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bearhunter [10]
2 years ago
7

15.0 mL of an unknown clear liquid is added to a 50 mL graduated cylinder. The mass of the liquid is determined to be 12.7 grams

. What is the density of this unknown liquid?
A :0.85 g/mL
B :01.18 g/mL
C :0.847 g/mL
D :1.1811 g/mL​
Chemistry
1 answer:
sveticcg [70]2 years ago
4 0

Answer:

\boxed {\tt A. \ d=0.85 \ g/mL}

Explanation:

Density is found by dividing the mass by the volume.

d=\frac{m}{v}

The mass of the liquid is 12.7 grams.

We know that 15 mL of this liquid was added to a 50 mL graduated cylinder. Therefore, the volume is 15 mL. The 50 mL is not relevant, it only tells us about the graduated cylinder.

m= 12.7 \ g\\v= 15 \ mL

Substitute the values into the formula.

d=\frac{12.7 \ g}{ 15 mL}

Divide.

d=0.846666667 \ g/mL

Round to the nearest hundredth. The 6 in the tenth place tells us to round the 4 to a 5.

d \approx 0.85 \ g/mL

The density of the liquid is about 0.85 grams per milliliter and choice A is correct.

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Dolomite is a mixed carbonate of calcium and magnesium. Calcium and magnesium carbonates both decompose upon heating to produce
Setler79 [48]

Answer:

72.03 %

Explanation:

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Let the mass of Magnesium carbonate = x g

The mass of calcium carbonate = 9.66 - x g

Calculation of the moles of Magnesium carbonate as:-

Molar mass of Magnesium carbonate = 122.44 g/mol

The formula for the calculation of moles is shown below:

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

Thus,

Moles= \frac{x\ g}{84.3139\ g/mol}=\frac{x}{84.3139}\ mol

Calculation of the moles of calcium carbonate as:-

Molar mass of calcium carbonate = 100.0869 g/mol

Thus,

Moles= \frac{9.66 - x\ g}{100.0869\ g/mol}=\frac{9.66 - x}{100.0869}\ mol

According to the reaction shown below:-

MgCO_3\rightarrow MgO+CO_2

CaCO_3\rightarrow CaO+CO_2

In both the cases, the oxides formed from the carbonates in the 1:1 ratio.

So, Moles of MgO = \frac{x}{84.3139}\ mol

Molar mass of MgO = 40.3044 g/mol

Thus, Mass = Moles*Molar mass = \frac{x}{84.3139}\times 40.3044 \ g

Moles of CaO = \frac{9.66 - x}{100.0869}\ mol

Molar mass of CaO = 56.0774 g/mol

Thus, Mass = Moles*Molar mass = \frac{9.66 - x}{100.0869}\times 56.0774 \ g

Given that total mass of the oxide = 4.84 g

Thus,

\frac{x}{84.3139}\times 40.3044 +\frac{9.66 - x}{100.0869}\times 56.0774=4.84

\frac{40.3044x}{84.3139}+56.0774\times \frac{-x+9.66}{100.0869}=4.84

-694.1618435x+45673.48749\dots =40843.38968\dots

x=\frac{4830.09780\dots }{694.1618435}

x=6.9582

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\%\ mass=\frac{Mass_{MgCO_3}}{Total\ mass}\times 100

\%\ mass=\frac{6.9582}{9.66}\times 100=72.03\ \%

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zaharov [31]

Answer:

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Explanation:

Hello!

In this case, since the chemical reaction between copper and nitric acid is:

2HNO_3+Cu\rightarrow Cu(NO_3)_2+H_2

By starting with 0.80 g of copper metal (molar mass = 63.54 g/mol) and considering the 1:1 mole ratio between copper and copper (II) nitrate (molar mass = 187.56 g/mol) we can compute that mass via stoichiometry as shown below:

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However, the real reaction between copper and nitric acid releases nitrogen oxide, yet it does not modify the calculations since the 1:1 mole ratio is still there:

4HNO_3+Cu\rightarrow Cu(NO_3)_2+2H_2O+2NO_2

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