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:
Bacteria are vital in keeping nitrogen cycling through the ecosystem, and nitrogen is vital to plant growth. Without bacteria around to break down biological waste, it would build up. And dead organisms wouldn't return their nutrients back to the system
Explanation:
Answer:
8.61 mL of the HCl solution
Explanation:
The reaction that takes place is:
- 2HCl + Mg(OH)₂ → MgCl₂ + 2H₂O
From the given mass of Mg(OH)₂, we can calculate <u>the moles of HCl that are neutralized</u>:
- 4x10² mg = 400 mg = 0.400g
- 0.400g Mg(OH)₂ ÷ 58.32g/1mol = 6.859*10⁻³ mol Mg(OH)₂
- 6.859*10⁻³ mol Mg(OH)₂ *
3.429x10⁻³ mol HCl
Finally, to calculate the volume of an HCl solution, we need both the moles and the concentration. We can <u>calculate the concentration using the pH value</u>:
= [H⁺]
- 0.0398 M = [H⁺] = [HCl] *Because HCl is a strong acid*
Thus, the volume is:
- 0.0398 M = 3.429x10⁻³mol HCl / Volume
- Volume = 8.616x10⁻³ L = 8.62 mL
Solution :
It is given that :
Weight of the antacid tablet = 5.4630 g
4.3620 gram of antacid is crushed and is added to the stomach acid of 200 mL and is reacted.
25 mL of the stomach acid that is partially neutralized required 13.6 mL of NaOH to be titrated for a red end point.
27.7 mL of 
solution is equivalent to 
of the original stomach acid. Therefore, 13.6 mL of NaOH will take x 
= 12.27 ml of the original stomach acid.
