Let us assume propane was the fuel
C3H8(g) + 5O2(g) ---> 3CO2(g) + 4H2O(g) = 2217kJ
1 mole ofpropane produces 3 moles of CO2
heat absorbed by pork = 0.11 x 2217
= 243.87 kJ/mol
number of moles of propane = 1700kJ / 243.87 kJ/mol
= 6.971 moles
1 mole of C3H8 = 3 moles ofCO2
6.971 moles of C3H8 = ?
3 x 6.971 = 20.913 moles of CO2
Convert to grams
mass = MW x mole
= 44 x 20.913
= 920.172g of CO2 emitted
Answer:
The unknown amount of potassium chloride is 13.6 grams.
Explanation:
The reaction of 14 grams of KNO₃ with KCl produces a total mass of 27.6 grams of the products.
The law of conservation of mass tells us that the total mass of the reactants must be the same that the total mass of the products. So, we can find the mass of KCl as follows:
Where <em>r</em> is for reactants and <em>p </em>is for products
Therefore, the unknown amount of potassium chloride is 13.6 grams.
I hope it helps you!
Answer:
45.3°C
Explanation:
Step 1:
Data obtained from the question.
Initial pressure (P1) = 82KPa
Initial temperature (T1) = 26°C
Final pressure (P2) = 87.3KPa.
Final temperature (T2) =.?
Step 2:
Conversion of celsius temperature to Kelvin temperature.
This is illustrated below:
T(K) = T(°C) + 273
Initial temperature (T1) = 26°C
Initial temperature (T1) = 26°C + 273 = 299K.
Step 3:
Determination of the new temperature of the gas. This can be obtained as follow:
P1/T1 = P2/T2
82/299 = 87.3/T2
Cross multiply to express in linear form
82 x T2 = 299 x 87.3
Divide both side by 82
T2 = (299 x 87.3) /82
T2 = 318.3K
Step 4:
Conversion of 318.3K to celsius temperature. This is illustrated below:
T(°C) = T(K) – 273
T(K) = 318.3K
T(°C) = 318.3 – 273
T(°C) = 45.3°C.
Therefore, the new temperature of the gas in th tire is 45.3°C
Formula of density is mass divided by volume.
