Answer:
1 mole of propane combines with 5 moles of oxygen gas to produce 3 mole of carbon dioxide and 4 moles of water.
Explanation:
The word equation for the combustion of propane can be obtained from the chemical equation;
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
The word equation is therefore:
1 mole of propane combines with 5 moles of oxygen gas to produce 3 mole of carbon dioxide and 4 moles of water.
For such a combustion reaction, carbon dioxide and water are produced in the process.
Explanation:
The hydronium ion concentration can be found from the pH by the reverse of the mathematical operation employed to find the pH. [H3O+] = 10-pH or [H3O+] = antilog (- pH) Example: What is the hydronium ion concentration in a solution that has a pH of 8.34? On a calculator, calculate 10-8.34, or "inverse" log ( - 8.34).
I think the answer is 7mm but I'm not sure.
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The mass of a NaCl solution that is required to prepare 0.40 L of a 0.75 M solution is 17.55g. Details about mass can be found below.
<h3>How to calculate mass?</h3>
The mass of a substance can be calculated by multiplying the number of moles by its molar mass.
However, the number of moles of a solution must be initially calculated by using the following formula:
molarity = no of moles ÷ volume
no of moles = 0.75 × 0.40
no of moles = 0.3 moles
mass of NaCl = 0.3 × 58.5 = 17.55g
Therefore, the mass of a NaCl solution that is required to prepare 0.40 L of a 0.75 M solution is 17.55g.
Learn more about mass at: brainly.com/question/19694949
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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>
