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marin [14]
3 years ago
10

Sugar dissolves when stirred into coffee. The coffee is the ________, the sugar is the ________, and the sweetened coffee is the

________.Question options:
A) solution; solvent; solute
B) solute; solvent; solution
C) solvent; solute; solution
D) solution; solute; solvent
Chemistry
1 answer:
inysia [295]3 years ago
4 0

Answer:

C

Explanation:

A solution is made when solute is dissolved in a solvent. Take a simple scenario for example, a salt solution typically is comprises of a an amount of salt mixed with a certain volume of water. The mixture of both is what makes a salt solution.

In this case, we need to make a solution if sweetened coffee. Prior to now, we know we have a particular coffee which is unsweetened that is has no sugar in it. Now, we are tasked with the responsibility of making a sweetened coffee. Our coffee is without sugar thus will serve as the solvent because it is the carrier medium. The sugar which obviously would dissolve in the solvent to form a solution would be our solute.

Now dissolving it in the unsweetened coffee thus will give us a sweetened coffee solution.

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3 0
3 years ago
Calculate the density of O2(g) at 415 K and 310 bar using the ideal gas and the van der Waals equations of state. Use a numerica
Lera25 [3.4K]

Answer:

Explanation:

From the given information:

The density of O₂ gas = d_{ideal} = \dfrac{P\times M}{RT}

here:

P = pressure of the O₂ gas = 310 bar

= 310 \ bar \times \dfrac{0.987 \ atm}{1 \ bar}

= 305.97 atm

The temperature T = 415 K

The rate R = 0.0821 L.atm/mol.K

molar mass of O₂  gas = 32 g/mol

∴

d_{ideal} = \dfrac{305.97 \ \times 32}{0.0821 \times 415}

d_{ideal} = 287.37 g/L

To find the density using the Van der Waal equation

Recall that:

the Van der Waal constant for O₂ is:

a = 1.382 bar. L²/mol²    &

b = 0.0319  L/mol

The initial step is to determine the volume = Vm

The Van der Waal equation can be represented as:

P =\dfrac{RT}{V-b}-\dfrac{a}{V^2}

where;

R = gas constant (in bar) = 8.314 × 10⁻² L.bar/ K.mol

Replacing our values into the above equation, we have:

310 =\dfrac{0.08314\times 415}{V-0.0319}-\dfrac{1.382}{V^2}

310 =\dfrac{34.5031}{V-0.0319}-\dfrac{1.382}{V^2}

310V^3 -44.389V^2+1.382V-0.044=0

After solving;

V = 0.1152 L

∴

d_{Van \ der \ Waal} = \dfrac{32}{0.1152}

d_{Van \ der \ Waal} = 277.77  g/L

We say that the repulsive part of the interaction potential dominates because the results showcase that the density of the Van der Waals is lesser than the density of ideal gas.

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Is 2AI + 3O2 = 2AI2O3 balanced or unbalanced?
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This equation is balanced.
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Can someone help me?
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Answer:

Explanation:

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