<u>Answer:</u> The moles of KHP in the sample is ![9.79\times 10^{-4}mol](https://tex.z-dn.net/?f=9.79%5Ctimes%2010%5E%7B-4%7Dmol)
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
![\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20moles%7D%3D%5Cfrac%7B%5Ctext%7BGiven%20mass%7D%7D%7B%5Ctext%7BMolar%20mass%7D%7D)
Given mass of KHP = 0.2000 g
Molar mass of KHP = 204.22 g/mol
Putting values in above equation, we get:
![\text{Moles of KHP}=\frac{0.2000g}{204.22g/mol}=9.793\times 10^{-4}mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20KHP%7D%3D%5Cfrac%7B0.2000g%7D%7B204.22g%2Fmol%7D%3D9.793%5Ctimes%2010%5E%7B-4%7Dmol)
Hence, the moles of KHP in the sample is ![9.79\times 10^{-4}mol](https://tex.z-dn.net/?f=9.79%5Ctimes%2010%5E%7B-4%7Dmol)
Answer:
Vitamins
Explanation:
Vitamins does not directly provide energy to the objects .
Answer:
30.34g (corrected to 4 significant figures).
Explanation:
Take the atomic mass of C=12.0, H=1.0, O=16.0.
no. of moles = mass / molar mass
So, no. of moles of butane reacted = 10 / (12x4 + 1x10)
= 0.172414 mol
Since O2 is in excess and butane is the limiting reagent, the no. of moles of carbon dioxide produced depends on the no. of moles of butane reacted.
From the equation, the mole ratio of butane:Carbon dioxide = 2: 8 = 1: 4,
meaning 1 mole of butane gives 4 moles of CO2.
Using this ratio,
we can deduce that the no. of moles of CO2 produced = 0.172414 x 4
=0.689655 mol
As mass = no. of moles x molar mass
mass of CO2 produced = 0.689655 x (12.0+16.0x2)
=30.34g (corrected to 4 significant figures).
Answer:VO(ClO4)3 i'm sure this is it
Explanation:
Answer:
0.585 mol
25.7 g
Explanation:
There is some info missing. I think this is the complete question.
<em>Carbon dioxide gas is collected at 27.0 °C in an evacuated flask with a measured volume of 30.0L. When all the gas has been collected, the pressure in the flask is measured to be 0.480atm.
</em>
<em>
Calculate the mass and number of moles of carbon dioxide gas that were collected. Be sure your answer has the correct number of significant digits.</em>
<em />
Step 1: Given data
T = 27.0°C + 273.15 = 300.2 K
V = 30.0 L
P = 0.480 atm
n = ?
m = ?
Step 2: Calculate moles of CO₂
We can calculate the moles of CO₂ using the ideal gas equation.
![P.V=n.R.T\\n=\frac{P.V}{R.T} =\frac{0.480 atm \times 30.0L }{(0.08206atm.L/mol.K) \times 300.2K} =0.585mol](https://tex.z-dn.net/?f=P.V%3Dn.R.T%5C%5Cn%3D%5Cfrac%7BP.V%7D%7BR.T%7D%20%3D%5Cfrac%7B0.480%20atm%20%5Ctimes%2030.0L%20%7D%7B%280.08206atm.L%2Fmol.K%29%20%5Ctimes%20300.2K%7D%20%3D0.585mol)
Step 3: Calculate mass of CO₂
We know that the molar mass is 44.01 g/mol. Then,
![0.585mol.\frac{44.01g}{mol} =25.7g](https://tex.z-dn.net/?f=0.585mol.%5Cfrac%7B44.01g%7D%7Bmol%7D%20%3D25.7g)