Answer:
Actual yield: 86.5 grams.
Explanation:
How many moles of formula units in 95 grams of calcium carbonate
?
Refer to a modern periodic table for relative atomic mass data:
- Ca: 40.078;
- C: 12.011;
- O: 15.999.
Formula mass of
:
.
.
How many moles of
will be produced?
The coefficient in front of
in the chemical equation is the same as that in front of
. That is:
.
.
What's the theoretical yield of calcium chloride? In other words, what's the mass of
of
?
Again, refer to a periodic table for relative atomic data:
.
.
What's the actual yield of calcium chloride?
.
.
The answer is A. As the first statement is a true statement. Hope this help you
Adding acid and and catching the solution that drains through.
Answer:
Explanation:
Combustion reaction is given below,
C₂H₅OH(l) + 3O₂(g) ⇒ 2CO₂(g) + 3H₂O(g)
Provided that such a combustion has a normal enthalpy,
ΔH°rxn = -1270 kJ/mol
That would be 1 mol reacting to release of ethanol,
⇒ -1270 kJ of heat
Now,
0.383 Ethanol mol responds to release or unlock,
(c) Determine the final temperature of the air in the room after the combustion.
Given that :
specific heat c = 1.005 J/(g. °C)
m = 5.56 ×10⁴ g
Using the relation:
q = mcΔT
- 486.34 = 5.56 ×10⁴ × 1.005 × ΔT
ΔT= (486.34 × 1000 )/5.56×10⁴ × 1.005
ΔT= 836.88 °C
ΔT= T₂ - T₁
T₂ = ΔT + T₁
T₂ = 836.88 °C + 21.7°C
T₂ = 858.58 °C
Therefore, the final temperature of the air in the room after combustion is 858.58 °C
Answer:
hello
Explanation:
second option is correct answer