Answer:
60 grams
Explanation:
We have the balanced equation (without state symbols):
6
H
2
O
+
6
C
O
2
→
C
6
H
12
O
6
+
6
O
2
So, we would need six moles of carbon dioxide to fully produce one mole of glucose.
Here, we got
88
g
of carbon dioxide, and we need to convert it into moles.
Carbon dioxide has a molar mass of
44
g/mol
. So here, there exist
88
g
44
g
/mol
=
2
mol
Since there are two moles of
C
O
2
, we can produce
2
6
⋅
1
=
1
3
moles of glucose
(
C
6
H
12
O
6
)
.
We need to find the mass of the glucose produced, so we multiply the number of moles of glucose by its molar mass.
Glucose has a molar mass of
180.156
g/mol
. So here, the mass of glucose produced is
1
3
mol
⋅
180.156
g
mol
≈
60
g
to the nearest whole number.
So, approximately
60
grams of glucose will be produced.
Answer:
838 torr
Step-by-step explanation:
To solve this problem, we can use the <em>Combined Gas Laws</em>:
p₁V₁/T₁ = p₂V₂/T₂ Multiply each side by T₁
p₁V₁ = p₂V₂ × T₁/T₂ Divide each side by V₁
p₁ = p₂ × V₂/V₁ × T₁/T₂
<em>Data:
</em>
p₁ = ?; V₁ = 2.42 L; T₁ = 27.0 °C
p₂ = 754 torr; V₂ = 2.37 L; T₂ = -8.8 °C
Calculations:
(a) Convert <em>temperatures to kelvins
</em>
T₁ = (27.0 + 273.15) K = 300.15 K
T₂ = (-8.8 + 273.15) K = 264.35 K
(b) Calculate the<em> pressure
</em>
p₁ = 754 torr × (2.37 L/2.42) × (300.15/264.35)
p₁ = 754 torr × 0.979 × 1.135
p₁ = 838 torr
The answer that would best complete the given statement above would be the term PHASE. Here is the complete statement. <span>A homogeneous portion of a mixture that is characterized by uniform properties and capable of being separated by mechanical means is called a PHASE. Hope this helps.</span>
