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3 years ago

In the reaction 2co ( g) + o2( g) → 2co2( g), what is the ratio of moles of oxygen used to moles of co2produced?

Chemistry

1 answer:

KIM [24]3 years ago

8 0

In the reaction 2co ( g) + o2( g) → 2co2( g), the ratio of moles of oxygen used to moles of co2produced is 1:2.

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3 years ago

A chemist has a 2.0 M solution of copper (II) sulfate. Is he takes 400 mL of this solution and dilutes it with 1200 mL of water,

Ksenya-84 [330]

0.66 M is the accurate molarity of the new solution of volume of 1200 ml.

Explanation:

Data given:

molarity of copper(II) sulphate, Mconc.= 2M

volume of 2M solution taken Vconc. = 400 ml

volume taken for dilution, Vdilute = 1200 ml

molarity of the diluted solution, Mdilute =?

We will use the formula for dilution as

Mconc Vconc = Mdilute x V dilute (conc is concentrated)

putting the values in the equation:

2 x 400 = Mdilute x 1200

Mdilute = $\frac{800}{1200}$

Mdilute = 0.66 M

When the solution is diluted to the volume of 1200 ml its molarity changes to 0.66 M.

4 0

3 years ago

How much heat is released when 1.4 mol of hydrogen fluoride are produced?

Assoli18 [71]

Answer:

+375.2 KJ.

Explanation:

The balanced equation for the reaction is given below:

H₂ + F₂ —> 2HF ΔH = +536 KJ

From the balanced equation above,

2 moles of HF required +536 KJ .

Finally, we shall determine the heat required to produce 1.4 mol of hydrogen fluoride, HF. This is illustrated below:

From the balanced equation above,

2 moles of HF required +536 KJ .

Therefore, 1.4 moles of HF will require = (1.4 × 536)/2 = +375.2 KJ

Thus, +375.2 KJ of heat energy is required.

7 0

2 years ago

Hydrogen reacts with nitrogen monoxide to form dinitrogen monoxide (laughing gas) according to the equation: H2O(g) + 2NO(g) rig

xxTIMURxx [149]

Answer:

A. $r=k[NO]^2[H_2]$

B. $k=0.121\frac{L^2}{mol^2*s}$

C. Second-order with respect to NO and first-order with respect to H₂

Explanation:

Hello,

In this case, for the reaction:

$H_2O(g) + 2NO(g) \rightarrow N_2O + H_2O(g)$

The rate law is determined by writing the following hypothetical rate laws:

$3.822x10^{-3}=k[0.3]^m[0.35]^n\\\\1.529x10^{-2}=k[0.6]^m[0.35]^n\\\\3.058x10^{-2}=k[0.6]^m[0.7]^n$

Whereas we can compute m as follows:

$\frac{3.822x10^{-3}}{1.529x10^{-2}} =\frac{[0.3]^m[0.35]^n}{[0.6]^m[0.35]^n} \\\\0.25=(0.5)^m\\\\m=\frac{log(0.25)}{log(0.5)} \\\\m=2$

Therefore, the reaction is second-order with respect to NO. Thus, for hydrogen, we find n:

$\frac{1.529x10^{-2}}{3.058x10^{-2}} =\frac{[0.6]^2[0.35]^n}{[0.6]^2[0.7]^n} \\\\0.5=(0.5)^n\\\\n=\frac{log(0.5)}{log(0.5)}\\ \\n=1$

A) Therefore, the reaction is first-order with respect to H₂. In such a way, we conclude that that the rate law is:

$r=k[NO]^2[H_2]$

B) Rate constant is computed from one kinetic data:

$k=\frac{1.529x10^{-2}\frac{mol}{L*s} }{(0.6\frac{mol}{L} )^2(0.35\frac{mol}{L})}\\\\k=0.121\frac{L^2}{mol^2*s}$

C. As mentioned before, reaction is second-order with respect to NO and first-order with respect to H₂.

Best regards.

3 0

2 years ago