Answer:
454.3 g.
Explanation:
1.0 mol of CaO liberates → – 64.8 kJ.
??? mol of CaO liberates → - 525 kJ.
∴ The no. of moles needed = (1.0 mol)(- 525 kJ)/(- 64.8 kJ) = 8.1 mol.
<em>∴ The no. of grams of CaO needed = no. of moles x molar mass</em> = (8.1 mol)(56.077 g/mol) = <em>454.3 g.</em>
Answer:
A sample of 830 kg would contain 830000000 mg
Explanation:
Answer:
40% of the ammonia will take 4.97x10^-5 s to react.
Explanation:
The rate is equal to:
R = k*[NH3]*[HOCl] = 5.1x10^6 * [NH3] * 2x10^-3 = 10200 s^-1 * [NH3]
R = k´ * [NH3]
k´ = 10200 s^-1
Because k´ is the psuedo first-order rate constant, we have the following:
b/(b-x) = 100/(100-40) ; 40% ammonia reacts
b/(b-x) = 1.67
log(b/(b-x)) = log(1.67)
log(b/(b-x)) = 0.22
the time will equal to:
t = (2.303/k) * log(b/(b-x)) = (2.303/10200) * (0.22) = 4.97x10^-5 s
(a) In this section, give your answers to three decimal places.
(i)
Calculate the mass of carbon present in 0.352 g of CO
2
.
Use this value to calculate the amount, in moles, of carbon atoms present in 0.240 g
of
A
.
(ii)
Calculate the mass of hydrogen present in 0.144 g of H
2
O.
Use this value to calculate the amount, in moles, of hydrogen atoms present in 0.240 g
of
A
.
(iii)
Use your answers to calculate the mass of oxygen present in 0.240 g of
A
Use this value to calculate the amount, in moles, of oxygen atoms present in 0.240 g
of
A
(b)
Use your answers to
(a)
to calculate the empirical formula of
A
thank you
hope it helpsss
