Answer:
The answer is endothermic as the heat flows into the system from the surroundings. The products are at higher energy than the reactants, as they have absorbed energy.
Answer:
1.60x10⁶ billions of g of CO₂
Explanation:
Let's calculate the production of CO₂ by a single human in a day. The molar mass of glucose is 180.156 g/mol and CO₂ is 44.01 g/mol. By the stoichiometry of the reaction:
1 mol of C₆H₁₂O₆ -------------------------- 6 moles of CO₂
Transforming for mass multiplying the number of moles by the molar mass:
180.156 g of C₆H₁₂O₆ ----------------- 264.06 g of CO₂
4.59x10² g ---------------- x
By a simple direct three rule:
180.156x = 121203.54
x = 672.77 g of CO₂ per day per human
So, in a year, 6.50 billion of human produce:
672.77 * 365 * 6.50 billion = 1.60x10⁶ billions of g of CO₂
Answer:
V₂ = 12.43 L
Explanation:
Given data:
Initial pressure = 650 KPa
Initial volume = 2.2 L
Final pressure = 115 KPa
Final volume = ?
Solution:
The given problem will be solved through the Boyles law,
"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"
Mathematical expression:
P₁V₁ = P₂V₂
P₁ = Initial pressure
V₁ = initial volume
P₂ = final pressure
V₂ = final volume
Now we will put the values in formula,
P₁V₁ = P₂V₂
650 KPa ×2.2 L = 115 KPa × V₂
V₂ = 1430 KPa. L/ 115 KPa
V₂ = 12.43 L
Answer:
Molar heat of solution of KBr is 20.0kJ/mol
Explanation:
Molar heat of solution is defined as the energy released (negative) or absorbed (Positive) per mole of solute being dissolved in solvent.
The dissolution of KBr is:
KBr → K⁺ + Br⁻
In the calorimeter, the temperature decreases 0.370K, that means the solution absorbes energy in this process. The energy is:
q = 1.36kJK⁻¹ × 0.370K
q = 0.5032kJ
Moles of KBr in 3.00g are:
3.00g × (1mol / 119g) = 0.0252moles
Thus, molar heat of solution of KBr is:
0.5032kJ / 0.0252moles = <em>20.0kJ/mol</em>
Answer:
The answer to your question is 2 molecules, or B(edg 2021).
Explanation:
edg 2021
