Answer:
Chemical reaction, compound, molecule, covalent bonds, ionic bonds, ions. (first page)
Oxygen and hydrogen, 1 hydrogen 2 oxygen, H2O, a covalent bond, then mark off is always a liquid. (Second page)
Explanation:
Given :
Number of moles , n = 36.25 mol .
Molarity , M = 1.25 M .
To Find :
The volume of water required .
Solution :
Moarity is given by :

So , 
Here , n is number of moles and M is molarity .
Putting all values in above equation , we get :

Therefore , volume of water required is 29 L .
Answer:
7.37 mL of KOH
Explanation:
So here we have the following chemical formula ( already balanced ), as HNO3 reacts with KOH to form the products KNO3 and H2O. As you can tell, this is a double replacement reaction,
HNO3 + KOH → KNO3 + H2O
Step 1 : The moles of HNO3 here can be calculated through the given molar mass ( 0.140 M HNO3 ) and the mL of this nitric acid. Of course the molar mass is given by mol / L, so we would have to convert mL to L.
Mol of NHO3 = 0.140 M
30 / 1000 L = 0.140 M
0.03 L = .0042 mol
Step 2 : We can now convert the moles of HNO3 to moles of KOH through dimensional analysis,
0.0042 mol HNO2
( 1 mol KOH / 1 mol HNO2 ) = 0.0042 mol KOH
From the formula we can see that there is 1 mole of KOH present per 1 moles of HNO2, in a 1 : 1 ratio. As expected the number of moles of each should be the same,
Step 3 : Now we can calculate the volume of KOH knowing it's moles, and molar mass ( 0.570 M ).
Volume of KOH = 0.0042 mol
( 1 L / 0.570 mol )
( 1000 mL / 1 L ) = 7.37 mL of KOH
Answer:
9.64g
Explanation:
The balanced equation for the reaction is given below:
2NH3 (g) —> 3H2 (g) + N2 (g)
Next, we need to calculate the mass NH3 that decomposed and the mass of H2 produced from the balanced equation. This is illustrated below:
Molar Mass of NH3 = 14 + (3x1) = 14 + 3 = 17g/mol
Mass of NH3 that decomposed from the balanced equation = 2 x 17 = 34g
Molar Mass of H2 = 2x1 = 2g/mol
Mass of H2 produced from the balanced equation = 3 x 2 = 6g.
Now, we can obtain the mass of H2 formed from 54.6g of NH3 as follow:
From the balanced equation above,
34g of NH3 decomposed to produce 6g of H2.
Therefore, 54.6g of NH3 will decompose to produce = (54.6x6)/34 = 9.64g of H2
Therefore, 9.64g of H2 can be obtained from 54.6g of NH3.