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4 years ago

Consider this reaction:

Chemistry

1 answer:

o-na [289]4 years ago

3 0

Answer:

$t=9.7s$

Explanation:

Hello,

In this case, we have a second order kinetics given the second power of the concentration of chlorine (V) oxide in the rate expression, thus, the integrated equation for the concentration decay is:

$\frac{1}{[Cl_2O_5]}=kt+\frac{1}{[Cl_2O_5]_0}$

Thus, the final concentration for a 94% decrease is:

$[Cl_2O_5]=0.600M-0.600M*0.94=0.036M$

Therefore, we compute the time for such decrease:

$kt=\frac{1}{[Cl_2O_5]}-\frac{1}{[Cl_2O_5]_0}=\frac{1}{0.036M}-\frac{1}{0.60M} =26.1M^{-1}$

$t=\frac{26.1M^{-1}}{k}= \frac{26.1M^{-1}}{2.7M^{-1}*s^{-1}}\\\\t=9.7s$

Regards.

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Whitney's lung capacity was measured as 3.3 l at a body temperature of 37 ∘c and a pressure of 746 mmhg . how many moles of oxyg

myrzilka [38]

the ideal gas equation is PV=nRT
where P=pressure
V=Volume
n=no. of moles
R=universal gas constant
T=temperature
The universal gas constant (R) is 0.0821 L*atm/mol*K
a pressure of 746 mmhg =0.98 atm= 1 atm (approx)
T=37 degrees Celsius =37+273=310 K (convert it to Kelvin by adding 273)
V=0.7 L (only getting oxygen, get 21% of 3.3L)
Solution:
(1 atm)(0.7 L)=n(0.0821 L*atm/mol*K)(310 K)
0.7 L*atm=n(25.451 L*atm/mol)
n=0.0275 mole
Answer:
n=0.0275 mole of oxygen in the lungs.

4 0

3 years ago

What are the concentrations of Cu2+, NH3, and Cu(NH3)42+ at equilibrium when 18.8 g of Cu(NO3)2 is added to 1.0 L of a 0.800 M s

Svetllana [295]

Answer:

Explanation:

Cu(NO₃)₂ + 4NH₃ = Cu(NH₃)₄²⁺ + 2 NO₃⁻

187.5 gm 4M 1 M

187.5 gm reacts with 4 M ammonia

18.8 g reacts with .4 M ammonia

ammonia remaining left after reaction

= .8 M - .4 M = .4 M .

187.5 gm reacts with 4 M ammonia to form 1 M Cu(NH₃)₄²⁺

18.8 g reacts with .4 M ammonia to form 0.1 M Cu(NH₃)₄²⁺

At equilibrium , the concentration of Cu²⁺ will be zero .

concentration of ammonia will be .4 M

concentration of Cu(NH₃)₄²⁺ formed will be 0.1 M

3 0

4 years ago

What is the diffrence between speed and avarege speed

Gekata [30.6K]

Speed is actually instantaneous speed, the speed an object is moving at that very instant.
But average speed is the total distance traveled divided by the total time.
Let's say you travelled 10 miles in 2 hours, your average speed would be 10 divided by 2, which is 10 miles / hr.

But during those 10 miles you may be accelerating, decelerating and be travelling 1 mile/hour and 20 miles/hour at another point. But your average speed would be total distance / total time. So your instantaneous speed can change throughout that period of time.

4 0

4 years ago

Read 2 more answers

A chemist wants to find Kc for the following reaction at 751 K: 2NH3(g) + 3 I2 (g) LaTeX: \Longleftrightarrow ⟺ 6HI(g) + N2(g) K

charle [14.2K]

Answer: The equilibrium constant for the total reaction is $4.09\times 10^{-6}$

Explanation:

We are given:

$K_{c_1}=0.282\\\\K_{c_2}=41$

We are given two intermediate equations:

Equation 1: $N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g);K_{c_1}=0.282$

The expression of $K_{c_1}$ for the above equation is:

$K_{c_1}=\frac{[NH_3]^2}{[N_2][H_2]^3}$

$0.282=\frac{[NH_3]^2}{[N_2][H_2]^3}$ .......(1)

Equation 2: $H_2(g)+I_2(g)\rightleftharpoons 2HI(g);K_{c_2}=41$

The expression of $K_{c_2}$ for the above equation is:

$K_{c_2}=\frac{[HI]^2}{[H_2][I_2]}$

$41=\frac{[HI]^2}{[H_2][I_2]}$ ......(2)

Cubing both the sides of equation 2, because we need 3 moles of HI in the main expression if equilibrium constant.

$(41)^3=\frac{[HI]^6}{[H_2]^3[I_2]^3}$

Now, dividing expression 1 by expression 2, we get:

$\frac{K_{c_1}}{K_{c_2}}=\left(\frac{\frac{[NH_3]^2}{[N_2][H_2]^3}}{\frac{[HI]^6}{[H_2]^3[l_2]^3}}\right)\\\\\\\frac{0.282}{68921}=\frac{[NH_3]^2[I_2]^3}{[N_2][HI]^6}$

$\frac{[NH_3]^2[I_2]^3}{[N_2][HI]^6}=4.09\times 10^{-6}$

The above expression is the expression for equilibrium constant of the total equation, which is:

$2NH_3(g)+3I_2(g)\rightleftharpoons 6HI(g)+N_2;K_c$

Hence, the equilibrium constant for the total reaction is $4.09\times 10^{-6}$

8 0

3 years ago

In the laboratory, a student combines 28.3 mL of a 0.489 M manganese(II) acetate solution with 12.5 mL of a 0.339 M manganese(II

Eddi Din [679]

Answer:

M = 0.441 M

Explanation:

In this case, we have two solutions that involves the Manganese II cation;

We have Mn(CH₃COOH)₂ and MnSO₄

In both cases, the moles of Mn are the same in reaction as we can see here:

Mn(CH₃COO)₂ <-------> Mn²⁺ + 2CH₃COO⁻

MnSO₄ <------> Mn²⁺ + SO₄²⁻

Therefore, all we have to do is calculate the moles of Mn in both solutions, do the sum and then, calculate the concentration with the new volume:

moles of MnAce = 0.489 * 0.0283 = 0.0138 moles

moles MnSulf = 0.339 * 0.0125 = 0.0042 moles

the total moles are:

moles of Mn²⁺ = 0.0138 + 0.0042 = 0.018 moles

Finally the concentration: 12.5 + 28.3 = 40.8 mL or 0.0408 L

M = 0.018 / 0.0408

M = 0.441 M

This would be the final concentration of the manganese after the mixing of the two solutions

7 0

4 years ago