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1 year ago

Earthquakes result from the build-up and pressure. True or false?

Chemistry

1 answer:

xxTIMURxx [149]1 year ago

4 0

The answer is True.... Earthquakes result from the build up and pressure

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What is the molarity of an intravenous glucose solution prepared from 108 g of glucose in 2.0 L of solution? 0.018 mol/L 0.30 mo

scZoUnD [109]

Answer:

0.30 mol/L

Explanation:

Mass = 108 g

Molar mass of glucose = 180.156 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{108\ g}{180.156\ g/mol}$

$Moles= 0.5995\ mol$

Given Volume = 2 L

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity=\frac{0.5995}{2}$

Molarity = 0.3 mol/L

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1 year ago

Chemistry skeletal structure

yan [13]

Answer:

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1 year ago

In the carbon-oxygen cycle, carbon dioxide is used for _____ and oxygen is used for _____.

LuckyWell [14K]

A.respiration, photosynthesis

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1 year ago

Read 2 more answers

Molality is a way to express the concentration of a solution that represents the number of moles of

GrogVix [38]

Answer:

Explanation:

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1 year ago

How many grams of H2O will be formed when 32.0 g H2 is mixed with 84.0 g of O2 and allowed to react to form water

Zarrin [17]

Answer:

94.58 g of $H_2O$

Explanation:

For this question we have to start with the reaction:

$H_2~+~O_2~->~H_2O$

Now, we can balance the reaction:

$2H_2~+~O_2~->~2H_2O$

We have the amount of $H_2$ and the amount of $O_2$ . Therefore we have to find the limiting reactive, for this, we have to follow a few steps.

1) Find the moles of each reactive, using the molar mass of each compound ( $H_2~=~2~g/mol~~O_2=~32~g/mol$ ).

2) Divide by the coefficient of each compound in the balanced reaction ("2" for $H_2$ and "1" for $O_2$ ).

Find the moles of each reactive

$32.0~g~H_2\frac{1~mol~H_2}{2~g~of~H_2}=15.87~mol~H_2$

$84.0~g~of~O_2\frac{1~mol~of~O_2}{32~g~of~O_2}=2.62~mol~of~O_2$

Divide by the coefficient

$\frac{15.87~mol~H_2}{2}=7.94$

$\frac{2.62~mol~of~O_2}{1}=2.62$

The smallest values are for $H_2$ , so hydrogen is the limiting reagent. Now, we can do the calculation for the amount of water:

$32.0~g~H_2\frac{1~mol~H_2}{2~g~of~H_2}\frac{2~mol~H_2O}{2~mol~H_2}\frac{18~g~H_2O}{1~mol~H_2O}=94.58~g~H_2O$

We have to remember that the molar ratio between $H_2O$ and $H_2$ is 2:2 and the molar mass of $H_2O$ is 18 g/mol.

6 0

1 year ago