P = 2.30 atm
Volume in liter = 2.70 mL / 1000 => 0.0027 L
Temperature in K = 30.0 + 273 => 303 K
R = 0.082 atm
molar mass O2 = 31.9988 g/mol
number of moles O2 :
P * V = n * R* T
2.30 * 0.0027 = n * 0.082 * 303
0.00621 = n * 24.846
n = 0.00621 / 24.846
n = 0.0002499 moles of O2
Mass of O2:
n = m / mm
0.0002499 = m / 31.9988
m = 0.0002499 * 31.9988
m = 0.008 g
Answer:
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Explanation:
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It is 0.5474 but you can put 0.5 hope this helps
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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