Answer:
18,1 mL of a 0,304M HCl solution.
Explanation:
The neutralization reaction of Ba(OH)₂ with HCl is:
2 HCl + Ba(OH)₂ → BaCl₂ + 2 H₂O
The moles of 17,1 mL≡0,0171L of a 0,161M Ba(OH)₂ solution are:
= 2,7531x10⁻³moles of Ba(OH)₂
By the neutralization reaction you can see that 2 moles of HCl reacts with 1 mole of Ba(OH)₂. For a complete reaction of 2,7531x10⁻³moles of Ba(OH)₂ you need:
= 5,5062x10⁻³moles of HCl.
The volume of a 0,304M HCl solution for a complete neutralization is:
= 0,0181L≡18,1mL
I hope it helps!
Answer:
9 L
Explanation:
According to the question , the given reaction is -
2NO(g) + O₂(g)------->2NO₂(g)
Since ,
At STP ,
One mole of a gas occupies the volume of 22.4 L.
Hence , as given in the question -
9 L of NO , i.e .
22.4 L = 1 mol
1 L = 1 / 22.4 mol
9 L = 1 / 22.4 * 9 L = 0.40 mol
From the chemical reaction ,
The Oxygen is in excess , hence NO becomes the limiting reagent , and will determine the moles of product .
Hence ,
2 moles of NO will produce 2 moles of NO₂.
Therefore ,
0.40 mol of NO will produce 0.40 mol of NO₂.
Hence , the volume of NO₂ can be calculated as -
1 mol = 22.4 L
0.40 mol = 0.40 * 22.4 L = 9 L
Answer:
7.32g of HNO3 are required.
Explanation:
1st) From the balanced reaction we know that 2 moles of HNO3 react with 1 mole of Ca(OH)2 to produce 2 moles of H2O and 1 mole of Ca(NO3)2.
From this, we find that the relation between HNO3 and Ca(OH)2 is that 2 moles of HNO3 react with 1 mole of Ca(OH)2.
2nd) This is the order of the relations that we have to use in the equation to calculate the grams of nitric acid:
• starting with the 4.30 grams of Ca(OH)2.
,
• using the molar mass of Ca(OH)2 (74g/mol).
,
• relation of the 2 moles of HNO3 that react with 1 mole of Ca(OH)2 .
,
• using the molar mass of HNO3 (63.02g/mol).
So, 7.32g of HNO3 are required.
Answer:
<h2>2.49 g/cm³</h2>
Explanation:
The density of a substance can be found by using the formula
From the question we have
We have the final answer as
<h3>2.49 g/cm³</h3>
Hope this helps you
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