Answer:
Concentration of the resulting solution = 2.04 M
Explanation:
<u>Data:</u>
M1 = 5.10 M
V1 = 200.0 mL
V2 = 500.0 mL
M2 = ?
By modifying the volume of solution, keeping the amount of solute constant, the concentration changes. To perform the calculations, the equation will be:
Where M1 is the initial concentration of the solution, M2 the final concentration and V is the value of the volumes of the initial and final solution.
Clearing the value of M2 from the equation and replacing the values we have:
You have molarity and you have volume. Use the formula :
Molarity(M)= Moles(N)/Liter(L) to get the solution.
150 ml= .150 L
7.7 = N/.150
N=.1.155 moles of NaOH.
And since you know the moles, use the molar mass to figure out the grams.
<span> (40g/mol NaOH) x (1.155mol) =
46.2 g of NaOH.</span>
Hey there!:
Number of moles = ( number of atoms / 6.023*10²³ atoms )
given number of atoms = 5.03*10²⁴
Therefore:
Number of moles B = 5.03*10²⁴ / 6.023*10²³
Number of moles B = 8.35 moles
Hope that helps!
Gregor Mendel
Hope this helped !
2NaCN(s) + H₂SO₄(aq) --> Na₂SO₄(aq) + 2HCN(g)
The molar ratio between NaCN : HCN is 2:2 or 1:1
Mass of HCN = 16.7 g
Molar mass of HCN = 1 + 12 + 14 = 27 g/mol
Molar mass of NaCN = 49 g/mol
Therefore, the mass of NaCN is
16.7 g of HCN x 49 g/mol of NaCN / 27 g/mol of HCN = 30.3 grams of NaCN
Therefore, 30.3 grams of NaCN gives the lethal dose in the room.
