The question mentions a change in temperature from 25 to 50 °C. With that, the aim of the question is to determine the change in volume based on that change in temperature. Therefore this question is based on Gay- Lussac's Gas Law which notes that an increase in temperature, causes an increase in pressure since the two are directly proportional (once volume remains constant). Thus Gay-Lussac's Equation can be used to solve for the answer.
Boyle's Equation:

=

Since the initial temperature (T₁) is 25 C, the final temperature is 50 C (T₂) and the initial pressure (P₁) is 103 kPa, then we can substitute these into the equation to find the final pressure (P₂).

=

∴ by substituting the known values, ⇒ (103 kPa) ÷ (25 °C) = (P₂) ÷ (50 °C)
⇒ P₂ = (4.12 kPa · °C) (50 °C)
=
206 kPa
Thus the pressure of the gas since the temperature was raised from 25 °C to 50 °C is
206 kPa
Answer: The given substances are placed in order from most soluble in water to least soluble in water as follows.
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Explanation:
It is known that like dissolves like which means a polar compound will be soluble in a polar solvent. And, a non-polar compound will be soluble in non-polar solvent.
Ionic compounds are soluble in polar solvent like water. For example,
is an ionic compound and it is readily soluble in water.
2-butanol (
) is also a polar compound and it will also dissolve in water but to a lesser extent as compared to magnesium chloride.
Whereas both methane and propane are non-polar in nature. But as propane is larger in size as compared to methane so, it will be slightly more soluble in water as compared to methane.
Thus, we can conclude that the given substances are placed in order from most soluble in water to least soluble in water as follows.
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The radon-222 sample has a half-life of 3.8 days, and we are asked how many times would the mass divide in half after 23 days. First we calculate the amount of times division occurs by taking the number of days and dividing that by the number of days for one half-life to occur: 23/3.8 = 6.05.
We have 198.6 grams of sample, and we are going to divide it in half 6 times to determine how much of it remains after 23 days:
198.6/2 = 99.3 grams
99.3/2 = 49.65 grams
49.65/2 = 24.83 grams
24.83/2 = 12.41 grams
12.41/2 = 6.21 grams
6.21/2 = 3.1 grams
Therefore, we are left with 3.1 grams of radon-222 after 23 days if one half-life equals to 3.8 days.
I believe the type of compound fe3n2 is ionic.
They are known as “American Central”