With the given components above, the chemical reaction required to solve the problem is <span>H2C2O4*2H2O + 2 NaOH = 4 H2O + Na2C2O<span>4 where 1 mole of diprotic acid needs 2 moles of NaOH to complete the reaction.In this case, given </span></span><span>0.4500 M NaoH and 0.02 L of it, the moles diprotic acid needed is 0.0045 moles. This is equivalent to 0.351 grams.</span>
Answer:
Acetic Acid (HC2H3O2) dissolves in water, but only partially dissociates into ions.
I believe the correct answer would be that b<span>oiling points and melting points are similar because they both involve the change in a state of a material, but they are different because boiling point involves a change from a liquid to a gas and melting point involves a change from a solid to a liquid. Boiling and melting are phase changes that can happen to a substance however they differ in the process that happens.</span>
Answer:
The answer is 1.06g.
Explanation:
Analysis of question:
1. Identify the information in the question given.
- volume of HCl is 2 dm3
- pH of HCl is 2.0
2. What the question want?
- mass of Na2CO3 is ?(unknown)
- 3. Do calculation.
- 1st-Write a balanced chemical equation:
Na2CO3 + 2HCl (arrow) 2NaCl + H20 + CO2
- 2nd-Determine the molarity of HCl with the value of 2.0.
pH= -log[H+]
2.0= -log[H+]
log[H+]= -2.0
[H+]= 10 to the power of negative 2(10-2)
=0.01 mol dm-3
molarity of HCl is 0.01 mol dm-3
- 3rd-Find the number of moles of HCl
n=MV
=0.01 mol dm-3 × 2 dm3
=0.02 mol of HCl
- 4th-Find the second mol of it.
Based on the chemical equation,
2.0 mol of HCl reacts with 1.0 mol of Na2CO3
0.02 mol of HCl reacts with 0.01 mol of Na2CO3
<u>N</u><u>a</u>2CO3>a=<u>1</u><u> </u>mol
<u>2</u><u>H</u>Cl>b=<u>2</u><u> </u>mol
mass= number of mole × molar mass
g=0.01 × [2(23)+ 12+ 3(16)]
g=0.01 × 106
# =1.06 g.
Answer:
Explanation:
MM: 30.01
N₂ + O₂ ⟶ 2NO; ΔH = +43 kcal/mol
m/g: 147
Treat the heat as if it were a reactant in the reaction. Then you can write
N₂ + O₂ + 43 kcal ⟶ 2NO
The conversion factor is then 43 kcal/2 mol NO.
1. Moles of NO
2. Amount of heat
