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3 years ago

4 Nitrogen monoxide reacts with oxygen like this:

Chemistry

1 answer:

olga2289 [7]3 years ago

6 0

a. mol O₂=0.5

b. volume O₂ = 25 cm³

c. i. the total volume of the two reactants = 75 cm³

c. ii. the volume of nitrogen dioxide formed = 50 cm³

<h3>Further explanation</h3>

Reaction

2NO(gas) + O₂(gas) ⇒ 2NO₂ (gas)

mol NO = 1

From the equation, mol ratio NO : O₂ = 2 : 1, so mol O₂ :

$\tt \dfrac{1}{2}\times 1=0.5$

From Avogadro's hypothesis, at the same temperature and pressure, the ratio of gas volume will be equal to the ratio of gas moles

Because mol ratio NO : O₂ = 2 : 1, so volume O₂ :

$\tt \dfrac{1}{2}\times 50~cm^3=25~cm^3$

i. total volume of reactants : 25 cm³+ 50 cm³=75 cm³

ii. the volume of nitrogen dioxide formed :

mol ratio NO : NO₂ = 2 : 2, so volume NO₂ = volume NO = 50 cm³

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If 9 moles of P203 are formed, how many<br> moles of O2 reacted?

masha68 [24]

9 x 3 = 27
27 moles of O reacted
27 / 2 = 13.5 O2 reacted
round up to 14 moles of O2

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2 years ago

Calculate the pH for the following 1.0M weak acid solutions:a. HCOOH Ka = 1.8 x 10-4 [

diamong [38]

Answer: pH=2.38

Explanation:

To calculate the pH, let's first write out the equation. Then, we will make an ICE chart. The I in ICE is initial quantity. In this case, it is the initial concentration. The C in ICE is change in each quantity. The E is equilibrium.

HCOOH ⇄ H⁺ + HCOO⁻

I 1.0M 0 0

C -x +x +x

E 1.0-x x x

<u>For the steps below, refer to the ICE chart above.</u>

1. Since we were given the initial of HCOOH, we can fill this into the chart.

2. Since we were not given the initial for H⁺ and HCOO⁻, we will put 0 in their place.

3. For the change, we need to add concentration to the products to make the reaction reach equilibrium. We would add on the products and subtract from the reactants to equalize the reaction. Since we don't know how much the change in, we can use variable x.

4. We were given the Kₐ of the solution. We know $K_{a} =\frac{product}{reactant}=\frac{[H^+][HCOO^-]}{[HCOOH]}$ .

5. The problem states that the Kₐ=1.8×10⁻⁴. All we have to so is to plug it in and to solve for x.

$1.8*10^-^4 =\frac{x^2}{0.1-x}$

6. Once we plug this into the quadratic equation, we get x=0.00415.

7. The equilibrium concentration of [H⁺]=0.00415. pH is -log(H⁺).

-log(0.00415)=2.38

Our pH for the weak acid solution is 2.38.

8 0

3 years ago

A 25.00 ml sample of an aqueous solution of ba(oh)2 requires 18.45 ml of 0.01500 m hcl (aq) for its neutralization. what is the

klio [65]

5.54 x 10⁻³ M

<h3>Further explanation</h3>

Given:

Neutralization reaction between:

25.00 ml of Ba(OH)₂
18.45 ml of 0.015 M HCl

Question:

What is the molarity of the Ba(OH)₂ solution?

The Process:

Let us say the molarity of the Ba (OH)₂ solution as x M.

Step-1: prepare moles for each reagent

$\boxed{ \ M = \frac{n}{V} \ }$ $\rightarrow \boxed{ \ n = MV \ }$

$\boxed{ \ Ba(OH)_2 \rightarrow n = (x \ \frac{mol}{L})(25 \ ml) = 25x \ mmol \ }$

$\boxed{ \ HCl \rightarrow n = (0.015 \ \frac{mol}{L})(18.45 \ ml) = 0.277 \ mmol \ }$

Step-2: neutralization

We use the ICE table to see how neutralization occurs between acid and base.

Balanced reaction:

$\boxed{ \ Ba(OH)_2_{(aq)} + 2HCl_{(aq)} \rightarrow BaCl_2_{(aq)} + 2H_2O_{(l)} \ }$

Initial: 25x 0.277 - -

Change: - ¹/₂ · (0.277) -0.277 +¹/₂ · (0.277) +0.277

Equlibrium: - - +¹/₂ · (0.277) +0.277

Neutralization causes no excess of hydrogen or hydroxide ions in solution. In the end, the number of acid and base reactions is balanced. In other words, the two reagents have run out with nothing left.
HCl acts as a limiting reagent.

Step-3: calculate the molarity of the Ba(OH)₂ solution.

We consider Ba (OH) from the initial, change, and equilibrium stages.

$\boxed{ \ 25x - \frac{1}{2}(0.277) = 0 \ }$

$\boxed{ \ 25x = \frac{1}{2}(0.277) \ }$

$\boxed{ \ 50x = 0.277 \ }$

$\boxed{ \ x = \frac{0.277}{50} \ }$

$\boxed{ \ x = 5.54 \times 10^{-3} \ }$

Thus, the molarity of the Ba(OH)₂ solution is 5.54 x 10⁻³ M.

_ _ _ _ _ _ _ _ _ _

Alternative Steps

Valence of base = the number of OH⁻ ions
Valence of acid = the number of H⁺ ions

Neutralization: $\boxed{ \ V_b \cdot M_b \cdot valence \ of \ base = V_a \cdot M_a \cdot valence\ of \ acid \ }$

$\boxed{ \ (25.00 \ ml) \cdot x \cdot 2 = (18.45 \ ml) \cdot (0.015 \ M) \cdot 1 \ }$

$\boxed{ \ x = \frac{(18.45 \ ml) \cdot (0.015 \ M)}{(25.00 \ ml) \cdot 2} \ }$

Thus the same results were obtained. The molarity of Ba (OH) ₂ solution is 5.54 x 10⁻³ M.

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