Answer:
P1 = 2.5ATM
Explanation:
V1 = 28L
T1 = 45°C = (45 + 273.15)K = 318.15K
V2 = 34L
T2 = 35°C = (35 + 273.15)K = 308.15K
P1 = ?
P2 = 2ATM
applying combined gas equation,
P1V1 / T1 = P2V2 / T2
P1*V1*T2 = P2*V2*T1
Solving for P1
P1 = P2*V2*T1 / V1*T2
P1 = (2.0 * 34 * 318.15) / (28 * 308.15)
P1 = 21634.2 / 8628.2
P1 = 2.5ATM
The initial pressure was 2.5ATM
25 g of NH₃ will produce 47.8 g of (NH₄)₂S
<u>Explanation:</u>
2 NH₃ + H₂S ----> (NH₄)₂S
Molecular weight of NH₃ = 17 g/mol
Molecular weight of (NH₄)₂S = 68 g/mol
According to the balanced reaction:
2 X 17 g of NH₃ produces 68 g of (NH₄)₂S
1 g of NH₃ will produce 
g of (NH₄)₂S
25g of NH₃ will produce 
of (NH₄)₂S
= 47.8 g of (NH₄)₂S
Therefore, 25 g of NH₃ will produce 47.8 g of (NH₄)₂S
Answer:
0.129g MgCl2
Explanation:
For this we need to understand the concept of molarity.
Molarity is number of moles of solute/litres of solution
M=n/L
Here we are given molarity of 0.054M and volume of 25ml. we just plug this in formula to find moles of MgCl2
0.054=x/(25/1000) (we divided 25 by 1000 to convert it to litres of solution)
x=0.00135 moles of MgCl2 (we are not done yet the question asks for grams so to convert to grams we multiply by molar mass of MgCl2.)
0.00135molesMgCl2 x 95.211g MgCl2/1molMgCl2
= 0.129g MgCl2
Your answer is right.
Important elements to consider:
- to use the balanced equation (which you did)
- divide the masses of each compound by the correspondant molar masses (which you did)
- compare the theoretical proportions with the current proportions
Theoretical: 2 mol of Na OH : 1 mol of CuSO4
Then 4 mol of NaOH need 2 mol of CUSO4.
Given that you have more than 2 mol of of CUSO4 you have plenty of it and the NaOH will consume first, being this the limiting reagent.
Answer:
2HCl(aq) + Na2SO3(aq) —> 2NaCl(aq) + H2SO3(aq)
Explanation:
HCl(aq) + Na2SO3(aq) —> NaCl(aq) + H2SO3(aq)
Let us balance the equation. This is illustrated below:
There are 2 atoms of Na on the left side of the equation and 1atom on the right side. It can be balance by putting 2 in front of NaCl as shown below:
HCl(aq) + Na2SO3(aq) —> 2NaCl(aq) + H2SO3(aq)
Now, we have 2 atoms of Cl on the right side and 1 atom on the left side. Thus, it can be balance by putting 2 in front of HCl as shown below:
2HCl(aq) + Na2SO3(aq) —> 2NaCl(aq) + H2SO3(aq)
A careful look at the equation proved that the equation is balanced as the numbers of the different atoms of the element on both side of the equation are the same.
