Answer:
in an ancient
Explanation:
when the word begins with a vowel you say an instead of a
Answer:
C₆H₈O₆
Explanation:
First off, the<u> percent of oxygen by mass</u> of vitamin C is:
- 100 - (40.9+4.58) = 54.52 %
<em>Assume we have one mol of vitamin C</em>. Then we would have <em>180 grams</em>, of which:
- 180 * 40.9/100 = 73.62 grams are of Carbon
- 180 * 4.58/100 = 8.224 grams are of Hydrogen
- 180 * 54.52/100 = 98.136 grams are of Oxygen
Now we <u>convert each of those masses to moles</u>, using the <em>elements' respective atomic mass</em>:
- C ⇒ 73.62 g ÷ 12 g/mol = 6.135 mol C ≅ 6 mol C
- H ⇒ 8.224 g ÷ 1 g/mol = 8.224 mol H ≅ 8 mol H
- O ⇒ 98.136 g ÷ 16 g/mol = 6.134 mol O ≅ 6 mol O
So the molecular formula for vitamin C is C₆H₈O₆
Answer:
b) 2.0 mol
Explanation:
Given data:
Number of moles of Ca needed = ?
Number of moles of water present = 4.0 mol
Solution:
Chemical equation:
Ca + 2H₂O → Ca(OH)₂ + H₂
now we will compare the moles of Ca and H₂O .
H₂O : Ca
2 : 1
4.0 : 1/2×4.0 = 2.0 mol
Thus, 2 moles of Ca are needed.
Answer:
P = 13.5 atm
Explanation:
Given that
No. of moles, n = 20 moles
Volume of nitrogen gas = 36.2 L
Temperature = 25°C = 298 K
We need to find the pressure of the gas. Using the ideal gas equation
PV = nRT
Where
R is gas constant, ![R=0.082057\ L-atm/K-mol](https://tex.z-dn.net/?f=R%3D0.082057%5C%20L-atm%2FK-mol)
So,
![P=\dfrac{nRT}{V}\\\\P=\dfrac{20\times 0.082057\times 298}{36.2 }\\\\P=13.5\ atm](https://tex.z-dn.net/?f=P%3D%5Cdfrac%7BnRT%7D%7BV%7D%5C%5C%5C%5CP%3D%5Cdfrac%7B20%5Ctimes%200.082057%5Ctimes%20298%7D%7B36.2%20%7D%5C%5C%5C%5CP%3D13.5%5C%20atm)
so, the pressure of the gas is equal to 13.5 atm.
Answer:
m = 50.74 kg
Explanation:
We have,
Initial temperature of water is 20 degrees Celsius
Final temperature of water is 46.6 degrees Celsius
Heat absorbed is 5650 J
It is required to find the mass of the sample. The heat absorbed is given by the formula ad follows :
![Q=mc\Delta T](https://tex.z-dn.net/?f=Q%3Dmc%5CDelta%20T)
c is specific heat of water, c = 4.186 J/g°C
So,
![m=\dfrac{Q}{c\Delta T}\\\\m=\dfrac{5650}{4.186\times (46.6-20)}\\\\m=50.74\ kg](https://tex.z-dn.net/?f=m%3D%5Cdfrac%7BQ%7D%7Bc%5CDelta%20T%7D%5C%5C%5C%5Cm%3D%5Cdfrac%7B5650%7D%7B4.186%5Ctimes%20%2846.6-20%29%7D%5C%5C%5C%5Cm%3D50.74%5C%20kg)
So, the mass of the sample is 50.74 kg.