Answer:
C8H17N
Explanation:
Mass of the unknown compound = 5.024 mg
Mass of CO2 = 13.90 mg
Mass of H2O = 6.048 mg
Next, we shall determine the mass of carbon, hydrogen and nitrogen present in the compound. This is illustrated below:
For carbon, C:
Molar mass of CO2 = 12 + (2x16) = 44g/mol
Mass of C = 12/44 x 13.90 = 3.791 mg
For hydrogen, H:
Molar mass of H2O = (2x1) + 16 = 18g/mol
Mass of H = 2/18 x 6.048 = 0.672 mg
For nitrogen, N:
Mass N = mass of unknown – (mass of C + mass of H)
Mass of N = 5.024 – (3.791 + 0.672)
Mass of N = 0.561 mg
Now, we can obtain the empirical formula for the compound as follow:
C = 3.791 mg
H = 0.672 mg
N = 0.561 mg
Divide each by their molar mass
C = 3.791 / 12 = 0.316
H = 0.672 / 1 = 0.672
N = 0.561 / 14 = 0.040
Divide by the smallest
C = 0.316 / 0.04 = 8
H = 0.672 / 0.04 = 17
N = 0.040 / 0.04 = 1
Therefore, the empirical formula for the compound is C8H17N
Amino acids join end to end to form proteins since they are monomers.
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Answer:
a) AgNO3 + KI → AgI + KNO3
b) Ba(OH)2 + 2HNO3 → Ba(NO3)2 + 2H2O
c) 2Na3PO4 + 3Ni(NO3)2 → Ni3(PO4)2 + 6NaNO3
d) 2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O
Explanation:
a) AgNO3 + KI → Ag+ + NO3- + K+ + I-
Ag+ + NO3- + K+ + I- → AgI + KNO3
AgNO3 + KI → AgI + KNO3
b) Ba(OH)2 + 2HNO3 → Ba^2+ + 2OH- + 2H+ + 2NO3-
Ba^2+ + 2OH- + 2H+ + 2NO3- → Ba(NO3)2 + 2H2O
Ba(OH)2 + 2HNO3 → Ba(NO3)2 + 2H2O
c) 2Na3PO4 + 3Ni(NO3)2 → 6Na+ + 2PO4^3- + 3Ni^2+ + 6NO3-
6Na+ + 2PO4^3- + 3Ni^2+ + 6NO3- → Ni3(PO4)2 + 6NaNO3
2Na3PO4 + 3Ni(NO3)2 → Ni3(PO4)2 + 6NaNO3
d) 2Al(OH)3 + 3H2SO4 → 2Al^3+ + 6OH- + 6H+ + 3SO4^2-
2Al^3+ + 3OH- + 3H+ + 3SO4^2- → Al2(SO4)3 + 6H2O
2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O
I don't know Im just Trying to get points somehow
Answer: 3.306 mmol
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.
where,
Molarity
n= moles of solute
= volume of solution in ml
According to stoichiometry:
1 mole of 
gives 2 moles of 
1.653 mmol of gives =
of
Thus number of millimoles of are 3.306
