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galben [10]
3 years ago
13

A diver has a lung capacity of 2.4 L when the pressure is 0.8 atm. What is the volume of the diver’s lungs when the pressure cha

nges to 1.2 atm?
a) 1.60 L

b) 2.50 L

c) 3.60 L

d) 1.40 L
Chemistry
1 answer:
tangare [24]3 years ago
7 0

Answer:

The final volume is 1.6 L.

Explanation:

It is given that,

A diver has a lung capacity of 2.4 L when the pressure is 0.8 atm. We need to find the volume of the diver’s lungs when the pressure changes to 1.2 atm. Let V₂ is volume.

It is based on Boyle's law. According to this law,

PV=K

K is constant

P_1V_1=P_2V_2

V_2=\dfrac{P_1V_1}{P_2}\\\\V_2=\dfrac{0.8\times 2.4}{1.2}\\\\V_2=1.6\ L

So, the final volume is 1.6 L.

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How many grams of Cl2 are in 1.20 x 1024 Cl atoms?
Vaselesa [24]
<h3>Answer:</h3>

70.906 g

<h3>Explanation:</h3>

We are given;

  • Atoms of Chlorine = 1.2 × 10^24 atoms

We are required to calculate the mass of Chlorine

  • We know that 1 mole of an element contains atoms equivalent to the Avogadro's number, 6.022 × 10^23.
  • That is , 1 mole of an element = 6.022 × 10^23 atoms
  • Therefore; 1 mole of Chlorine = 6.022 × 10^23 atoms

But since Chlorine gas is a molecule;

  • 1 mole of Chlorine gas = 2 × 6.022 × 10^23 atoms

But, molar mass of Chlorine gas = 70.906 g/mol

Then;

70.906 g Of chlorine gas = 2 × 6.022 × 10^23 atoms

                                          = 1.20 × 10^24 atoms

Thus;

For 1.2 × 10^24 atoms ;

= ( 70.906 g/mol × 1.2 × 10^24 atoms ) ÷ (1.20 × 10^24 atoms)

<h3>=  70.906 g </h3>

Therefore, 1.20 × 10^24 atoms of chlorine contains a mass of 70.906 g

=  

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The elementary reaction 2H2O(g)↽−−⇀2H2(g)+O2(g) 2H2O(g)↽−−⇀2H2(g)+O2(g) proceeds at a certain temperature until the partial pres
Dima020 [189]

Answer:

6.25\times 10^{-6} is the value of the equilibrium constant at this temperature.

Explanation:

Equilibrium constant in terms of partial pressure is defined as the ratio of partial pressures of products to the partial pressures  of reactants each raised to the power equal to their stoichiometric ratios. It is expressed as K_{p}

2H_2O(g)\rightleftharpoons 2H_2(g)+O_2(g)

Partial pressures at equilibrium:

p^o_{H_2O}=0.070 atm

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p^o_{O_2}=0.0025 atm

The equilibrium constant in terms of pressures is given as:

K_p=\frac{(p^o_{H_2})^2\times (p^o_{O_2})}{(p^o_{H_2O})62}

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