<span>Smooth ice reduces the unbalanced forces that would slow the hockey puck. So the answer is C.
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<h3>
Answer:</h3>
The pressure increases by 10% of the original pressure
Thus the new pressure is 1.1 times the original pressure.
<h3>
Explanation:</h3>
We are given;
- Initial temperature as 30°C, but K = °C + 273.15
- Thus, Initial temperature, T1 =303.15 K
- Final temperature, T2 is 333.15 K
We are required to state what happens to the pressure;
- We are going to base our arguments to Pressure law;
- According to pressure law, the pressure of a gas and its temperature are directly proportional at a constant volume
- That is; P α T
- Therefore, at varying pressure and temperature

Assuming the initial pressure, P1 is P
Rearranging the formula;
[tex]P2=\frac{P1T2}{T1}[/tex]


= 1.10 P
The new pressure becomes 1.10P
This means the pressure has increased by 10%
We can conclude that, the new pressure will be 1.1 times the original pressure.
The increase of the boling point of a solution is a colligative property.
The formula for the increase of the normal boiling point of water is:
ΔTb = Kb * m
Where m is the molallity of the solution and Kb is the molal boiling constant in °C/mol.
ΔTb = 0.51 °C / m * 0.100 m = 0.051 °C.
So, the new boiling temperature is Tb = 100°C + 0.051°C = 100.051 °C.
Answer: 100.051 °C
12gHe/1 × 1molHe/4.0026g × 6.02x10^23atomHe/1mol = 1.8 atoms