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Jobisdone [24]
2 years ago
14

Describe an example that could be an observational study

Chemistry
1 answer:
Valentin [98]2 years ago
4 0
This is an observational study, because the researcher is simply observing the answers of the survey without influencing the outcome in any way. Another example of an observational study would be if a researcher was trying to determine the effects that eating strictly organic foods has on overall health.
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List three examples of a carbon reservoir.
romanna [79]

Answer:

Oceans, Fossil fuels, atmosphere

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In just thomsons experiments with electricity he showed that an electrical current can be
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Which of the following processes rearranges atoms during a reaction?
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A) Burning
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3 years ago
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28.7 l of propane, c3h8, are consumed in a combustion reaction. how many grams of water are produced?
lora16 [44]
The chemical reaction would be:

C3H8 + 5O2 = 3CO2 + 4H2O

For this case, we assume that gas is ideal thus in every 1 mol the volume would be 22.41 L. We calculate as follows:

28.7 L C3H8 ( 1 mol / 22.41 L ) ( 4 mol H2O / 1 mol C3H8 ) ( 18.02 g / mol ) = 92.31 g H2O produced

Hope this answers the question.
7 0
3 years ago
Acetylene gas, C2H2, can be produced by the reaction of calcium carbide and water. CaC2(s) + 2H2O(l) --> C2H2(g) + Ca(OH)2(aq
nekit [7.7K]

Answer:

1.0 L

Explanation:

Given that:-

Mass of CaC_2 = 2.54\ g

Molar mass of CaC_2 = 64.099 g/mol

The formula for the calculation of moles is shown below:

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

Thus,

Moles= \frac{2.54\ g}{64.099\ g/mol}

Moles_{CaC_2}= 0.0396\ mol

According to the given reaction:-

CaC_2_{(s)} + 2H_2O_{(l)}\rightarrow C_2H_2_{(g)} + Ca(OH)_2_{(aq)}

1 mole of CaC_2 on reaction forms 1 mole of C_2H_2

0.0396 mole of CaC_2 on reaction forms 0.0396 mole of C_2H_2

Moles of C_2H_2 = 0.0396 moles

Considering ideal gas equation as:-

PV=nRT

where,

P = pressure of the gas = 742 mmHg  

V = Volume of the gas = ?

T = Temperature of the gas = 26^oC=[26+273]K=299K

R = Gas constant = 62.3637\text{ L.mmHg }mol^{-1}K^{-1}

n = number of moles = 0.0396 moles

Putting values in above equation, we get:

742mmHg\times V=0.0396 mole\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 299K\\\\V=\frac{0.0396\times 62.3637\times 299}{742}\ L=1.0\ L

<u>1.0 L of acetylene  can be produced from 2.54 g CaC_2.</u>

4 0
3 years ago
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