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2 years ago

What can cause an increase in the density of ocean water?

2 answers:

3 0

Your answer would be A because Ocean water gets more dense with a decrease in water temp. also an increase in salinity could increase the density.

posledela2 years ago

3 0

Answer: The correct answer is option(A).

Explanation:

Salinity is defined as measure of all the salts dissolved in water.

Salinity of the ocean gets effected by the temperature:

When the temperature increases the the density of the water decrease by density of the ocean water decrease.
When the temperature decreases the density of the water increases by which density if the ocean water ls increases.

So, in order to increase the density of the ocean water temperature should be decreases.

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Which reacts faster, rock salt or grains of salt? Why?

sweet-ann [11.9K]

Answer:

salt

Explanation:

because it has a much larger volume ratio

5 0

3 years ago

What is HC2H3O2 in science

Leona [35]

The chemical name of Hc2h3o2 is Acetic Acid.

3 0

3 years ago

Read 2 more answers

If you combine 27.1 g of a solute that has a molar mass of 27.1 g/mol with 100.0 g of a solvent, what is the molality of the res

baherus [9]

Answer: The molality of the solution is 0.1 m.

Explanation:

To calculate the molality of solution, we use the equation:

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

Where,

$m_{solute}$ = Given mass of solute = 27.1 g

$M_{solute}$ = Molar mass of solute = 27.1 g/mol

$W_{solvent}$ = Mass of solvent = 100 g

Putting values in above equation, we get:

$\text{Molality of solution}=\frac{27.1\times 1000}{27.1\times 100}\\\\\text{Molality of solution}=0.1m$

Hence, the molality of the solution is 0.1 m.

8 0

3 years ago

Read 2 more answers

A solution of NaOH is titrated with H2SO4. It is found that 20.05 mL of 0.3564 M H2SO4 solution is equivalent to 43.42 mL of NaO

Darya [45]

Answer : The concentration of NaOH is, 0.336 M

Explanation:

To calculate the concentration of base, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=0.3564M\\V_1=20.05mL\\n_2=1\\M_2=?\\V_2=43.42mL$

Putting values in above equation, we get:

$2\times 0.3564M\times 20.05mL=1\times M_2\times 43.42mL$

$M_2=0.336M$

Thus, the concentration of NaOH is, 0.336 M

3 0

3 years ago

The piece of iron that miguel measured had a mass of 51.1 g and a volume of 6.63 cm 3 . what did miguel calculate to be the dens

likoan [24]

Given:
Mass, m = 51.1 g
Volume, V = 6.63 cm³

By definition,
Density = Mass/Volume
= (51.1 g)/(6.63 cm³)
= 7.7074 g/cm³

In SI units,
Density = (7.7074 g/cm³)*(10⁻³ kg/g)*(10² cm/m)³
= 7707.4 kg/m³

Answer: 7.707 g/cm³ or 7707.4 kg/m³

4 0

3 years ago