Answer:
The correct answer is 4.16 grams.
Explanation:
Based on the given information, the concentration of KCl solution is 16 % m/v, which means that 100 ml of the solution will contain 16 grams of KCl.
The molarity of the solution can be determined by using the formula,
M = weight/molecular mass × 1000/Volume
The molecular mass of KCl is 74.6 grams per mole.
M = 16/74.6 × 1000/100
M = 16/74.6
M = 2.14 M
Now the weight of KCl present in the solution of 26 ml will be,
2.14 = Wt./74.6 × 1000 /26
Wt. = 4.16 grams
Answer:
When aqueous solutions of silver(I) acetate and manganese(II) iodide are combined, an insoluble precipitate of silver(I) iodide is formed
Explanation:
- It is an example of precipitation reaction.
- When aqueous solutions of silver(I) acetate and manganese(II) iodide are combined, an insoluble precipitate of silver(I) iodide is formed.
- Precipitation of silver(I) iodide is confirmed by it's yellow color.
- Hence a reaction is observed.
- Molecular reaction:

Hotter less dense then colder air as a result cold air sinks and hot air rises
CH4 + 2O2 → CO2 + 2H2O + 890 kJ
MM of CH4 = (12.01 + 4x1.008) g/mol = 16.04 g/mol
Moles of CH4 = 45.5 g CH4 x (1 mol CH4/16.04 g CH4) = 2.837 mol CH4
q = 2.837 mol CH4 x (890 kJ/1 mol CH4) = 2520 kJ
Answer:
<u>37.5 g of H₂</u>
Explanation:
N₂ (28g) + 3H₂ (6g) => 2NH₃ (34g)
Every 28g of N₂ needs 6g of H₂
=> Every 7g of N₂ needs 1.5g of H₂
=> 175g of N₂ needs 1.5 x 25 g of H₂
=> <u>37.5 g of H₂</u>