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Aleonysh [2.5K]
3 years ago
6

A salad is an example of a mixture that is ... Mixed

Chemistry
1 answer:
melomori [17]3 years ago
8 0
Heterogeneous Mixture
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The mass of an atom of element x is equivalent to the total mass of 7 hydrogen atoms
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Hydrogen mass is 1, the atom with mass number 7 is Lithium
4 0
2 years ago
Calculate the molarity of a solution with 114.95 grams of sodium dissolved in 2 L of water. ​
koban [17]

Answer:

The molarity of a solution is 2.5 M

Explanation:

Molarity is a concentration unit that describes how much of a solution is dissolved in solution.

Molarity of a solution can found by using the formula,

Molarity (M) = (moles of solute)/(Liters of Solution).

Given, mass of Sodium = 114.95 grams.

Volume of water = 2 L.

Here, Sodium is solute as it is dissolved in water, which is the solvent.

Moles of Sodium(solute) can be found by using the formula,

Number of Moles = mass/Molecular weight.

mass of Sodium = 114.95 grams.

Molecular weight = 22.989 grams

Number of Moles of Sodium(solute) =114.95/22.989 = 5.

Substituting the values in the formula, we get,

M = 5/2 = 2.5 M

3 0
3 years ago
A 3.4 g sample of an unknown monoprotic organic acid composed of C,H, and O is burned in air to produce 8.58 grams of carbon dio
Pavlova-9 [17]

Answer:

C_7H_6O_2

Explanation:

Hello there!

In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.

(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:

n_C=8.58gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.195molC

And the moles of hydrogen due to the produced 1.50 grams of water:

n_H=1.50gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O}  =0.166molH

Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:

m_O=3.4g-0.195molC*\frac{12.01gC}{1molC}-0.166molH*\frac{1.01gH}{1molH} =0.89gO\\\\n_O=0.89gO*\frac{1molO}{16.0gO}=0.0556molO

Thus, the subscripts in the empirical formula are:

C=\frac{0.195}{0.0556}=3.5 \\\\H=\frac{0.166}{0.0556}=3\\\\O=\frac{0.0556}{0.0556}=1\\\\C_7H_6O_2

As they cannot be fractions.

(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:

n_{KOH}=0.279L*0.1mol/L\\\\n_{KOH}=0.0279mol

Which are equal to the moles of the acid:

n_{acid}=0.0279mol

And the molar mass:

MM_{acid}=\frac{3.4g}{0.0279mol} =121.86g/mol

(3) Finally, since the molar mass of the empirical formula is:

7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol

Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:

C_7H_6O_2

Best regards!

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