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scoray [572]
2 years ago
9

What mass of NH3 (g) is produced when 2 x 10^23 molecules of N2 reacts

Chemistry
1 answer:
Ronch [10]2 years ago
7 0

Answer:

m=8.58

Explanation:

n =  \frac{m}{mr}

N=n÷la

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Explanation:

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Describe happens when you move any atom toward the positive nuclear charge.
bekas [8.4K]

Answer:

Across a period, effective nuclear charge increases as electron shielding remains constant. A higher effective nuclear charge causes greater attractions to the electrons, pulling the electron cloud closer to the nucleus which results in a smaller atomic radius. ... This results in a larger atomic radius.

Explanation:

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2 years ago
Calculate the volume in mL of a 0.708 M KOH solution containing 0.098 mol of solute
Olenka [21]
Molarity = mol/liter

0.708M = 0.098mol/L
Rearrange to find L:
0.098mol/0.708M = .138L

For every liter there is 1000 mL:
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7 0
3 years ago
Consider the following reaction between mercury(II) chloride and oxalate ion.
Alina [70]

<u>Answer:</u> The rate law of the reaction is \text{Rate}=k[HgCl_2][C_2O_4^{2-}]^2

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

2 HgCl_2(aq.)+C_2O_4^{2-}(aq.)\rightarrow 2Cl^-(aq.)+2CO_2(g)+Hg_2Cl_2(s)

Rate law expression for the reaction:

\text{Rate}=k[HgCl_2]^a[C_2O_4^{2-}]^b

where,

a = order with respect to HgCl_2

b = order with respect to C_2O_4^{2-}

Expression for rate law for first observation:

3.2\times 10^{-5}=k(0.164)^a(0.15)^b  ....(1)

Expression for rate law for second observation:

2.9\times 10^{-4}=k(0.164)^a(0.45)^b  ....(2)

Expression for rate law for third observation:

1.4\times 10^{-4}=k(0.082)^a(0.45)^b  ....(3)

Expression for rate law for fourth observation:

4.8\times 10^{-5}=k(0.246)^a(0.15)^b  ....(4)  

Dividing 2 from 1, we get:

\frac{2.9\times 10^{-4}}{3.2\times 10^{-5}}=\frac{(0.164)^a(0.45)^b}{(0.164)^a(0.15)^b}\\\\9=3^b\\b=2

Dividing 2 from 3, we get:

\frac{2.9\times 10^{-4}}{1.4\times 10^{-4}}=\frac{(0.164)^a(0.45)^b}{(0.082)^a(0.45)^b}\\\\2=2^a\\a=1

Thus, the rate law becomes:

\text{Rate}=k[HgCl_2]^1[C_2O_4^{2-}]^2

3 0
2 years ago
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