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prisoha [69]
3 years ago
6

A system of knowledge based on facts and principles

Chemistry
1 answer:
Ymorist [56]3 years ago
8 0
I’m pretty sure the answer you’re looking for is science...but I’m not quite sure because your question is kinda vague
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A coffee cup calorimeter will not be used to directly measure the enthalpy of magnesium combustion because
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Answer:

A coffee cup calorimeter is great for measuring heat flow in a solution, but it can't be used for reactions that involve gases since they would escape from the cup. The coffee cup calorimeter can't be used for high-temperature reactions, either, because they would melt the cup.

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Sedimentary rocks are described as small pieces of rock and animal remains that are weathered, swept downstream and then settle,
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What type of bond would occur between carbon (C) and nitrogen (N)
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Answer:

covalent

Explanation:

The carbon and the nitrogen very often form bonds in nature, carbon-nitrogen bonds, which are covalent types of bonds. In fact, the bonds between the carbon and nitrogen are one of the most abundant in the biochemistry and the organic chemistry. The bonds between these two can be double bonds, as well as triple bonds. The carbon-nitrogen bonds have the tendency to be strongly polarized toward the nitrogen.

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If 0.580 moles of a monoprotic weak acid (ka = 7.4 10-5 is titrated with naoh, what is the ph of the solution at the half-equiva
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7 0
3 years ago
Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 2.7 g of ethane is m
Bond [772]

Answer:

m_{H_2O}=4.86gH_2O

Explanation:

Hello,

In this case, the described chemical reaction is:

C_2H_6+\frac{7}{2} O_2\rightarrow 2CO_2+3H_2O

Thus, for the given reacting masses, we must identify the limiting reactant for us to determine the maximum mass of water that could be produced, therefore, we proceed to compute the available moles of ethane:

n_{C_2H_6}=2.7gC_2H_6*\frac{1molC_2H_6}{30gC_2H_6} =0.09molC_2H_6

Next, we compute the moles of ethane consumed by 13.0 grams of oxygen by using the 1:7/2 molar ratio between them:

n_{C_2H_6}^{consumed\ by \ O_2}=13.0gO_2*\frac{1molO_2}{32gO_2}*\frac{1molC_2H_6}{\frac{7}{2} molO_2}=0.116molC_2H_6

Thus, we notice there are less available moles of ethane, for that reason, it is the limiting reactant, thereby, the maximum amount of water is computed by considering the 1:3 molar ratio between ethane and water:

m_{H_2O}=0.09molC_2H_6*\frac{3molH_2O}{1molC_2H_6} *\frac{18gH_2O}{1molH_2O} \\\\m_{H_2O}=4.86gH_2O

Best regards.

3 0
3 years ago
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