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2 years ago

4. Why can't the subscripts be changed in a chemical equation in chemistry

1 answer:

4 0

If you change the subscripts it would change the reactants or products and then you would be solving a different formula, you would change what the chemical is

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Mintu needs to determine the missing ecosystem in the set.

alexandr1967 [171]

The answer is Wetland and Stream

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3 years ago

Read 2 more answers

The actual yield of a certain reaction is 44.0 g, while the theoretical yield is 50.0 g. Calculate the percent yield.

Goshia [24]

Percent yield represents to what extent the reaction runs to completion. In this, the theoretical yield is 50 grams (100% completion).

To calculate percent yield, divide the actual by the theoretical. In doing so, the percent yield is 88% (44/50).

3 0

3 years ago

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James decided to do an experiment with a can of soda. He opened a 340 g can of soda and left it open for a day. The next day, he

amm1812

Answer:

A - The can was open and gases were released.

8 0

3 years ago

The activation energy of a certain uncatalyzed reaction is 64 kJ/mol. In the presence of a catalyst, the Ea is 55 kJ/mol. How ma

Ksivusya [100]

Answer:

About 5 times faster.

Explanation:

Hello,

In this case, since the Arrhenius equation is considered for both the catalyzed reaction (1) and the uncatalized reaction (2), one determines the relationship between them as follows:

$\frac{k_1}{k_2}=\frac{Aexp(-\frac{Ea_1}{RT} )}{Aexp(-\frac{Ea_2}{RT})} \\\frac{k_1}{k_2}=\frac{exp(-\frac{Ea_1}{RT} )}{exp(-\frac{Ea_2}{RT})}$

By replacing the corresponding values we obtain:

$\frac{k_1}{k_2}=\frac{exp(-\frac{55000J/mol}{8.314J/molK*673.15K} )}{exp(-\frac{64000J/mol}{8.314J/molK*673.15K} )} =4.8$

Such result means that the catalyzed reaction is about five times faster than the uncatalyzed reaction.

Best regards.

4 0

3 years ago

If 3 moles of a compound use 12 J of energy in a reaction, what is the Hreaction in kJ/mol

igomit [66]

Answer:

$\Delta _RH=4x10^{-3}\frac{kJ}{mol}$

Explanation:

Hello,

In this case, the molar enthalpy of reaction is obtained by dividing the involved energy by the reacting moles:

$\Delta _RH=\frac{12J}{3mol} =4\frac{J}{mol}$

Thus, it is important to notice that the compound "uses" the energy, it means that it absorbs the energy, for that reason the sign is positive. Moreover, computing the result in kJ/mol we finally obtain:

$\Delta _RH=4\frac{J}{mol}*\frac{1kJ}{1000J} =4x10^{-3}\frac{kJ}{mol}$

Best regards.

5 0

2 years ago