<u>Answer:</u> The amount of pure antifreeze that must be added is 6.4 gallons
<u>Explanation:</u>
We are given:
Percent of antifreeze given = 10 %
Amount of solution = 8 gallons
Let the amount of pure antifreeze be 'x'
Calculating the amount of pure antifreeze:

Let the amount of pure antifreeze added in the solution be 'y'
Percent of solution to be made = 50 %
So,

Hence, the amount of pure antifreeze that must be added is 6.4 gallons
Answer:

Explanation:
We are asked to find the pressure of a gas given a change in volume. Since the temperature remains constant, we are only concerned with volume and pressure. We will use Boyle's Law, which states the volume is inversely proportional to the pressure. The formula for this law is:

Initially, the oxygen gas occupies a volume of 18.7 liters at a pressure of 1.19 atmospheres.

The gas expands to a volume of 29.4 liters, but the pressure is unknown.

We are solving for the new pressure, so we must isolate the variable
. It is being multiplied by 29.4 liters. The inverse operation of multiplication is division. Divide both sides of the equation by 29.4 L.


The units of liters cancel.



The original measurements all have 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandth place. The 9 in the ten-thousandth place to the right of this place tells us to round the 6 up to a 7.

The pressure of the gas sample is approximately <u>0.757 atmospheres.</u>
Answer:
The number following the name of the element is the number of subatomic particles inside the nucleus of the atom. This means that it is the mass number of the isotope. The average atomic mass of the element is the sum of the products of the percentage abundance and mass number of the naturally occurring isotopes.
Since, the average atomic mass of the hydrogen is nearest to 1 then, the most abundant isotope should be hydrogen-1.
Answer:
1 mol SO2 contains 6.0213*10^23 molecules
6.023*10^24 molecules = 10 mol SO2
Equation
S(s) + O2(g) → SO2(g)
1 mol S reacts with 1 mol O2 to prepare 1 mol SO2
To prepare 10 mol SO2 you require : 10 mol S plus 10 mol O2
And that is the answer to the question
If you want a mass :
Molar mass S = 32 g/mol You require 10 mol = 320 g
Molar mass O2 = 32 g/mol :You require 10 mol = 320 g