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luda_lava [24]
4 years ago
13

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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Balancing oxidation-reduction reactions <br> Mg+ N2—&gt;Mg3N2
BartSMP [9]

Answer:

{ \sf{3Mg_{(s)} + N_{2(g)} →Mg _{3}N_{2(s)}}}

3 0
3 years ago
What is the theory that can explain the model of Pangaea?
NISA [10]
The bones of the same animal found out continents far away from each other
4 0
3 years ago
Given the following heats of combustion. CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(l) ΔH°rxn = -726.4 kJ C(graphite) + O2(g) CO2(g) ΔH
sergeinik [125]

Answer:

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation:

The formation reaction of CH_3OH will be,

C(s)+2H_2(g)+\frac{1}{2}O_2\rightarrow CH_3OH(g),\Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

C(graphite)+O_2(g)\rightarrow CO_2(g), \Delta H_1=-393.5kJ/mole..[1]

H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l), \Delta H_2=-285.8kJ/mole..[2]

CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l) , \Delta H_3=-726.4kJ/mole..[3]

Now we will reverse the reaction 3, multiply reaction 2 by 2  then adding all the equations, Using Hess's law:

We get :

C(graphite)+O_2(g)\rightarrow CO_2(g) , \Delta H_1=-393.5kJ/mole..[1]

2H_2(g)+2O_2(g)\rightarrow 2H_2O(l) ,\Delta H_2=2\times (-285.8kJ/mole)=-571.6kJ/mol..[2]

CO_2(g)+2H_2O(l)\rightarrow CH_3OH(g)+\frac{3}{2}O_2(g) ,\Delta H_3=726.4kJ/mole [3]

The expression for enthalpy of formation of C_2H_4 will be,

\Delta H_{formation}=\Delta H_1+2\times \Delta H_2+\Delta H_3

\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)

\Delta H=-238.7kJ/mole

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

4 0
3 years ago
A compound composed of 3. 3 % h, 19. 3 % c, and 77. 4 % o has a molar mass of approximately 60 g/mol. What is the molecular form
Brut [27]

The molecular formula of the given compound is $\mathrm{H}_{2} \mathrm{CO}_{3}$ also known as  Carbonic acid.

<h3>What is empirical formula and molecular formula?</h3>

The simplest whole-number ratio of the various atoms in a compound is represented by an empirical formula.

The precise number of various atom types present in a compound's molecule is indicated by the molecular formula.

Given that,

H = 3.3%

C = 19.3%

O = 77.4%

No. of moles of H = 3.3/1

No. of moles of H = 3.3

No. of moles of C = 19.3 / 12

No. of moles of C = 1.60

No. of moles of O = 77.4/16

No. of moles of O = 4.83

Therefore, the ratio of the atoms of C, H and O = 3.3 : 1.60 : 4.83

Divide by smallest value which you get =3.3 / 1.60 : 1.60 / 1.60 : 4.83 / 1.60

The ratio of the atoms of C, H and O = 2 : 1 : 3

So, the empirical formula is $\mathrm{H}_{2} \mathrm{CO}_{2}$

Let the molecular formula is $\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right) \mathrm{n}$

Then, molar mass $=(2 \times 1+1 \times 12+3 \times 16) n\\

Molar mass = 62n

As the question, 62 n = 60

n = 0.96 or n = 1 (rounded off to nearest ones)

So, the molecular formula is $\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right) 1=\mathrm{H}_{2} \mathrm{CO}_{3}$ i.e., the compound is Carbonic acid.

To know more about molecular formula visit:

brainly.com/question/14425592

#SPJ4

7 0
2 years ago
0.500 L of a gas is collected at 2911 MM and 0°C. What will the volume be at STP?
ioda

Answer:

V₂ =  1.92 L

Explanation:

Given data:

Initial volume = 0.500 L

Initial pressure =2911 mmHg (2911/760 = 3.83 atm)

Initial temperature = 0 °C (0 +273 = 273 K)

Final temperature = 273 K

Final volume = ?

Final pressure = 1 atm

Solution:

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

by putting values,

V₂ = P₁V₁ T₂/ T₁ P₂  

V₂ = 3.83 atm × 0.500 L × 273 K / 273 K × 1 atm

V₂ = 522.795 atm .L. K / 273 K.atm

V₂ =  1.92 L

4 0
3 years ago
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