Answer:
True it is a metaphor
Explanation:
b. radioactivity
c. half-life
d. fission
(2) 35.0% (4) 60.0%
Answer:
The correct option is (2) 35.0%
Explanation:
If we want to calculate the percent composition by mass of nitrogen in NH4NO3 we first need to calculate the mass of nitrogen in one mole of NH4NO3.
Because of the molecular form of NH4NO3 (it has 2 ''N'') we know that one mole of NH4NO3 contains 2 moles of N.
The molar mass of N (nitrogen) is
This means that 1 mole of N contains 14.0067 grams of N.
Because we have 2 moles of N ⇒
We know that in 1 mole of NH4NO3 we have 2 moles of N(28.0134 g of N).
To calculate the percent composition by mass of nitrogen we need to divide the mass of nitrogen in one mole of NH4NO3 by the mass of one mole of NH4NO3 and then multiply it by 100.
We know that in 1 mole of NH4NO3 there are 80.0 g of NH4NO3 (Because of the molar mass data) ⇒
% ≅ 35.0%
We find that the percent composition by mass of nitrogen in NH4NO3 is 35.0%
We conclude that the correct option is (2) 35.0%
Using the combined gas law, we find that the new pressure of the gas sample, after it's compressed and heated, is approximately 353.8 kPa.
The subject of this question is gas laws, specifically the combined gas law which states that the ratio of the product of pressure and volume and the absolute temperature of a gas is constant. We apply this law to calculate the new pressure of the gas sample. Starting from the conditions of STP (Standard Temperature and Pressure, defined as 273.15 K and 1 atm i.e., 101.325 kPa), the volume of gas is decreased from 700.0 mL to 200.0 mL and the temperature is increased from 273.15 K to 30.0 degrees Celsius (or 303.15 K in absolute terms).
We set up the equation P1*V1/T1 = P2*V2/T2, where P1 = 101.325 kPa, V1 = 700.0 mL, T1 = 273.15 K, V2 = 200.0 mL, and T2 = 303.15 K. Plugging in these numbers and solving for P2 (the new pressure), we get P2 = P1*V1*T2 / (T1*V2) = 101.325 kPa * 700.0 mL * 303.15 K / (273.15 K * 200.0 mL) = approximately 353.8 kPa.
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