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Tresset [83]
3 years ago
14

CuCl2 + 2NaNO3 mc023-1.jpg Cu(NO3)2 + 2NaCl What is the percent yield of NaCl if 31.0 g of CuCl2 reacts with excess NaNO3 to pro

duce 21.2 g of NaCl?
Chemistry
2 answers:
MatroZZZ [7]3 years ago
3 0
The stoichiometric ratio of CuCl2 to NaCl is 1 is to 2. The stoichiometric ratio of 31.0 g CuCl2 is 26.95 grams of NaCl by converting the amount of CuCl2 to mole and multiplying by 0.5 and molar mass of NaCl.This amount is equal to 78.65% yield.
Lera25 [3.4K]3 years ago
3 0
The answer is D 78.7%
just finished the test
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Water is heated to boiling and water vapor (steam) is released. Chemical change or physical change?
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"You measure 48.9 mL of a solution of sulfuric acid with an unknown concentration, and carefully titrate this solution using a 1
Blizzard [7]

Answer:

C= 0.532M

Explanation:

The equation of reaction is

H2SO4 + 2KOH = K2SO4+ H2O

nA= 1, nB= 2, CA= ?, VA= 48.9ml, CB= 1.5M, VB= 34.7ml

Applying

CAVA/CBVB = nA/nB

(CA× 48.9)/(1.5×34.7)= 1/2

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The following data were obtained in a kinetics study of the hypothetical reaction A + B + C → products. [A]0 (M) [B]0 (M) [C]0 (
Vladimir [108]

Answer:

B. First order, Order with respect to C = 1

Explanation:

The given kinetic data is as follows:

A + B + C → Products

     [A]₀     [B]₀    [C]₀       Initial Rate (10⁻³ M/s)

1.   0.4      0.4     0.2       160

2.  0.2      0.4      0.4       80

3.   0.6     0.1       0.2       15

4.   0.2     0.1       0.2        5

5.   0.2     0.2      0.4       20

The rate of the above reaction is given as:

Rate = k[A]^{x}[B]^{y}[C]^{z}

where x, y and z are the order with respect to A, B and C respectively.

k = rate constant

[A], [B], [C] are the concentrations

In the method of initial rates, the given reaction is run multiple times. The order with respect to a particular reactant is deduced by keeping the concentrations of the remaining reactants constant and measuring the rates. The ratio of the rates from the two runs gives the order relative to that reactant.

Order w.r.t A : Use trials 3 and 4

\frac{Rate3}{Rate4}= [\frac{[A(3)]}{[A(4)]}]^{x}

\frac{15}{5}= [\frac{[0.6]}{[0.2]}]^{x}

3 = 3^{x} \\\\x =1

Order w.r.t B : Use trials 2 and 5

\frac{Rate2}{Rate5}= [\frac{[B(2)]}{[B(5)]}]^{y}

\frac{80}{20}= [\frac{[0.4]}{[0.2]}]^{y}

4 = 2^{y} \\\\y =2

Order w.r.t C : Use trials 1 and 2

\frac{Rate1}{Rate2}= [\frac{[A(1)]}{[A(2)]}]^{x}[\frac{[B(1)]}{[B(2)]}]^{y}[\frac{[C(1)]}{[C(2)]}]^{z}

we know that x = 1 and y = 2, substituting the appropriate values in the above equation gives:

\frac{160}{80}= [\frac{[0.4]}{[0.2]}]^{1}[\frac{[0.4]}{[0.4]}]^{2}[\frac{[0.2]}{[0.4]}]^{z}

1 = (0.5)^{z}

z = 1

Therefore, order w.r.t C = 1

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