Answer:
As nouns the difference between impression and imprint
is that impression is the indentation or depression made by the pressure of one object on or into another while imprint is an impression; the mark left behind by printing something.
Answer:
The initial temperature of the metal is 84.149 °C.
Explanation:
The heat lost by the metal will be equivalent to the heat gain by the water.
- (msΔT)metal = (msΔT)water
-32.5 grams × 0.365 J/g°C × ΔT = 105.3 grams × 4.18 J/g °C × (17.3 -15.4)°C
-ΔT = 836.29/12.51 °C
-ΔT = 66.89 °C
-(T final - T initial) = 66.89 °C
T initial = 66.89 °C + T final
T initial = 66.89 °C + 17.3 °C
T initial = 84.149 °C.
I cannot fully remember but the charge of a compound always has to be the same. However, I believe that the sign of the Y2 means that it is positive 2.
so in my opinion it is in between b and c. use your best judgment.
Answer:
ΔH°rxn = - 433.1 KJ/mol
Explanation:
- CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g)
⇒ ΔH°rxn = 4ΔH°HCl(g) + ΔH°CCl4(g) - 4ΔH°Cl2(g) - ΔH°CH4(g)
∴ ΔH°Cl2(g) = 0 KJ/mol.....pure element in its reference state
∴ ΔH°CCl4(g) = - 138.7 KJ/mol
∴ ΔH°HCl(g) = - 92.3 KJ/mol
∴ ΔH°CH4(g) = - 74.8 KJ/mol
⇒ ΔH°rxn = 4(- 92.3 KJ/mol) + (- 138.7 KJ/mol) - 4(0 KJ/mol) - (- 74.8 KJ/mol)
⇒ ΔH°rxn = - 369.2 KJ/mol - 138.7 KJ/mol - 0 KJ/mol + 74.8 KJ/mol
⇒ ΔH°rxn = - 433.1 KJ/mol
The question is incomplete, here is the complete question:
What volume (mL) of the partially neutralized stomach acid having concentration 2 M was neutralized by 0.1 M NaOH during the titration? (portion of 25.00 mL NaOH sample was used; this was the HCl remaining after the antacid tablet did it's job)
<u>Answer:</u> The volume of HCl neutralized is 1.25 mL
<u>Explanation:</u>
To calculate the volume of acid, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of stomach acid which is HCl
are the n-factor, molarity and volume of base which is NaOH.
We are given:
Putting values in above equation, we get:
Hence, the volume of HCl neutralized is 1.25 mL
