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USPshnik [31]
3 years ago
6

Which gas law is considered when scuba diving?

Chemistry
2 answers:
jarptica [38.1K]3 years ago
5 0
<h2>Answer:</h2>

The law is <u>Boyel's law</u>.

<h2>Explanation:</h2>

One of the primary laws of material science that impacts the submerged jumping condition for scuba jumpers is Boyle's law.  

The law states that as weight changes, the volume of gases in a jumper's body pits and adaptable gear changes as well. Where the water weight builds, 'air spaces' reduces in size, yet as water weight diminishes, the 'air spaces' increments in size. Where the two changes are in direct extent to the weight increment or lessening, the temperature is held consistent.  

m_a_m_a [10]3 years ago
3 0

Answer:

Combined Gas Law

Explanation:

PV/T=PV/T

The pressure and temperature fluctuates depending on how deep to travel and the volume depends on the size of the oxygen tank.

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So among the difference plasma will have high kinetic energy and solids will have high density in comparison with each other.

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Which of the following is an example of a MOLECULE? *
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Convert the boiling temperature of liquid ammonia, -28.1 degree fahrenheit, into degrees celsius and kelvin.
vova2212 [387]

Answer:

°C = -33.39 °C

K = 239.76 K

Explanation:

  • °C = ( °F -32 ) / 1.8
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⇒ °C = ( -28.1 - 32 ) / 1.8

⇒ °C = - 33.39 °C

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6 0
4 years ago
An impure sample of solid Na₂CO₃ is allowed to react with 0.1755 M HCl.
jasenka [17]

Answer: Thus percentage purity of sodium carbonate is 60.83%.

Explanation:

Na_2CO_3 + 2HCl \longrightarrow 2NaCl + CO_2 + H_2O

Volume of the HCl solution ,V = 15.55 mL= 0.01555 L

Concentration of HCl solution ,C= 0.1755 M

Moles of HCl in 0.1755 M solution = n

Molarity=\frac{n}{V(L)}

n=C\time V=0.1755 M\times 0.01555 L = 2.729\times 10^{-3} mol

According o reaction 2 moles of HCl reacts with 1 mole of sodium carbonate.

Then n moles of HCl will react with:

\frac{1}{2}\times n=\frac{1}{2}\times 2.729\times 10^{-3} mol=1.3645\times 10^{-3} mol of sodium carbonate.

Mass of 1.364\times 10^{-3} mol of sodium carbonate:

1.364\times 10^{-3} mol\times 106 g/mol=0.144584 g

Mass of the sample given= 0.2377 grams

Mass used up in reaction = 0.144585 g

percentage purity= \frac{0.144585}{0.2337}\times 100=60.83\%

Thus percentage purity of sodium carbonate is 60.83%.

4 0
4 years ago
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