Given:
n = 0.0021 mol
M = 44 gram/mol
Required: m or the mass of the molecule
Solution:
M = m/n
Rearranging the expression,
m = M x n
m = 44 g/mol x 0.0021 mol
Evaluating the expression and cancelling units, we obtain,
m = 0.0924 grams
M is the molar mass. This value is very useful in chemistry calculations especially in calculating the amounts of reactants and products for a chemical reaction.
Answer:
Mass of SO₂ can be made from 25.0 g of Na₂SO₃ and 22 g of HCl = 12.672 g
Explanation:
SO₂( sulfur dioxide) can be produced in the lab. by the reaction of hydrochloric acid & sulphite salt such as sodium.
the balanced chemical equation is as follows
Na₂SO₃ + 2 HCl → 2 NaCl + SO₂ + H₂O
Moles of Na₂SO₃ =
Moles of HCl =
using mole ratio method to find limiting reagent
For sodium sulfite
for HCl
since <u>sodium sulfite</u> is <u>limiting reactant</u> for above chemical reaction
1 mole of Na₂SO₃ produce 1 mole of SO₂
0.198 mole of Na₂SO₃ produce 0.198 mole of SO₂
∴ Mass of SO₂ produce = mole x molar mass of SO₂
= 0.198 x 64
= 12.672 g
Answer: The correct statements are ; A and B.
Explanation:
The strength of the magnetic field(B) in an electromagnet can be calculated using formula:
= Relative permeability
N = number of turns
I = Current in the wire of the solenoid
L = length of the solenoid or electromagnetic
As we can see from the formula:
So, by increasing the turns and increasing current flowing through wire one can increase the strength of an electromagnet.
Hence, the correct statements are ; A and B.
Answer:
50
Explanation:
We will need a balanced equation with masses, moles, and molar masses of the compounds involved.
1. Gather all the information in one place with molar masses above the formulas and masses below them.
Mᵣ: 30.01 32.00 46.01
2NO + O₂ ⟶ 2NO₂
Mass/g: 80.00 16.00
2. Calculate the moles of each reactant
3. Calculate the moles of NO₂ we can obtain from each reactant
From NO:
The molar ratio is 2 mol NO₂:2 mol NO
From O₂:
The molar ratio is 2 mol NO₂:1 mol O₂
4. Identify the limiting and excess reactants
The limiting reactant is O₂ because it gives the smaller amount of NO₂.
The excess reactant is NO.
5. Mass of excess reactant
(a) Moles of NO reacted
The molar ratio is 2 mol NO:1 mol O₂
(b) Mass of NO reacted
(c) Mass of NO remaining
Mass remaining = original mass – mass reacted = (80.00 - 30.01) g = 50 g NO