Answer:
1. 77.5g,
2. 1.78 mol,
3. 281 g
4. add 80 mL of water
Explanation:
1. Molar mass (Na2O) = 2*23.0 + 16.0 = 62.0 g/mol
1.25 mol *62.0 g/1mol = 77.5 g Na2O
2. Molar mass (KOH) = 39.1 +16.0+1.0 = 56.1 g/mol
100 g * 1 mol/56.1 g = 1.78 mol KOH
3. Molar mass (KOH) = 39.1 +16.0+1.0 = 56.1 g/mol
5 mol * 56.1g/1 mol = 281 g KOH
4. M1V1 =M2V2
M - molarity, V- volume of solution
2.0M*20mL = 0.4M*V2
V2 = 2.0*20/0.4 = 100 mL
Volume of diluted solution = 100 mL
We have 20 mL of solution, we need 100 mL of solution,
so we need to add (100-20) =80mL of water.
Take 20 ml 2.0 Molar of solution of HCl , add 80 mL of water, then we get 100 mL 0.4 M solution of HCl.
Answer:
A) 9.60 g
B) 19.2 g
C) 28.8 g
Explanation:
A) Elemental oxygen mass is 16.00 g/mol, so in 0.600 mol of elementar oxygen there is 9.60 g
0.600 mol x 16.00 g/mol=9.60 g
B) Oxygen molecular mass is 32.00 g/mol, so in 0.600 mol of oxygen molecules there is 19.2 g
0.600 mol x 32.00 g/mol=19.2 g
C) Ozone molecular mass is 48.00 g/mol, so in 0.600 mol of oxygen molecules there is 28.8 g
0.600 mol x 48.00 g/mol=28.8 g
False, changes in matter indicate a physical change.
Hello!
The set of compounds that have the same empirical formula is b) N₂O₄ and NO₂.
The Empirical Formula is the most simple representation of the atom ratio in a chemical compound. Of the listed sets of compounds, the only in which the atom ratio is the same for both compounds is the pair N₂O₄ and NO₂ in which the atom ratio N:O is 1:2. The Empirical Formula for this pair is NO₂.
N₂O₄ has the same atom ratio as NO₂ but this formula has each atom multiplied by two. However, its Empirical Formula is the same.
Have a nice day!
Answer:
49°C
Explanation:
Let's apply the Ideal Gases Law in order to solve this question:
P . V = n . R . T
Pressure = 1 atm
Volume = 6 L
n = number of moles → 10 g. 1mol /44g = 0.227 moles
R = Ideal Gases Constant
We replace data: 1 atm . 6 L = 0.227 mol . 0.082 . T
6 atm.L / ( 0.227 mol . 0.082) = T
T° = 322 K
We convert T° from K to °C → 322 K - 273 = 49°C
