Let us assume propane was the fuel
C3H8(g) + 5O2(g) ---> 3CO2(g) + 4H2O(g) = 2217kJ
1 mole ofpropane produces 3 moles of CO2
heat absorbed by pork = 0.11 x 2217
= 243.87 kJ/mol
number of moles of propane = 1700kJ / 243.87 kJ/mol
= 6.971 moles
1 mole of C3H8 = 3 moles ofCO2
6.971 moles of C3H8 = ?
3 x 6.971 = 20.913 moles of CO2
Convert to grams
mass = MW x mole
= 44 x 20.913
= 920.172g of CO2 emitted
Answer:
a) 7.0.
b) Nickel sulfate hepta hydrate.
c) 280.83 g/mol.
d) 44.9%.
Explanation:
<u><em>a) What is the formula of the hydrate?</em></u>
The mass of the hydrated sample (NiSO₄.xH₂O) = 5.0 g,
The mass of the anhydrous salt (NiSO₄) = 2.755 g,
The mass of water = 5.0 g - 2.755 g = 2.245 g.
∴ no. of moles of water = mass/molar mass = (2.245 g)/(18.0 g/mol) = 0.1247 mol.
∴ no. of moles of anhydrous salt (NiSO₄) = mass/molar mass = (2.755 g)/(154.75 g/mol) = 0.0178 mol.
∴ water of crystallization in the sample (x) = no. of moles of water/no. of moles of anhydrous salt (NiSO₄) = (0.1247 mol)/(0.0178 mol) = 7.0.
<u><em>b) What is the full chemical name for the hydrate?</em></u>
The name of the salt (NiSO₄.7H₂O) is Nickel sulfate hepta hydrate.
<u><em>c) What is the molar mass of the hydrate? </em></u>
(NiSO₄.7H₂O)
The molar mass = molar mass of NiSO₄ + 7(molar mass of H₂O) = (154.75 g/mol) + 7(18.0 g/mol) = 280.83 g/mol.
<em><u>d) What is the mass % of water in the hydrate?</u></em>
The mass % of water = (mass of water)/(mass of hydrated sample) x 100 = (2.245 g)/(5.0 g) x 100 = 44.9%.
Answer:
9.
a. NH2CH2COOH
b. The function group is what I put in bold.
c. carboxylic acid and amine
10.
a. NH2CH(CH3)COOH
b. The functional group is in bold.
c. carboxylic acid and amine
Explanation:
NH2 is amine (amino acid)
COOH is the carboxylic acid (acetic acid)
Answer:
B. Ba2+ and Mn (nevermind its not B)
Explanation:
Answer:
expected configuration for Cr is
1s2 2s2 2p6 3s2 3p6 4s2 3d4
but in real, it is
1s2 2s2 2p6 3s2 3p6 4s1 3d5
electron from 4s orbital jumps to 3d orbital to get stable configuration.
so the last electron comes in 3d orbital as filling of 3d takes place after filling of 4s orbital.
Hence,quantum numbers for last electron in Cr is :-
n = 3
l = 2
m = +2
s = +1/2
Explanation:
