The amount of water that must be added to 6.0 M silver nitrate to make 500mL of 1.2 M solution is : 2000 mL
<u>Given data :</u>
Concentration of siilver nitrate ( M₁ ) = 6.0 M
volume of solution ( V₁ ) = 500 mL
Conc of solution ( M₂ )= 1.2 M
<h3>Determine the amount of water that must be added</h3>
we will apply the equation below
M₁V₁ = M₂V₂ ---- ( 1 )
where : V₂ = V₁ + water added ---- ( 2 )
V₂ ( Final volume ) = ( M₁V₁ ) / M₂
= ( 6 * 500 ) / 1.2
= 2500 mL
Back to eqaution ( 2 )
2500 mL = 500 mL + added water
therefore ; added water = 2500 - 500
= 2000 mL
Hence we can conclude that The amount of water that must be added to 6.0 M silver nitrate to make 500mL of 1.2 M solution is : 2000 mL.
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There is a high density on water and not on gas
1:3
The ratio of Al3+ ions to Cl− ions in the chemical formula is 1:3.
I’m sorry if this didn’t help I’m new to Brainly
Molecular weight of P (Phosphorous) = 30.97g/mol
Hydrogen is just 1 g/mol.
How many moles is 34g of PH3?
Get the weight of PH3 (30.97 + (3X1)) = 33.97g/mol
So 34g/33.97g/mol = 1.0009 moles.
I bet for this problem it's easier to round this to 1.
If you look at just the moles in the equation:
P4(s) + 6 H2(g) → 4 PH3(g)
OR
1 + 6 → 4
If 1 (P4) gives us 4 (PH3), what gives us 1 (PH3)?
1/4 = x/1
solve for x