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klio [65]
3 years ago
7

A sample of seawater consists of water, sodium ions, chloride ions, and several other dissolved salts. These ions and salts are

evenly distributed throughout the water.
Which term or terms could be used to describe this sample of seawater?
a. Mixture
b. Heterogeneous mixture
c. Homogeneous mixture
d. Solution
e. Pure chemical substance
f. Compound
g. Element
Chemistry
1 answer:
wolverine [178]3 years ago
4 0

Answer:

The correct option is c

Explanation:

An homogeneous mixture is a mixture (either solid, liquid or gas) in which the constituents of the mixture are evenly distributed. Examples are pure sample of air, salt solution and sugar solution. Homogeneous mixture can be explained from the example given in the question; if 5 ml sample of a sea water is collected, the quantity of water, sodium ions, chloride ions and other dissolved salts will be equal in each 1 ml of the sample. This means the water, ions and salts are equally distributed.

When they are not equally distributed, it is known as heterogeneous mixture. Generally, a mixture is a combination of two or more substances that are not chemically combined or are physically combined.

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hope it helps

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3 0
4 years ago
A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required 10
masya89 [10]

<u>Answer:</u> The molar mass of unknown gas is 367.12 g/mol

<u>Explanation:</u>

Rate of a gas is defined as the amount of gas displaced in a given amount of time.

\text{Rate}=\frac{V}{t}

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation:

\text{Rate of effusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}

So,

\left(\frac{\frac{V_{X}}{t_{X}}}{\frac{V_{O_2}}{t_{O_2}}}\right)=\sqrt{\frac{M_{O_2}}{M_{X}}}

We are given:

Volume of unknown gas (X) = 1.0 L

Volume of oxygen gas = 1.0 L

Time taken by unknown gas (X) = 105 seconds

Time taken by oxygen gas = 31 seconds

Molar mass of oxygen gas = 32 g/mol

Molar mass of unknown gas (X) = ? g/mol

Putting values in above equation, we get:

\left(\frac{\frac{1.0}{105}}{\frac{1.0}{31}}\right)=\sqrt{\frac{32}{M_X}}\\\\M_X=367.12g/mol

Hence, the molar mass of unknown gas is 367.12 g/mol

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