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3 years ago

Ignore the answer I have selected but please help!! Ty in advance, also please explain

Chemistry

1 answer:

weeeeeb [17]3 years ago

4 0

Answer:

Iodine is most reactive because it is very close to having a "full shell" which is 8 electrons so they are "eager" to gain the last electron to became balanced, so that makes it the most reactive. Hope that helps:)

Explanation:

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When shopping for major appliances, people should look for high energy-rating numbers because those items _____.

madam [21]

The answer would be :Use less energy

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3 years ago

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Explain the water cycle in your own words. make sure u tell the steps in order

gregori [183]

Answer:

Collection - Evaporation - Condensation - Precipitation

Explanation:

The Water cycle is a series of continuous events that cause rain

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2 years ago

Explain the reaction of non-metals with oxygen

Ray Of Light [21]

Non-metals and oxygen produce non-metal oxides. Such as sulphur dioxide and nitrogen oxide which are responsible for acidic rain. Hope this helps

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3 years ago

What is the molarity of a solution made by dissolving 10.305 g of calcium carbonate, CaCO3, in water and diluting with water to

Gemiola [76]

Answer:

0.9359 M

Explanation:

M(CaCO3)=100.1 g/mol

10.305 g * 1 mol/100.1 g = 10.305/100.1 mol CaCO3

(10.305/100.1 mol )/0.1100 L = 0.936 mol/L = 0.9359 M

8 0

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

Ratling [72]

<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

8 0

3 years ago