Answer: 0.082 atm L k^-1 mole^-1
Explanation:
Given that:
Volume of gas (V) = 62.0 L
Temperature of gas (T) = 100°C
Convert 100°C to Kelvin by adding 273
(100°C + 273 = 373K)
Pressure of gas (P) = 250 kPa
[Convert pressure in kilopascal to atmospheres
101.325 kPa = 1 atm
250 kPa = 250/101.325 = 2.467 atm]
Number of moles (n) = 5.00 moles
Gas constant (R) = ?
To get the gas constant, apply the formula for ideal gas equation
pV = nRT
2.467 atm x 62.0L = 5.00 moles x R x 373K
152.954 atm•L = 1865 K•mole x R
To get the value of R, divide both sides by 1865 K•mole
152.954 atm•L / 1865 K•mole = 1865 K•mole•R / 1865 K•mole
0.082 atm•L•K^-1•mole^-1 = R
Thus, the value of gas constant is 0.082 atm L k^-1 mole^-1
A.) baking soda is the answer
Answer:
1.25 moles
Explanation:
First, we need to balance the equation. Essentially, this means making sure we have the same number of each atom on each side.
On the left side, we currently have:
- 1 Co atom
- 2 F aomts
On the right side, we have:
- 1 Co atom
- 3 F atoms
To balance it, add a 2 to Co on the left, 3 to F2 on the left, and 2 to CoF3 on the right:
→ 
Now, we have 1.25 moles of Co, and since the ratio between Co and CoF3 is 1:1, we also have 1.25 moles of CoF3.
Thus, the answer is 1.25 moles.
Answer: Thus 44 g of 
will be produced if 30 g of 
were reacted with an excess of oxygen
Explanation:
To calculate the moles :
The balanced chemical equation is:
As is the limiting reagent as it limits the formation of product and 
is the excess reagent.
According to stoichiometry :
1 mole of produce = 6 moles of
Thus 0.17 moles of will produce=
of
Mass of 
Thus 44 g of will be produced if 30 g of were reacted with an excess of oxygen.
