Answer:
6.7 x 10²⁶molecules
Explanation:
Given parameters
Mass of CO₂ = 4.9kg = 4900g
Unknown:
Number of molecules = ?
Solution:
To find the number of molecules, we need to find the number of moles first.
Number of moles = 
Molar mass of CO₂ = 12 + 2(16) = 44g/mol
Number of moles = 
= 111.36mole
A mole of substance is the quantity of substance that contains the avogadro's number of particles.
1 mole = 6.02 x 10²³molecules
111.36 moles = 111.36 x 6.02 x 10²³molecules = 6.7 x 10²⁶molecules
Answer
5
Explanation:
We can go about this using the percentage compositions.
First, we calculate the percentage composition of the copper sulphate. This is obtainable by using the mass.
0.96/1.5 * 100 = 64%
Hence the percentage by mass of the water present is 36%
The molar mass of the anhydrous sulphate is 64 + 32 +4(16) = 160g/mol
The molar mass of the water is 2(1) + 16 = 18g/mol
Not forgetting that it is in multiples of x, the total molar mass of the water is 18x moles
The total mass of the copper sulphate hydrate is 160+ 18x
Now how do we get x? Like it is said earlier, the percentage composition is constant.
Hence, 64/100 * (160 + 18x) = 160
16000 = 64(160 + 18x)
16000 = 10,240 + 1152x
16,000 - 10,240 = 1152x
1152x = 5760
x = 5760/1152
x = 5
<span>
• </span>Volume of the marshmallow:
V = 2.75 in^3 (but, 1 in^3 = 16.39 cm^3)
V = 2.75 × 16.39 cm^3
V = 2.75 × 16.39 cm^3
V = 45.0725 cm^3
• Density:
d = 0.242 g/cm^3
<span>• </span>Mass:
m = d × V
m = (0.242 g/cm^3) × (45.0725 cm^3)
m = (0.242 g/cm^3) × (45.0725 cm^3)
m = 10.907545 g
m ≈ 10.9 g <——<span>— this is the answer.
I hope this helps. =)
</span>
Answer:
A. PO43-
Explanation:
in the given is buffer . so H2PO4- and HPO42- both are present with equal concentration . Na+ is spectator ion it is also present in the concentration higher than the given species above .
but PO4-3 is not present . so it is lowest concentration
Volume in liters:
5.00x10² mL / 1000 => 0.5 L
Molar mass KI => 166.0028 g/mol
Mass KI = volume x molar mass x molarity
Mass KI = 0.5 x 166.0028 x 2.80
= 232.40392 g of KI
hope this helps!
