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3 years ago

True or False Gases do have a definite shape and volume.

Chemistry

2 answers:

____ [38]3 years ago

8 0

Answer:

false

Explanation:

Mumz [18]3 years ago

8 0

Answer:

False

Explanation:

It is false as the particles of gases keep moving. They have very weak molecular bonds. They are so weak that they can move freely anywhere and in any direction. Even if you trap some gases and a closed bottle, they keep moving. We can't measure what the shape is or what the volume is. Therefore, it is false.

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3 years ago

Which statement is correct about the theory of combustion? (1 point)

Vadim26 [7]

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2 years ago

For the following reaction, 5.61 grams of carbon monoxide are mixed with excess water . Assume that the percent yield of carbon

stealth61 [152]

<u>Answer:</u> The ideal yield of carbon dioxide is 7.506 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of carbon monoxide = 5.61 g

Molar mass of carbon monoxide = 28 g/mol

Putting values in equation 1, we get:

$\text{Moles of carbon monoxide}=\frac{5.61g}{28g/mol}=0.200mol$

The chemical equation for the reaction of carbon monoxide and water follows:

$CO(g)+H_2O(l)\rightarrow CO_2(g)+H_2(g)$

By Stoichiometry of the reaction:

1 mole of carbon monoxide produces 1 mole of carbon dioxide

So, 0.200 moles of carbon monoxide will produce = $\frac{1}{1}\times 0.200=0.200mol$ of carbon dioxide

Now, calculating the mass of carbon dioxide from equation 1, we get:

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = 0.200 moles

Putting values in equation 1, we get:

$0.200mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(0.200mol\times 44g/mol)=8.8g$

To calculate the experimental yield of carbon dioxide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Percentage yield of carbon dioxide = 85.3 %

Theoretical yield of carbon dioxide = 8.8 g

Putting values in above equation, we get:

$85.3=\frac{\text{Experimental yield of carbon dioxide}}{8.8g}\times 100\\\\\text{Experimental yield of carbon dioxide}=\frac{85.3\times 8.8}{100}=7.506g$