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gladu [14]
4 years ago
7

you are studying an element that is a gas at room temperature. Is it most likely a metal, nonmetal, or metalloid?

Chemistry
2 answers:
konstantin123 [22]4 years ago
8 0
Nonmetal, for example, the Noble Gases, or Hydrogen.
Grace [21]4 years ago
3 0
It is 100% Nonmetal, why? Because gas is not metallic (so that crosses out metals and metalloids), and can't conduct, in any way.
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1(a). What is the solubility of calcium chloride (CaCl2) at 5°C?
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The overall reaction is given by:

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The slow step reaction is given as:

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Now, the expression for the rate of reaction of fast reaction is:

r_{1}=k_{1}[NO][Br_{2}]-k_{-1}[NOBr_{2}]

The expression for the rate of reaction of slow reaction is:

r_{2}=k_{2}[NOBr_{2}] [NO]

Slow step is the rate determining step. Thus, the overall rate of formation is the rate of formation of slow reaction as [NOBr_{2}] takes place in this reaction.

The expression of rate of formation is:

\frac{d(NOBr)}{dt}=r_{2}

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Now, consider that the fast step is always is in equilibrium. Therefore, r_{1}=0

k_{1}[NO][Br_{2}]= k_{-1}[NOBr_{2}]

[NOBr_{2}] = \frac{k_{1}}{k_{-1}}[NO][Br_{2}]

Substitute the value of [NOBr_{2}] in equation (1), we get:

\frac{d(NOBr)}{dt}=k_{2}[NOBr_{2}][NO]

=k_{2} \frac{k_{1}}{k_{-1}}[NO][Br_{2}][NO]

= \frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}]

Thus, rate law of formation of NOBr in terms of reactants is given by \frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}].









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3 years ago
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