8 moles of H 2O are produced.
First, we need to figure out the chemical equation for producing water with oxygen which is H 2 + O2 = H 2O. Then, we need to balance the equation, resulting in 2H 2 + O2 = 2H 2O.
<h3>How many moles of H2 are required to make one mole of NH3?</h3>
Calculate 0.88074 mol H2's mass. If N2 is too much, 1.776 g H2 is needed to create 10.00 g of NH3. To create 8.2 moles of ammonia, 2 moles of NH3 are created when 1 mole of N2 and 3 moles of H2 mix. 4.1 moles of N2 Fast are consequently needed to make 8.2 moles of NH3.
<h3>
How many moles of h2 are needed to produce a solution?</h3>
An O-H bond has a bond energy of 1 09 Kcal. 3.6. A 38.0mL 0.026M HCl solution and a 0.032M NaOH solution react. Thus, 10 moles of NH 3 are obtained by dividing 15 moles of H2 by the 1.5 moles of H2 required for the product. and 9.3 x 10-3 moles of bromobutane (1.27/137 =.00927moles).
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Well as far as I know to make one ton of sulfuric acid takes 1,000,000 grams, so the answer should be 98,000,000 grams
Answer:
146 kJ
Explanation:
There are two heat flows in this question.
Heat lost on cooling + heat lost on solidifying = 0
q₁ + q₂ = 0
mCΔT + nΔHsol = 0
Data:
m = 575 g
C = 0.449 J·K⁻¹g⁻¹
T_i = 1825 K
T_f = 1811 K
ΔHsol = -13.8 kJ·mol⁻¹
Calculations:
(a) Heat lost on cooling
ΔT = T_f - T_i = 1811 K - 1825 K = -14 K
q₁ = mCΔT = 575 g × 0.449 J·K⁻¹g⁻¹ × (-14 K) = -361 J = -3.61 kJ
(b) Heat lost on solidifying
(c) Total heat lost
q = q₁ + q₂ = -3.61 kJ - 142.1 kJ = -146 kJ
The heat lost was 146 kJ.
<span>1.
London forces. 4. dipole - dipole. Due to a small hydrogen atom and a
much large fluorine atom, with a large , positive nuclus and large,
negative, p orbitals . This makes it very polar.</span>
