Which of the following best describes seawater?

2 answers:

4 0

The correct answer to this question is B. The best description for seawater would be a homogeneous mixture. A homogeneous mixture is one in which the composition remains the same, so you cannot tell the difference between different parts of it. If the question is making a reference to pure seawater, as in the mixture between water and salt, then the answer is A.

gogolik [260]2 years ago

3 0

Answer:

Homogeneous Mixture

Explanation:

No real explanation. I just took the test and got this question correct! :)

You might be interested in

An irregularly shaped solid which has a mass of 10.283g was placed in a graduated cylinder containing an inert liquid. The initi

Gwar [14]

Liquid + Solid = 8.89 mL
V ( Solid ) = 8.89 mL - 6.26 mL = 2.63 mL
The density of the solid = m / V = 10.283 g / 2.63 mL =
= 3.9 g/mL = 3.9 g / cm³

3 0

2 years ago

12. The most common factors that cause chemical reactions to occur are all the following except

ruslelena [56]

Transfer of electrons

6 0

2 years ago

The molecular weight of a gas is ________ g/mol if 6.7 g of the gas occupies 6.3 l at stp.

PIT_PIT [208]

At STP, also known as standard temperature and pressure, 1 mole of a gas occupies 22.4 L. Since we are given with the volume of 6.3L, we calculate the amount of gas in mol.
n = (6.3L)/ (22.4L/mol) = 0.28125 mol
We are given with the mass of 6.7 g. Therefore, the molar mass or molecular weight of the gas is equal to,
6.7g/0.28125 mol = 23.82 g/mol

6 0

2 years ago

6.0 g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of 13

vodomira [7]

Answer:

The molecular formula of X is given as $C_7 H_6 O_3$

Explanation:

$Moles $C O_{2}=\frac{\text { mass }}{\text { molar mass }}=\frac{13.39 \mathrm{g}}{44.01 \mathrm{g} \text { per mole }}=0.304 \mathrm{mol}$\\\\moles $\mathrm{C}=$ moles $\mathrm{CO}_{2}=0.304 \mathrm{mol}$$

$mass $C=$ moles $\times$ molar mass $=0.304 \mathrm{mol} \times 12 \frac{g}{m o l}=3.65g$\\\\moles $\mathrm{H}_{2} \mathrm{O}=\frac{2.35 \mathrm{g}}{18.02 \mathrm{g} \text { permole }}=0.130 \mathrm{mol}$\\\\moles $\mathrm{H}=2 \times$ moles $\mathrm{H}_{2} \mathrm{O}=0.130 \times 2=0.260 \mathrm{mol}$\\\\Mass $\mathrm{H}=0.260 \mathrm{mol} \times 1.008 \frac{g}{\mathrm{mol}}=0.262 \mathrm{g}$$