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hammer [34]
4 years ago
8

I'm having trouble balancing this chemical equation, please help

Chemistry
1 answer:
vovikov84 [41]4 years ago
6 0
2Ba3N2+12H20=6Ba(OH)2+4NH3
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Determine how many grams of N2 are produced from the reaction of 8.37 g of H2O2 and 5.29 g of N2H4.
Dmitriy789 [7]
N₂H₄  +  2H₂O₂    →   N₂  +  4H₂O

mol = mass ÷ molar mass

If mass of hydrazine (N₂H₄)  = 5.29 g 
then mol of hydrazine           = 5.29 g ÷ ((14 ×2) + (1 × 4))
                                              = 0.165 mol

mole ratio of hydrazine to Nitogen is     1   :  1
  ∴ if moles of hydrazine = 0.165 mol
     then moles of nitrogen = 0.165 mol

Mass = mol × molar mass

Since mol of nitrogen (N₂)  = 0.165
then mass of hydrazine      = 0.165 × (14 × 2)
                                           = 4.62 g
5 0
3 years ago
Find the ph of a buffer that consists of 0.18 m ch3nh2 and 0.73 m ch3nh3cl (pkb of ch3nh2 = 3.35)?
ozzi
Hello!

First, we need to determine the pKa of the base. It can be found applying the following equation:

pKa=14-pKb=14-3,35=10,65

Now, we can apply the Henderson-Hasselbach's equation in the following way:

pH=pKa+log( \frac{[CH_3NH_2]}{[CH_3NH_3Cl]} )=10,65+log( \frac{0,18M}{0,73M} )=10,04

So, the pH of this buffer solution is 10,04

Have a nice day!
8 0
3 years ago
Which of the following compounds will undergo an Sn2 reaction most readily?
Bad White [126]

Answer:

nekwkz8 waxed

Explanation:

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4 0
3 years ago
10 Points! Which type of fossil is formed when a rock hardens inside a mold fossil?
irina [24]

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I'm sure it's option A. Cast Fossil

3 0
3 years ago
Calculate the mass of water produced when 6.97 g of butane reacts with excess oxygen
andrew-mc [135]
The balanced reaction equation for the combustion of butane is as follows;
C₄H₁₀ + 13/2O₂ ---> 4CO₂ + 5H₂O 
the limiting reactant in this reaction is C₄H₁₀  This means that all the butane moles are consumed and amount of product formed depends on the amount of C₄H₁₀ used up.
stoichiometry of C₄H₁₀ to H₂O is 1:5
mass of butane used - 6.97 g
number of moles - 6.97 g / 58 g/mol = 0.12 mol
then the number of water moles produced - 0.12 mol x 5 = 0.6 mol
Therefore mass of water produced - 0.6 mol x 18 g/mol = 10.8 g
7 0
4 years ago
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