Question

What is the mass of calcium carbonate used in part a.?

Do not help me out with question 24a, as i have already worked out the answer for that question. I'm confused with question 24b, i need help with that question only. The answer for q24a is 25l
Thank You

Answer 1

`"45 g CaCO"_3`

Explanation:

The idea here is that you need to use the ideal gas law equation to find the number of moles of hydrogen gas used in the reaction (or the number of moles of methane produced by the reaction).

As you know, the ideal gas law equation looks like this

`color(blue)(ul(color(black)(PV = nRT)))`

Here

• `P` is the initial pressure of the gas
• `V` is the volume it occupies
• `n` is the number of moles of gas present in the mixture
• `R` is the universal gas constant, equal to `0.0821("atm L")/("mol K")`
• `T` is the absolute temperature of the gas

Rearrange to solve for `n`, the number of moles of hydrogen gas consumed by the reaction.

`PV = nRT implies n = (PV)/(RT)`

Plug in the values you have for the pressure, the temperature, and the volume of hydrogen gas, but do not forget to convert the pressure to atmospheres and the temperature to degrees Celsius!

`n = (100/101.325 color(red)(cancel(color(black)("atm"))) * 100 color(red)(cancel(color(black)("L"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 400)color(red)(cancel(color(black)("K"))))`

`n = "1.786 moles"`

The balanced chemical equation tells you that the reaction consumes `1` mole of calcium carbonate for every `4` moles of hydrogen gas that take part in the reaction, so use this `1:4` mole ratio to find the number of moles of calcium carbonate that took part in the reaction.

`1.786 color(red)(cancel(color(black)("moles H"_2))) * "1 mole CaCO"_3/(4color(red)(cancel(color(black)("moles H"_2)))) = "0.4465 moles CaCO"_3`

Now all you have to do to find the mass of calcium carbonate is to use the compound's molar mass.

`0.4465 color(red)(cancel(color(black)("moles CaCO"_3))) * "100.09 g"/(1color(red)(cancel(color(black)("mole CaCO"_3)))) = color(darkgreen)(ul(color(black)("45 g")))`

I'll leave the answer rounded to two sig figs, but keep in mind that you have only one significant figure for the pressure of the gas and the volume occupied by the hydrogen gas.

Answer 2

`0 < x < 2a`

Explanation:

.

Let's let:

`AD=x`, and `AB=y`. If you connect `B` and `D` triangle `Delta ABD` is a right angle triangle with `/_A=90^@` and `BD` is the hypotenuse of the triangle.

The center `Q` is on `BD` which means `BD` is the diameter of the circle and is equal to twice the radius:

`BD=2a`

From the Pythagorean formula:

`(AD)^2+(AB)^2=(BD)^2`

`x^2+y^2=(2a)^2` (Equation `1`)

`x^2+y^2=4a^2`

The area of the rectangle is the product of its length and width:

`A=xy` (Equation `2`)

Let's solve for `y` from equation `1` and substitute it in equation `2`:

`y^2=4a^2-x^2`

`y=sqrt(4a^2-x^2)`

`A=xsqrt(4a^2-x^2`

This is the Area as a function of `x`. Area can only be positive. As such, we can not have negative values of `x`. And since we have a radical in the equation, we have to make sure we do not have a negative value under the radical. This means:

`x > 0`, and

`4a^2-x^2 > 0`, or `x^2 < 4a^2` or `x < 2a`

This makes sense because if `x=2a` side `AD` becomes the diameter of the circle and side `AB` collapses to `0`, and the rectangle becomes a line and Area becomes `0`.

The same thing happens if `x=0`. Then `y` becomes the diameter and Area becomes `0`.

Therefore, we can define the domain of the Area function as:

`0 < x < 2a`

which means `x` can only be a value larger than `0` but smaller than `2a`.