Answer: why do I think that water boils when the vapor pressure is equal to the atmospheric pressure?
because The pressure level of a liquid lowers the number of pressure exerted on the liquid by the atmosphere. ... pressure level may be increased by heating a liquid and causing more molecules to enter the atmosphere. At the purpose where the pressure level is up to the air pressure boiling will begin.
Explanation: I hope this helps, have a good day :)
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Answer:
0.00500M of Na₂C₂O₄
Explanation:
<em>When are dissolved in 150 mL of 1.0 M H2SO4.</em>
<em />
We can solve this problem finding molarity of sodium oxalate: That is, moles of Na2C2O4 per liter of solution. Thus, we need to convert the 0.1005g to moles using molar mass of sodium oxalate (134g/mol) and dividing in the 0.150L of the solution:
0.1005g * (1mol / 134g) = 7.5x10⁻⁴ moles of Na₂C₂O₄
In 0.150L:
7.5x10⁻⁴ moles of Na₂C₂O₄ / 0.150L =
<h3>0.00500M of Na₂C₂O₄</h3>
<u>Answer:</u>
<u>For A:</u> The average molecular speed of Ne gas is 553 m/s at the same temperature.
<u>For B:</u> The rate of effusion of
gas is 
<u>Explanation:</u>
<u>For A:</u>
The average molecular speed of the gas is calculated by using the formula:

OR

where, M is the molar mass of gas
Forming an equation for the two gases:
.....(1)
Given values:

Plugging values in equation 1:

Hence, the average molecular speed of Ne gas is 553 m/s at the same temperature.
<u>For B:</u>
Graham's law states that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. The equation for this follows:

Where, M is the molar mass of the gas
Forming an equation for the two gases:
.....(2)
Given values:

Plugging values in equation 2:

Hence, the rate of effusion of
gas is 
The entropy of the reaction can be calculated alike the enthalpy of the reaction which is equal to the difference between the summation of entropies of the products multiplied by their corresponding stoich coeff. and the summation of the entropies of the reactants multiplied by their corresponding stoich coeff. In this case, the answer is -146.8 J / mol K