TIME REMAINING

A study of the decomposition reaction 3RS2  3R + 6S yields the following initial rate dat

shutvik [7]

Missing question: What is the rate constant for the reaction?
<span>[RS2](mol L-1) Rate (mol/(L·s))
0.150 0.0394
0.250 0.109
0.350 0.214
0.500 0.438</span>
Chemical reaction: 3RS₂ → 3R + 6S.
Compare second and fourth experiment, when concentration is doubled, rate of concentration is increaced by four. So rate is:
rate = k·[RS₂]².
k = 0,438 ÷ (0,500)².
k = 1,75 L/mol·s.

3 0

3 years ago

2) Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.800 mole of CO and 2.40 mole of H2

nadezda [96]

Answer:

Kc = 3.90

Explanation:

CO reacts with $H_2$ to form $CH_4$ and $H_2O$ . balanced reaction is:

$CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g) + H_2O(g)$

No. of moles of CO = 0.800 mol

No. of moles of $H_2$ = 2.40 mol

Volume = 8.00 L

Concentration = $\frac{Moles}{Volume\ in\ L}$

Concentration of CO = $\frac{0.800}{8.00} = 0.100\ mol/L$

Concentration of $H_2$ = $\frac{2.40}{8.00} = 0.300\ mol/L$

$CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g) + H_2O(g)$

Initial 0.100 0.300 0 0

equi. 0.100 -x 0.300 - 3x x x

It is given that,

at equilibrium $H_2O (x)$ = 0.309/8.00 = 0.0386 M

So, at equilibrium CO = 0.100 - 0.0386 = 0.0614 M

At equilibrium $H_2$ = 0.300 - 0.0386 × 3 = 0.184 M

At equilibrium $CH_4$ = 0.0386 M

$Kc=\frac{[H_2O][CH_4]}{[CO][H_2]^3}$

$Kc=\frac{0.0386 \times 0.0386}{(0.184)^3 \times 0.0614} =3.90$

8 0

3 years ago

PLEASE I NEED HELP ASAP!!!!

miv72 [106K]

The percent yield of the calcium hydroxide is 84.5%.

<h3>What is stoichiometry?</h3>

Stoichiometry enables us to obtain the mass of a substance form the equation of the reaction.

The equation of the reaction is;

CaCO3 + 2HCl -----> CaCl2 + CO2 + H2O

Number of moles of X = 40.0 grams/100 g/mol = 0.4 moles

Number of moles of HCl = 850/1000 * 1 M = 0.85 moles

If 1 mole of CaCO3 reacts with 2 moles of HCl

0.4 moles of CaCO3 reacts with 0.4 moles * 2 moles/1 mole

= 0.8 moles of HCl

Hence X is the limiting reactant.

The reaction is 1:1 then the amount of CO2 produced is 0.4 moles

Mass of CO2 = 0.4 CO2 * 44 g/mol = 17.6 g

2) The reaction equation is; 2NaOH + CaCO3 ---> Ca(OH)2 + Na2CO3

Number of moles of X = 25.0 grams/100 g/mol = 0.25 moles

Number of moles of NaOH= 40/1000 L * 2 M = 0.08 moles

If 1 mole of X reacts with 2 moles of NaOH

0.25 moles reacts with 0.25 moles * 2 moles /1 mole

= 0.5 moles

NaOH is the limiting reactant

2 moles of NaOH produces 1 mole of CO2

0.08 moles of NaOH produces 0.08 moles * 1 mole/2 moles

= 0.04 moles of CO2

Theoretical yield of CO2 = 0.04 moles of CO2 * 74 g/mol = 2.96 g

Percent yield = 2.5 g/ 2.96 g * 100

= 84.5%

Learn ore about percent yield:brainly.com/question/17042787

#SPJ1

5 0

2 years ago

Express the pressure 545 mm Hg in kilopascals.

Arlecino [84]

545mm Hg in Kilopascals is 72.6607

I hope this helps you. Good luck stay safe, healthy and, happy!<3

6 0

2 years ago

A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli

Diano4ka-milaya [45]

Answer:

$\large \boxed{109.17 \, ^{\circ}\text{C}}$

Explanation:

Data:

50/50 ethylene glycol (EG):water

V = 4.70 gal

ρ(EG) = 1.11 g/mL

ρ(water) = 0.988 g/mL

Calculations:

The formula for the boiling point elevation ΔTb is

$\Delta T_{b} = iK_{b}b$

i is the van’t Hoff factor — the number of moles of particles you get from 1 mol of solute. For EG, i = 1.

1. Moles of EG

$\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}$

2. Kilograms of water

$m = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{998 g}}{\text{1 L}} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{8.88 kg}$

3. Molal concentration of EG

$b = \dfrac{\text{159 mol}}{\text{8.88 kg}} = \text{17.9 mol/kg}$

4. Increase in boiling point

$\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 17.9 \cdot mol \cdot kg^{-1} = 9.17 \, ^{\circ}\text{C}$

5. Boiling point

$\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 9.17 \, ^{\circ}\text{C} = \mathbf{109.17 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}$

7 0

3 years ago