16)/100] + [(60.93×1.25) /100] + [(61.93×3.66)/100] + [(63.93×1.16)/100]
RAM = 39.253 + 15.678 + 0.762 + 2.267 + 0.742
RAM = 58.702
Therefore, the relative atomic mass (RAM) of the element is 58.702
Calculate ΔHrxnΔHrxn for the following reaction: CH4(g)+2O2(g)→CO2(g)+2H2O(l)CH4(g)+2O2(g)→CO2(g)+2H2O(l) Use the following reactions and given ΔHΔH values. CH4(g)+O2(g)→CH2O(g)+H2O(g)CH4(g)+O2(g)→CH2O(g)+H2O(g), ΔH=−ΔH=−284 kJkJ CH2O(g)+O2(g)→CO2(g)+H2O(g)CH2O(g)+O2(g)→CO2(g)+H2O(g), ΔH=−ΔH=−527 kJkJ H2O(l)→H2O(g)H2O(l)→H2O(g), ΔH=ΔH= 44.0 kJ
Answer: the enthalpy of reaction is, -155 kJ
Explanation:-
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.
The final reaction is:
[tex]CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)[/tex] [tex]\Delta H_{rxn}=?[/tex]
The intermediate balanced chemical reaction will be,
(1) [tex]CH_4(g)+O_2(g)\rightarrow CH_2O(g)+H_2O(g)[/tex] [tex]\Delta H_1=-284kJ[/tex]
(2) [tex]CH_2O(g)+O_2(g)\rightarrow CO_2(g)+H_2O(g)[/tex] [tex]\Delta H_2=-527kJ[/tex]
(3) [tex]H_2O(l)\rightarrow H_2O(g)[/tex] [tex]\Delta H_3=44.0kJ[/tex]
Now multiplying (3) by 2 and adding all the equations, we get :
[tex]\Delta H_{rxn}=\Delta H_1+\Delta H_2+2\times \Delta H_3[/tex]
[tex]\Delta H_{rxn}=(-284)+(-527)+2\times (44)[/tex]
[tex]\Delta H_{rxn}=-155kJ[/tex]
Therefore, the enthalpy of reaction is, -155 kJ
When 50.0 g of nitrogen react with excess hydrogen to form ammonia gas, 164.5 kJ of heat are liberated (released) at standard state conditions. Calculate the standard enthalpy of formation (in kJ/mol) for ammonia gas.
Answer:
THE STANDARD ENTHALPY OF FORMATION OF AMMONIA GAS IS 293.75kJ OF HEAT.
Explanation:
To solve this question, you must first write out the equation for the reaction.
Equation:
N2 (g) + 3H2(g) <-------> 2NH3(g)
So therefore, when 50 g of N2 reacts, 164.5 kJ of Heat was liberated.
First equate the number of moles of Nitrogen and ammonia gas
1 mole of N2 produces 2 moles of ammonia
Calculate the molar mass of each variables:
Molar mass of N2 = 14*2 = 28 g/mol
Molar mass of ammonia = ( 14 + 1*3) = 17 g/mol
So, 1 mole of N2 = 2 moles of NH3
28 g/mol of N2 = 17 * 2 g/mol of NH3
If 50 g of nitrogen was used to react with excess hydrogen, the mass of ammonia formed is;
28 g of N2 = 34 g/mol of NH3
50 g of N2 = ( 50 * 34 / 28 ) g of NH3
= 1700 / 28
= 60 .71 g of ammonia.
At standard conditions, 34 g of ammonia will liberate 164.5 kJ of heat. What amonut would be generated by 60.71 g of ammonia?
34 g of ammonia = 164.5 kJ of heat
60.71 g of ammonia = ( 60.71 * 164.5 / 34) kJ of heat
= 9987.5 / 34
= 293.75 kJ of heat.
In other words, the standard enthalpy of formulation for ammonia gas is 293.75 kJ of heat.
