Answer:
The final temperature of the iron is:- 38.0 °C
Explanation:
The initial temperature = 24.7 °C
Let Final temperature = T °C
Using,
Q = m C ×ΔT
Where,
Q is the heat absorbed by the iron = 85.7 J
C gas is the specific heat of the iron = 0.45 J/g °C
m is the mass of iron = 14.3 g
ΔT is the change in temperature. ( T - 24.7 ) °C
Applying the values as:
<u>Q = m C ×ΔT </u>
85.7 J = 14.3 g × 0.45 J/g °C × ( T - 24.7 ) °C
Solving for T, we get that:-
T = 38.0 °C
<u>The final temperature of the iron is:- 38.0 °C</u>
A gas with a vapor density greater than that of air, would be most effectively displaced out off a vessel by ventilation.
The two following principles determine the type of ventilation: Considering the impact of the contaminant's vapour density and either positive or negative pressure is applied.
Consider a vertical tank that is filled with methane gas. Methane would leak out if we opened the top hatch since its vapour density is far lower than that of air. A second opening could be built at the bottom to greatly increase the process' efficiency.
A faster atmospheric turnover would follow from air being pulled in via the bottom while the methane was vented out the top. The rate of natural ventilation will increase with the difference in vapour density. Numerous gases that require ventilation are either present in fairly low concentrations or have vapor densities close to one.
Answer:
When popcorn is popped, liquid inside the kernel is changed to steam. Pressure from the steam builds up inside the kernel. When the pressure reached a critical stage the kernel pops turning itself inside out. This is a physical change.
Explanation:
Answer:
The answer to your question is 23.4 moles of Glucose
Explanation:
Data
moles of Oxygen = ?
moles of glucose = 3.9
-Balanced chemical reaction
1C₆H₁₂O₆ + 6O₂ ⇒ 6H₂O + 6CO₂
Process
1.- To solve this problem, use the coefficient of the balanced chemical equation, and use proportions and cross multiplication.
1 mol of C₆H₁₂O₆ -------------------- 6 moles of O₂
3.9 moles of C₆H₁₂O₆ ------------------ x
x = (3.9 x 6 ) / 1
x = 23.4 moles
2.- Conclusion
3.9 moles of glucose consume 23.4 moles of oxygen
