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4 years ago

Why does the high pressure air being released into the ballast tanks of a submarine cause it to rise quickly?

Chemistry

1 answer:

Diano4ka-milaya [45]4 years ago

6 0

Answer:

Initially, the ballast tanks are filled with water. The weight of the submarine is equal to the upthrust of the water at the position of the submarine under water. When high pressure air is released into the ballast tanks displacing the water, the weight of the submarine becomes less than the upthrust of the water thus the net force is is upwards and it forces the sub to resurface. This is according to the Archimedes principle which states that a a body partially or wholly immersed in water displaces its own weight of the fluid in which it is immersed.

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How do atoms form new substances?

geniusboy [140]

Atoms combine as the electrons from each atom are attracted to the nuclei of the atoms. This results in bonds ranging from 100% covalent to bonds with high ionic character. The combination of atoms to form compounds occurs when the compounds being formed are at lower energy than the original atoms

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3 years ago

What is solubility? .

blsea [12.9K]

Solubility is the ability of a solid, liquid, or gaseous chemical substance (referred to as the solute) to dissolve in solvent (usually a liquid) and form a solution. The solubility of a substance fundamentally depends on the solvent used, as well as temperature and pressure.

4 0

3 years ago

Ammonia, NH3NH3 , can react with oxygen to form nitrogen gas and water. 4NH3(aq)+3O2(g)⟶2N2(g)+6H2O(l) 4NH3(aq)+3O2(g)⟶2N2(g)+6H

solmaris [256]

Answer:

36.37% is the percent yield of the reaction.

Explanation:

$4NH_3(aq)+3O_2(g)\rightarrow 2N_2(g)+6H_2O(l)$

1)0.650 L nitrogen gas , at 295 K and 1.01 bar.

Let the moles of nitrogen gas be n.

Pressure of the gas ,P= 1.01 bar = 0.9967 atm (1 bar = 0.9869 atm)

Temperature of the gas = T = 295 K

Volume of the gas = V = 0.650 L

Using an ideal gas equation:

$PV=nRT$

$n=\frac{PV}{RT}=\frac{0.9967 atm\times 0.650 L}{0.0821 atm L/mol K\times 295 K}=0.0267 mol$

2) Moles of ammonia gas= $\frac{2.53 g}{17 g/mol}=0.1488 mol$

Moles of oxygen gas = $\frac{3.53 g}{32 g/mol}=0.1101 mol$

According to reaction ,3 mol of oxygen reacts with 4 mol of ammonia.

Then,0.1101 mol of oxygen will react with:

$\frac{4}{3}\times 0.1101 mol=0.1468 mol$ of ammonia.

Hence, oxygen gas is in limiting amount and act as limiting reagent.

3) Theoretical yield of nitrogen gas :

According to reaction, 3 mol of oxygen gas gives 2 moles of nitrogen gas.

Then 0.1101 mol of oxygen will give:

$\frac{2}{3}\times 0.1101 mol=0.0734 mol$ of nitrogen.

Theoretical yield of nitrogen gas = 0.0734 mol

Experimental yield of nitrogen as calculated in part (1) = 0.0267 mol

Percentage yield:

$\frac{\text{Experiential yield}}{\text{Theoretical yield}}\times 100$

Percentage yield of the reaction:

$\frac{ 0.0267 mol}{0.0734 mol}\times 100=36.37\%$

36.37% is the percent yield of the reaction.

3 0

3 years ago

If a mixture looks smooth and the same throughout it is probably what

gulaghasi [49]

Answer:
If a mixture looks smooth and the same throughout it is probably Homogeneous.

Explanation:
Mixture is the combination of different compounds which are unreactive to each other.

Mixture are classified as ...

Solutions; in which the mixed compounds are thoroughly mixed and cannot be distinguished from each other and are said to be homogeneous. In solutions the size of solute is very small (i.e. Less than 1 nm).

Colloids; in which the solute is homogeneous visually but heterogeneous microscopically. The size of particles in this case is between 1 nm to 1 μm.

Suspensions; in which the mixture is heterogeneous, the particle size is greater than 1 μm and settles down (precipitation) under the influence of gravity.

4 0

3 years ago

Question 1 1 pts How many mols of bromine are present in 35.7g of Tin(IV) bromate?

sleet_krkn [62]

Answer:

n = 0.0814 mol

Explanation:

Given mass, m = 35.7g

The molar mass of Tin(IV) bromate, M = 438.33 g/mol

We need to find the number of moles of bromine. We know that,

No. of moles = given mass/molar mass

So,

$n=\dfrac{35.7}{438.33}\\\\n=0.0814\ mol$

So, there are 0.0814 moles of bromine in 35.7g of Tin(IV) bromate.

3 0

3 years ago