This is the symbols for protons, neutrons, and electrons
To solve this we use the equation,
M1V1 = M2V2
where M1 is the concentration of the stock
solution, V1 is the volume of the stock solution, M2 is the concentration of
the new solution and V2 is its volume.
65 x V1 = 2 x 200 L
V1 = 6.15 L
Answer: C(s) + O2(g) --> CO2(g)12g (C) .... 50.8g (O2)................. initial amounts0g(C) .........18.8g(O2) ................. amounts when reaction completeThat means that C was the limiting reactant, and the amount of CO2 is based on the amount of carbon that burned. Covert 12 grams of carbon to moles. The moles of CO2 will be the same, since they are in a 1:1 mole ratio. Then convert the moles of CO2 to grams.12g C x (1 mol C / 12.0 g C) x (1 mol CO2 / 1 mol C) x (44.0g CO2 / 1 mol CO2) =44 g of CO2
Answer: 36.53g
Explanation:
First we need to find the amount of NaCl that dissolves in 1L of the solution that produced 5M of NaCl
Molarity = 5M
MM of NaCl = 58.45
Molarity = Mass conc (g/L) / MM
Mass conc. (g/L) of NaCl = Molarity x MM
= 5 x 58.45 = 292.25g
Next, we need to find the amount that will dissolve in 125mL(i.e 0.125L)
From the calculations above,
292.25g of NaCl dissolved in 1L
Therefore Xg of NaCl will dissolve in 0.125L of the solution i.e
Xg of NaCl = 292.25 x 0.125 = 36.53g.
Therefore 36.53g of NaCl will dissolve in 125mL of the solution
