Answer:
4) 0.5507 Mol Ar
5) 44.548 Mol AgNO3
6) 2.16107 Mol Li
Explanation:
Divide = grams to mol and mol to molecules
Multiply = molecules to mol and mol to grams.
3) 22 <em>g of Ar</em> 1 mol 22
-------------------- * ---------------------- = ------------ (turn to decimal) = 0.5507 Mol Ar
39.948 <em>g of Ar</em> 39.948
4) AgNO3 = (Ag; 39.948) (N; 14.007) (O^3; 48)
39.948 + 14.007 + 48 = 101.955
7.4 x 10^23 Mc 6.02 x 10^23 Mc
---------------------- * ------------------------- = 44.548 Mol AgNO3
1 mol
5)
15g 1 Mol 15
------- * ------------------- = --------- (turn to decimal) = 2.16107 Mol Li
6.941 g of Li 6.941
Answer:
For a volume of 1 liter, we have a total mass of 220 grams of CaCl2 ( Option A)
Explanation:
<u>Step 1: </u>Given data
The concentration of the solution we want to make is 2M
2M = 2 moles/ L So the volume = 1L
The molar mass of CaCl2 is = 40 + 2*35 = 110g/mole
<u>Step 2:</u> Calculate number of moles
If we consider the volume = 1 L then for a concentration of 2M, this means the number of moles is 2
<u>Step 3:</u> Calculate mass
The molar mass of CaCl2 = 110 g/mole
This means 110 grams per 1 mole
for 2 moles the mass is 220 grams
This means that for a volume of 1 liter, we have a total mass of 220 grams of CaCl2
Answer:
a) Black
b) Non-magnetic
c) No reaction with carbon disulphide.
d) i don't know sry
But i had to go through my 6th grade notes for this .____.
Explanation:
this is your aur answer thank you for calcium test of tetraoxosulphate
Answer:
Mass of seawater = 512.5 gram
Explanation:
Given:
Volume of seawater = 500 ml
Density of seawater = 1.025 g/ml
Find:
Mass of seawater
Computation:
Mass of seawater = Volume of seawater × Density of seawater
Mass of seawater = 500 × 1.025
Mass of seawater = 512.5 gram
