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maxonik [38]
3 years ago
14

For the reaction PCl5(g) + heat PCl3(g) + Cl2(g), what will happen when the volume is increased?

Chemistry
1 answer:
butalik [34]3 years ago
7 0
I think that it’s b
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If the percent yield for the following reaction is 75.0%, and 25.0 g of NO₂ are consumed in the reaction, how many grams of nitr
victus00 [196]

Answer:

17.1195 grams of nitric acid are produced.

Explanation:

3NO_2+H_2O\rightarrow 2HNO_3+NO

Moles of nitrogen dioxide :

\frac{25.0 g}{56 g/mol}=0.5434 mol

According to reaction 3 moles of nitrogen dioxides gives 2 moles of nitric acid.

Then 0.5434 moles of nitrogen dioxides will give:

\frac{2}{3}\times 0.5434 mol=0.3623 mol of nitric acid.

Mass of 0.3623 moles of nitric acid :

0.3623 mol\times 63 g/mol=22.8260 g

Theoretical yield = 22.8260 g

Experimental yield = ?

\%Yield=\frac{\text{Experimental yield}}{\text{theoretical yield}}\times 100

75\%=\frac{\text{Experimental yield}}{22.8260 g}

Experimental yield of nitric acid = 17.1195 g

7 0
3 years ago
A temperature change in a reaction indicates_____.
natita [175]

A temperature change in a reaction indicates a chemical change

8 0
2 years ago
Read 2 more answers
so they are asking me to find the empirical formula for a compound that is 7.70% carbon and 92.3% chlorine. Can you show me step
Nikolay [14]

Answer:

Empirical formula is CCl₄

Explanation:

Given data:

Percentage of carbon = 7.70%

Percentage of chlorine = 92.3%

Empirical formula = ?

Solution:

Number of gram atoms of Cl = 92.3 / 35.5 = 2.6

Number of gram atoms of C = 7.70 / 12 = 0.64

Atomic ratio:

            C                      :              Cl            

           0.64/0.64        :             2.6/0.64

            1                      :                4        

C : Cl  = 1 : 4

Empirical formula is CCl₄.

5 0
3 years ago
Find the initial concentration of the weak acid or base in each of the following aqueous solutions: (a) a solution of HClO with
Luda [366]

Answer:

a) 0.021 M

b) 0.019 M

Explanation:

To do this, you need to calculate the concentration of ions in solution with the given value of pH for each solution, then, write the chemical equation for both solutions, Set an ICE chart, use the value of Ka and Kb reported for both solutions, and solve for the initial concentration.

This is the general procedure to do it, now let's do it by parts.

<em><u>a) Concentration of HClO pH = 4.6</u></em>

With the given pH, we use the following expression:

pH = -log[H₃O⁺]      From here, we solve for [H₃O⁺]

[H₃O⁺] = 10^(-pH)   (1)

Let's calculate first the hydronium concentration:

[H₃O⁺] = 10^(-4.6) = 2.51x10⁻⁵ M

This value indicates the equilibrium concentration of this ion in solution. Now, to know the initial concentration of the acid, we need to do an ICE chart and write the chemical equation. This is an acid - base reaction, so we need the value of Ka of the acid.

         HClO + H₂O <---------> H₃O⁺ + ClO⁻       Ka = 3x10⁻⁸

I:            Y                                 0          0

C:          -x                                +x         +x

E:           Y - x                            x          x

With this chart, we need to write the expression for Ka which is:

Ka = [H₃O⁺] * [ClO⁻] / [HClO] = x² / Y-x

But we already know the concentration of [H₃O⁺], which is the same for [ClO⁻], and the value of Ka, so all we have to do is replace the values in the above expression and solve for Y:

3x10⁻⁸ = (2.51x10⁻⁵)² / Y - 2.51x10⁻⁵

We can round to Y because "x" is a very small value as it's value of Ka so:

3x10⁻⁸ = (2.51x10⁻⁵)²/Y

Y = (2.51x10⁻⁵)²/3x10⁻⁸

<h2><em>Y = [HClO] = 0.021 M</em></h2>

<em>And this is the initial concentration of the acid.</em>

<u><em>b) Solution of hidrazine pH = 10.2</em></u>

We do the same procedure as part a) with the difference that instead of using Ka , we use Kb and concentration of [OH⁻]. The Kb for hydrazine is 1.3x10⁻⁶

Let's calculate the [OH⁻]:

pOH = 14 - pH

pOH = 14 - 10.2 = 3.8

[OH⁻] = 10^(-3.8) = 1.58x10⁻⁴ M

The chemical equation:

          N₂H₄ + H₂O <---------> N₂H₅⁺ + OH⁻    Kb = 1.3x10⁻⁶

I:            Y                                  0           0

C:          -x                                +x           +x

E:         Y-x                                 x           x

Kb = x²/(Y-x)

1.3x10⁻⁶ = (1.58x10⁻⁴)²/Y

Y = (1.58x10⁻⁴)²/1.3x10⁻⁶

<h2><em><u>Y = [OH⁻] = 0.019 M</u></em></h2>

And this is the initial concentration of hydrazine

4 0
2 years ago
When the outer electrons of substances are removed, they become statically charged.. . True. False.
elena55 [62]
It is true that when the outer electrons of substances are removed, they become statically charged. The correct answer is True.
6 0
3 years ago
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