Answer:
-800 kJ/mol
Explanation:
To solve the problem, we have to express the enthalpy of combustion (ΔHc) in kJ per mole (kJ/mol).
First, we have to calculate the moles of methane (CH₄) there are in 2.50 g of substance. For this, we divide the mass into the molecular weight Mw) of CH₄:
Mw(CH₄) = 12 g/mol C + (1 g/mol H x 4) = 16 g/mol
moles CH₄ = mass CH₄/Mw(CH₄)= 2.50 g/(16 g/mol) = 0.15625 mol CH₄
Now, we divide the heat released into the moles of CH₄ to obtain the enthalpy per mole of CH₄:
ΔHc = heat/mol CH₄ = 125 kJ/(0.15625 mol) = 800 kJ/mol
Therefore, the enthalpy of combustion of methane is -800 kJ/mol (the minus sign indicated that the heat is released).
Answer:
0.077 M
Explanation:
Data Given :
The concentration of half normal (NaCl) saline = 0.45g / 100 g
So,
Volume of Solution = 100 g = 100 mL
Volume of Solution in Liter = 100 mL / 1000
Volume of Solution = 0.1 L
molar mass of NaCl = 58.44 g/mol
Molarity:
Molarity is the representation of the solution. It is amount of solute in moles per liter of solution and represented by M
Formula used for Molarity
M = moles of solute / Liter of solution . . . . . . . . . . (1)
Now to find number of moles of Nacl
no. of moles of NaCl = mass of NaCl / molar mass
no. of moles of NaCl = 0.45g / 58.44 g/mol
no. of moles of NaCl = 0.0077 g
Put values in the eq (1)
M = moles of solute / Liter of solution . . . . . . . . . . (1)
M = 0.0077 g / 0.1 L
M = 0.077 M
So the molarity of half-normal saline solution (0.45% NaCl) = 0.077 M
<span>2 KClO3(s) → 3 O2(g) + 2 KCl(s)
</span><span>Note: MnO2 (Manganese Dioxide) is not part of the reaction. A catalyst lowers the activation energy and increases both forward and reverse reactions at equal rates.
</span>
molar mass of KClO3 = 122.5
Moles of KClO3 = 3.45 / 122.55 = 0.028
Moles of O2 produce =
= 0.042 moles
molar mass of O2 = 32
so, mass of O2 = 32 x 0.042 = 1.35 g
Answer:
Installing new technology, such as scrubbers, in factories will decrease their harmful emissions. This helps improve the safety of the surrounding community and the workers. But this technology is expensive and requires time and effort to install.
Explanation:
Hope this helped :)
