Answer:
2 393.3 mL
Explanation:
(256.3 mL) + (2 L) + (137 mL) =
2 393.3 mL
Answer:
456 kJ
Step-by-step explanation:
CH₄ + 2Cl₂ ⟶ CCl₄ + 2H₂ ; ΔH = 45.6 kJ
Treat the heat as if it were a product in the equation. Then use the molar ratio (45.6 kJ/2 mol Cl₂) in the usual way.
Amount of energy = 20 mol Cl₂ × (45.6 kJ/2 mol Cl₂) = 456 kJ
You must add 456 kJ to react 20 mol of chlorine with excess methane.
Answer:
(b) Calculate the molarity of a solution of 4.8 mole of HCl in 600 mL of solution. ... (g) Calculate the mass of Na2CO3 that must be used to make 700 mL of a 0.136 M Na2CO3 ... (h) What mass of NaOH is needed to make 200 mL of a 0.300 M NaOH solution? ... However, when we are reacting solutions we have to convert.
Explanation:
<span>We are given the volume of the solution which is 80 mL or
0.08 L and the density which is equal to water, density = 1 kg / L. Therefore calculate total mass:</span>
total mass = (1 kg / L) * (0.08 L) = 0.08 kg = 80 g
So the mass of NaCl is:
mass NaCl = 0.0092 * 80 g = 0.736 g
The molar mass of NaCl is 58.44 g/mol, so the number of
moles is:
moles NaCl = 0.736 g / (58.44 g/mol) = 0.0126 mol
SO the molarity is mol / L:
<span>Molarity = 0.0126 mol / 0.08 L = 0.157 mol/L = 0.157 M</span>
The percentage of dissolved salt in saltwater
is 3.5%. We are given the amount seawater sample. <span>We would just multiply the two values together.
750 g (0.035) = 26.25 g NaCl
26.25 g NaCl ( 1 mol / 59.44 g ) ( 1 mol Na / 1 mol NaCl ) ( 22.99 g / 1 mol ) = 10.15 g Na
Hope this answers the question. Have a nice day.</span>
