Answer: 4.1 g of barium precipitated.
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number 
of particles.
To calculate the moles, we use the equation:
Given : moles of barium = 0.030
Molar mass of barium = 137 g/mol
x= 4.1 g
Thus there are 4.1 g of barium that precipitated.
Answer:
B 1,3
Explanation:
Count the atoms on each side of the arrow, they have to be equal
After arrow
2x C
6x O
4x H
Before the arrow
2x C - correct, as it is the only compound with C then there must only be 1 mole of it.
This is also the correct amount of H
We need 6x O total so 6/2 = 3
So there are 3 moles of O
If he was 30.8% too low, it means that he was at 69.2% of the boiling point needed. So 50o C is 69.2% of total.
In order to know what 100% is, you can divide the number by it's percentage and then multiply it by a hundred.
So: 50/30.8=1.623
1.623*100=162.3
So the correct boiling point of the liquid he was working with in the lab is 162.3 oC
Answer:
Mass = 12.276 g
Explanation:
Given data:
Number of molecules of H₂S = 2.16 × 10²³
Mass in gram = ?
Solution:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
For example,
18 g of water = 1 mole = 6.022 × 10²³ molecules of water
Number of moles of H₂S:
2.16 × 10²³ molecules × 1 mole /6.022 × 10²³ molecules
0.36 moles
Mass = number of moles × molar mass
Mass = 0.36 mol × 34.1 g/mol
Mass = 12.276 g
<h3><u>Answer</u>;</h3>
= 226 Liters of oxygen
<h3><u>Explanation</u>;</h3>
We use the equation;
LiClO4 (s) → 2O2 (g) + LiCl, to get the moles of oxygen;
Moles of LiClO4;
(500 g LiClO4) / (106.3916 g LiClO4/mol)
= 4.6996 moles
Moles of oxygen;
But, for every 1 mol LiClO4, two moles of O2 are produced;
= 9.3992 moles of Oxygen
V = nRT / P
= (9.3992 mol) x (8.3144621 L kPa/K mol) x (21 + 273) K / (101.5 kPa)
= 226 L of oxygen
