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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Answer:
Kinetic energy: 
Velocity: 
Explanation:
From the equations of the photo-electric effect,
We know:
Where,
1.
is the Planck's constant which is 
2.
are the frequency of light emitted and threshold frequencies respectively.
3.
is the kinetic energy of the electrons emitted.
By fact, we come to know that the threshold frequency of Zn is 300nm
And also 
Where ,
1. 
is the speed of light 
2.
is the wavelength.
Thus,
Now to find velocity:
BiO₃⁻ → Bi³⁺
+5 +3
Balance oxidation state:
BiO₃⁻ + 2 e⁻ → Bi³⁺
Balancing oxygen by adding water and balance H
BiO₃⁻ + 6 H⁺ + 2 e⁻ → Bi³⁺ + 3 H₂O
balance charge:
BiO₃⁻ + 6 H⁺ + 2 e⁻ → Bi³⁺ + 3 H₂O
+ 3 + 3
Final:
BiO₃⁻ + 6 H⁺ + 2 e⁻ → Bi³⁺ + 3 H₂O
We convert the masses of our reactants to moles and use the stoichiometric coefficients to determine which one of our reactants will be limiting.
Dividing the mass of each reactant by its molar mass:
(10 g C2H6)(30.069 g/mol) = 0.3326 mol C2H6
(10 g O2)(31.999 g/mol) = 0.3125 mol O2.
Every 2 moles of C2H6 react with 7 moles of O2. So the number of moles of O2 needed to react completely with 0.3326 mol C2H6 would be (0.3326)(7/2) = 1.164 mol O2. That is far more than the number of moles of O2 that we are given: 0.3125 moles. Thus, O2 is our limiting reactant.
Since O2 is the limiting reactant, its quantity will determine how much of each product is formed. We are asked to find the number of grams (the mass) of H2O produced. The molar ratio between H2O and O2 per the balanced equation is 6:7. That is, for every 6 moles of H2O that is produced, 7 moles of O2 is used up (intuitively, then, the number of moles of H2O produced should be less than the number of moles of O2 consumed).
So, the number of moles of H2O produced would be (0.3125 mol O2)(6 mol H2O/7 mol O2) = 0.2679 mol H2O. We multiply by the molar mass of H2O to convert moles to mass: (0.2679 mol H2O)(18.0153 g/mol) = 4.826 g H2O.
Given 10 grams of C2H6 and 10 grams of O2, 4.826 g of H2O are produced.
