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3 years ago

3. Calculate the volume, in mL, of 100.0 g of alcohol, if the density is 0.79 g/mL.

Chemistry

1 answer:

Naya [18.7K]3 years ago

4 0

Answer:

The answer is

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

<h3> $volume = \frac{mass}{density}$ </h3>

From the question

mass of alcohol = 100 g

density = 0.79 g/mL

The volume is

$volume = \frac{100}{0.79} \\ = 126.58227848...$

We have the final answer as

Hope this helps you

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Chemicals used in a laboratory investigation must be safely discarded. Which method is the safest way to dispose of chemicals at

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Answer:

Disposing of the discarded chemicals can be done by certain sfae methods mention as follows:

Explanation:

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Recycling of these chemicals and incineration is also use for the industrial chemical waste under professional guidance. Incineration is the process of burning chemicals in to ash through high thermal burning.

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3 years ago

18. What is the percent by mass sugar of a solution made by dissolving 12.45 grams of sugar in 100 grams of water?

Fantom [35]

The percent by mass sugar of a solution : 11.07%

<h3>Further explanation</h3>

Given

mass of sugar = 12.45 g

mass of water = 100 g

Required

The percent by mass

Solution

Mass of solution :

$\tt mass~sugar+mass~water=12.45+100=112.45~g$

Percent mass of sugar :

$\tt \%mass=\dfrac{mass~sugar}{mass~solution}\times 100\%\\\\\%mass=\dfrac{12.45}{112.45}\times 100\%=11.07\%$

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3 years ago

In a lab experiment 80.0 g of ammonia [NH3] and 120 g of oxygen are placed in a reaction vessel. At the end of the reaction 72.2

valentinak56 [21]

The percent yield of the reaction : 89.14%

<h3>Further explanation</h3>

Reaction of Ammonia and Oxygen in a lab :

mass NH₃ = 80 g

mol NH₃ (MW=17 g/mol):

$\dfrac{80}{17}=4.706$

mass O₂ = 120 g

mol O₂(MW=32 g/mol) :

$\tt \dfrac{120}{32}=3.75$

Mol ratio of reactants(to find limiting reatants) :

$\tt \dfrac{4.706}{4}\div \dfrac{3.75}{5}=1.1765\div 0.75\rightarrow O_2~limiting~reactant(smaller~ratio)$

mol of H₂O based on O₂ as limiting reactants :

mol H₂O :

$\tt \dfrac{6}{5}\times 3.75=4.5$

mass H₂O :

4.5 x 18 g/mol = 81 g

The percent yield :

$\tt \%yield=\dfrac{actual}{theoretical}\times 100\%\\\\\%yield=\dfrac{72.2}{81}\times 100\%=89.14\%$

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3 years ago

Write a balanced equation for the following:

Ket [755]

(ANS1)— P4 + 5O2 ---> 2P2O5

(ANS2)— C3H8 + 5O2---> 3CO2 + 4H20

(ANS3)— Ca2Si + 4Cl2 ---> 2CaCl2 + SiCl4

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3 years ago

The standard internal energy change for a reaction can be symbolized as Δ U ∘ rxn or Δ E ∘ rxn . For each reaction equation, cal

rosijanka [135]

Answer : The internal energy change is, -506.3 kJ/mol

Explanation :

Formula used :

$\Delta H=\Delta U+\Delta n_gRT$

or,

$\Delta U=\Delta H-\Delta n_gRT$

where,

$\Delta H$ = change in enthalpy = $-511.3kJ/mol=-511300.0J/mol$

$\Delta U$ = change in internal energy = ?

$\Delta n_g$ = change in moles

Change in moles = Number of moles of product side - Number of moles of reactant side

According to the reaction:

Change in moles = 0 - 2 = -2 mole

That means, value of $\Delta n_gRT$ = 0

R = gas constant = 8.314 J/mol.K

T = temperature = $25^oC=273+25=298K$

Now put all the given values in the above formula, we get

$\Delta U=\Delta H-\Delta n_gRT$

$\Delta U=(-511300.0J/mol)-[-2mol\times 8.314J/mol.K\times 298K$

$\Delta U=-511300.0J/mol+4955.144J/mol$

$\Delta U=-506344.856J/mol=-506.3kJ/mol$

Therefore, the internal energy change is -506.3 kJ/mol

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3 years ago