Answer:
mCO2= 49.6932 kgCO2
Explanation:
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First we propose the balanced equation C3H8 + 5O2 ---> 3CO2 + 4H2O
We see that each mole of C3H8 (propane) we get 3 moles of CO2
From the propane volume we can obtain the grams of propane used.
molpropane = 26.5L * (1000mL / 1L) * (0.621g / 1mL) * (1mol / 44g) = 374.01mol propane
mCO2 = 374.01molC3H8 * (3molCO2 / 1molC3H8) * (44gCO2 / 1molCO2) = 49369.32g * (1kg / 1000g) = 49.6932 kgCO2
mCO2= 49.6932 kgCO2
Explanation:
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Answer:
572 g
Explanation:
Molar mass is the mass of 1 mol of an element or compound
molar mass of Li₂SO₄ is the sum of the products of the molar masses of the elements by the number of atoms in the compound
molar masses of each element making up lithium sulphate
Li - 7 g/mol
S - 32 g/mol
O - 16 g/mol
molar mass of Li₂SO₄ - (7 g/mol x 2) + ( 32 g/mol x 1) + ( 16 g/mol x 4 )
molar mass = 110 g/mol
mass of 1 mol of Li₂SO₄ is 110 g
therefore mass of 5.2 mol of Li₂SO₄ is - 110 g/mol x 5.2 mol = 572 g
mass is 572 g
Answer:
The ΔH is 5.5 kJ/mol and the reaction is endothermic.
Explanation:
To calculate the ∆H (heat of reaction) of the combustion reaction, that is, the heat that accompanies the entire reaction, you must make the total sum of all the heats of the products and of the reagents affected by their stoichiometric coefficient ( number of molecules of each compound participating in the reaction) and finally subtract them:
Combustion enthalpy = ΔH = ∑H products - ∑Hreactants
In this case:
ΔH = 15.7 kJ/mol - 10.2 kJ/mol= 5.5 kJ/mol
An endothermic reaction is one whose enthalpy value is positive, that is, the system absorbs heat from the environment (ΔH> 0).
<u><em>The ΔH is 5.5 kJ/mol and the reaction is endothermic.</em></u>
