Answer:
50 g of K₂CO₃ are needed
Explanation:
How many grams of K₂CO₃ are needed to make 500 g of a 10% m/m solution?
We analyse data:
500 g is the mass of the solution we want
10% m/m is a sort of concentration, in this case means that 10 g of solute (K₂CO₃) are contained in 100 g of solution
Therefore we can solve this, by a rule of three:
In 100 g of solution we have 10 g of K₂CO₃
In 500 g of solution we may have, (500 . 10) / 100 = 50 g of K₂CO₃
Answer:
3.00 mol
Explanation:
Given data:
Mass of P₄ = 211 g
Mass of oxygen = 240 g
Moles of P₂O₅ = ?
Solution:
Chemical equation:
P₄ + 5O₂ → 2P₂O₅
Number of moles of P₄:
Number of moles = mass/ molar mass
Number of moles = 211 g / 123.88 g/mol
Number of moles = 1.7 mol
Number of moles of O₂ :
Number of moles = mass/ molar mass
Number of moles = 240 g / 32g/mol
Number of moles = 7.5 mol
Now we will compare the moles of product with reactant.
O₂ : P₂O₅
5 : 2
7.5 : 2/5×7.5 = 3.00
P₄ : P₂O₅
1 : 2
1.7 : 2×1.7 = 3.4 mol
Oxygen is limiting reactant so the number of moles of P₂O₅ are 3.00 mol.
Mass of P₂O₅:
Mass = number of moles × molar mass
Mass = 3 mol ×283.9 g/mol
Mass = 852 g
Answer:
20 km north that's the answer hope it helped
Answer:
0.295 L
Explanation:
It seems your question lacks the final concentration value. But an internet search tells me this might be the complete question:
" A chemist must dilute 47.2 mL of 150. mM aqueous sodium nitrate solution until the concentration falls to 24.0 mM. He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in liters. Be sure your answer has the correct number of significant digits. "
Keep in mind that if your value is different, the answer will be different as well. However the methodology will remain the same.
To solve this problem we can<u> use the formula</u> C₁V₁=C₂V₂
Where the subscript 1 refers to the concentrated solution and the subscript 2 to the diluted one.
- 47.2 mL * 150 mM = 24.0 mM * V₂
And <u>converting into L </u>becomes:
- 295 mL *
= 0.295 L
Answer:
Atoms gain or lose electrons
Explanation: