Given :
Molarity of sulfuric acid solution is 3.0 M.
Amount of sulfuric acid present in solution is 9.809 g.
To Find :
The volume of solution.
Solution :
We know, molarity is given by :

Therefore, volume required is 33.33 ml .
B. 35.45 ! Hope this helps

Inside the stomach, Hydrochloric acid kills micro-organisms in the food. Stomach juices begin to break down <u>proteins</u> to amino acids.
✤ So, Fill the blank with proteins.
<h3>
<u>Explanation:-</u></h3>
- Inside the stomach, the digestion of proteins starts due to the action of pepsin enzyme.
- But this enzyme remains inactive and is activated by the Hydrochloric acid(HCl).
- The Hydrochloric acid also helps in killing the germs and microbes which entered along with food.
- The mucous lines the wall of stomach to protect it from the harm caused by HCl because HCl is a strong acid.
- In stomach, The partial digestion of proteins occur and rest is digested in the small intestine by Trypsin(Pancreatic enzyme) and Intestinal juices.
<u>━━━━━━━━━━━━━━━━━━━━</u>
Answer:
The enthalpy of the solution is -35.9 kJ/mol
Explanation:
<u>Step 1:</u> Data given
Mass of lithiumchloride = 3.00 grams
Volume of water = 100 mL
Change in temperature = 6.09 °C
<u>Step 2:</u> Calculate mass of water
Mass of water = 1g/mL * 100 mL = 100 grams
<u>Step 3:</u> Calculate heat
q = m*c*ΔT
with m = the mass of water = 100 grams
with c = the heat capacity = 4.184 J/g°C
with ΔT = the chgange in temperature = 6.09 °C
q = 100 grams * 4.184 J/g°C * 6.09 °C
q =2548.1 J
<u>Step 4:</u> Calculate moles lithiumchloride
Moles LiCl = mass LiCl / Molar mass LiCl
Moles LiCl = 3 grams / 42.394 g/mol
Moles LiCl = 0.071 moles
<u>Step 5:</u> Calculate enthalpy of solution
ΔH = 2548.1 J /0.071 moles
ΔH = 35888.7 J/mol = 35.9 kJ/mol (negative because it's exothermic)
The enthalpy of the solution is -35.9 kJ/mol
Answer:
1 mol of water is produced in those conditions.
Explanation:
The reaction to produce water between H₂ and O₂ is this:
2H₂ + O₂ → 2H₂O
We don't have the amount of hydrogen, so we have to think that is in excess.
Let's work with oxygen.
Ratio is 1:2
For 0.5 mole of oxygen, I will make the double of moles of water.