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3 years ago

9

Which part of a chemical formula indicates the number of copies of molecules or compounds? A. plus signs B. subscripts C. coeffi

cients D. chemical symbols

1 answer:

nadya68 [22]3 years ago

3 0

I think it is C. coefficients but I don't know what subscripts are

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When the products of a reaction are hotter than the reactants: The reaction is exothermic. The reaction is endothermic. The reac

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<span>Exothermic reaction evolves energy due to which products get hot...</span>

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Which equation demonstrates that nuclear fusion forms elements that are heavier than helium

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Answer: The correct option is 2.

Explanation: There are two types of nuclear reactions:

1) Nuclear fission: These reactions are defined as the reactions in which a heavier unstable nuclei breaks into two or more smaller stable nuclei.

2) Nuclear fusion: These reactions are the ones where two smaller nuclei fuse together or combine together to form a larger nuclei.

In the question, we need to find the fusion reaction which forms elements heavier than helium.

Option 1: In this fusion reaction occurs but the nuclei is Helium itself.

Option 2: In this also fusion reaction occurs and the nuclei is heavier than Helium which is Neon.

$_8^{16}\textrm{O}+_2^4\textrm{He}\rightarrow _{10}^{20}\textrm{Ne}$

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Hence, the correct option is 2.

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Perform the following calculation, 39 / 24.2 . The answer rounded to the correct number of significant figures is which of the f

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Answer:

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39 / 24.2 = 943.8

Explanation:

943.8 to 2 s.f (best option for rounding)

= 940

ANSWER = 940

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3 years ago

Chlorine atoms react with methane, forming HCl and CH3. The rate constant for the reaction was determined to be 3.600×107 at 278

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<u>Answer:</u> The activation energy for the reaction is 40.143 kJ/mol

<u>Explanation:</u>

To calculate activation energy of the reaction, we use Arrhenius equation for two different temperatures, which is:

$\ln(\frac{K_{317K}}{K_{278K}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_{317K}$ = equilibrium constant at 317 K = $3.050\times 10^{8}M^{-1}s^{-1}$

$K_{278K}$ = equilibrium constant at 278 K = $3.600\times 10^{7}M^{-1}s^{-1}$

$E_a$ = Activation energy = ?

R = Gas constant = 8.314 J/mol K

$T_1$ = initial temperature = 278 K

$T_2$ = final temperature = 317 K

Putting values in above equation, we get:

$\ln(\frac{3.050\times 10^8}{3.600\times 10^{7}})=\frac{E_a}{8.314J/mol.K}[\frac{1}{278}-\frac{1}{317}]\\\\E_a=40143.3J/mol=40.143kJ/mol$

Hence, the activation energy for the reaction is 40.143 kJ/mol

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3 years ago