Answer:
% Ca = 24.69%
% H = 1.2%
% C = 14.8%
% O = 59.25%
Explanation:
The percentage by mass of each element can be calculated by dividing the mass of each element in the compound by the molar mass of the compound.
Molar mass of Ca(HCO3)2
Where; (Ca= 40, H = 1, O = 16, C= 12)
= 40 + {1 + 12 + 16(3)}2
= 40 + {13 + 48}2
= 40 + {61}2
= 40 + 122
= 162g/mol
- % mass of Ca = 40/162 × 100
= 0.2469 × 100
= 24.69%
- % mass of H = 2/162 × 100
= 0.012 × 100
= 1.2%
- % mass of C = 24/162 × 100
= 0.148 × 100
= 14.8%
- % mass of O = 96/162 × 100
= 0.5925 × 100
= 59.25%
A. The longest carbon chain is eight, and it has two methyl groups attached to carbon three, and a special group attached to carbon five. Its two names could be:
3-dimethyl-5-(1-methylethyl)octane
3-dimethyl-5-isopropyloctane
Both of these are correct. This is an alkane, because it has all single bonds.
B. This has a triple bond contained between carbons 2 and 3, and has a methyl group off carbon 4. The longest chain is 5. It’s name is:
4-methyl-2-pentyne
This is an alkene, because of the double bond.
C. This has a double bond contained between carbons 2 and 3, and has a methyl off of four and an methyl off of six. The longest chain is eight (follow the longest chain of carbons).
4,6-dimethyl-2-octene
This is an alkene, because of the double bond.
D. This has an ethyl group at 1 and a methyl group at 2 (rotate the compound to make it as clean as possible, in this case, the ring is flipped and rotated to make it alphabetical with the smallest numbers possible). The two names are:
1-ethyl-2-methylbenzene
ortho-ethylmethylbenzene
Both are correct, the ortho prefix telling the location of the ethyl and methyl groups. This is an aromatic structure because of its double bonded ring.
E. The longest chain is nine, and has methyls at three, five, and seven, along with a propyl at five. The name is:
3,5,7-trimethyl-5-propylnonane
This is an alkane, due to the single bonds.
Hope this helps!
Volume of Argon V1 = 5.0 L
Pressure of Argon P1 = 2 atm
Final temperature T2 = 30 C = 30 + 273 = 303 K
Volume at final temperature V2= 6 L
Pressure at final temperature P2 = 8 atm
We know that (P1 x V1) / T1 = (P2 x V2) / T2
(2 x 5)/ T1 = (8 x 6)/ 303 => T1 = (10 x 303) / 48
Initial Temperature T1 = 3030 / 48 = 63.12
Initial Temperature = -209. 8 C
Answer:
Molar mass→ 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol
Explanation:
Let's apply the formula for freezing point depression:
ΔT = Kf . m
ΔT = 74.2°C - 73.4°C → 0.8°C
Difference between the freezing T° of pure solvent and freezing T° of solution
Kf = Cryoscopic constant → 5.5°C/m
So, if we replace in the formula
ΔT = Kf . m → ΔT / Kf = m
0.8°C / 5.5 m/°C = m → 0.0516 mol/kg
These are the moles in 1 kg of solvent so let's find out the moles in our mass of solvent which is 0.125 kg
0.0516 mol/kg . 0.125 kg = 6.45×10⁻³ moles. Now we can determine the molar mass:
Molar mass (mol/kg) → 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol
