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11111nata11111 [884]
2 years ago
10

The combustion of propane gas is used to fuel barbeque grills. If 4.65 moles of propane, C3H8, are burned in a grilling session,

how many moles of carbon dioxide gas are formed? C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)
Chemistry
1 answer:
Varvara68 [4.7K]2 years ago
8 0

Answer:

<h3>moles of carbon dioxide=13.95mol</h3>

Explanation:

First wrie down the balance chemical reaction:

C_3H_8(g) + 5 O_2(g) \rightarrow 3 CO_2(g) + 4 H_2O(l)

Combustion reaction: The reacion in which hydrocarbon is burnt in the presence of oxygen gas and it releases heat and this reaction exothermic because heat of cumbustion is negative.

eg. burning of methane

By using unitry method,

From the above balanced reaction it is clearly that,

from 1 mole of propane 3 moles of carbon dioxide is formed

there fore,

from 4.65 mole of propane 3 \times 4.65 moles of carbon dioxide will form

moles of carbon dioxide=13.95mol

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Many plants are poisonous because their stems and leaves contain oxalic acid, H2C2O4, or sodium oxalate, Na2C2O4. When ingested,
Neporo4naja [7]

Answer:

Explanation:

Calcium chloride is a soluble salt which dissociates into calcium and chloride ions when dissolved in water.

CaCl₂(aq) ----> Ca²⁺(aq) + 2Cl⁻(aq)

Similarly, sodium oxalate when dissolved in water dissociates into sodium and oxalate ions.

Na₂CO₄(aq) ----> 2Na⁺(aq) + C₂O₄²⁻(aq)

However, in a double displacement reaction where the two solutions of the salts are mixed, the insoluble salt calcium oxalate is precipitated. The net ionic equation for the reaction is shown below:

Ca²⁺(aq) + C₂O₄²⁻(aq) ----> CaC₂O₄(s)

8 0
2 years ago
Ordinary table sugar is primarily sucrose, which has the chemical formula c12h22o11. calculate the mass percentages of carbon, h
tester [92]
<span>carbon = 42.1% hydrogen = 6.5% oxygen = 51.4% First lookup the molar mass of carbon, hydrogen, and oxygen. mass of carbon = 12.0107 mass of hydrogen = 1.00794 mass of oxygen = 15.999 Now calculate the molar mass of each element in sucrose multiplying the atomic weight of each element by the number of times the element is used. carbon = 12 * 12.0107 = 144.1284 hydrogen = 22 * 1.00794 = 22.17468 oxygen = 11 * 15.999 = 175.989 Calculate the molar mass of sucrose by adding the mass of each element used. 144.1284 + 22.17468 + 175.989 = 342.2921 Finally, calculate the percentage by mass of each element by dividing the mass used for each element by the total mass of sucrose. carbon = 144.1284 / 342.2921 = 0.421068 = 42.1% hydrogen = 22.17468 / 342.2921 = 0.064783 = 6.5% oxygen = 175.989 / 342.2921 = 0.514149 = 51.4%</span>
6 0
2 years ago
When 10 g of diethyl ether is converted to vapor at its boiling point, about how much heat is absorbed? (c4h10o, δhvap = 15.7 kj
vagabundo [1.1K]

Answer:

ΔH= 3KJ

Explanation:

The total heat absorbed is the total energy in the process, and that is in form of entalpy.

ΔH = q + ΔHvap, where q is the heat necessary for elevate the temperature of dietil ether. Suppose the initial temperature is room temperature (25ºC=298 K), then

q= 10g x2.261 J/gK x(310 K - 298K)= 271.32 J= 0.3 kJ

Then

ΔHvap = 10g C4H10O x (1 mol C4H10O/74.12 g C4H10O) x( 15.7 KJ/ 1 mol C4H10O) = 2.12 KJ

ΔH= 2.5KJ ≈ 3KJ

5 0
2 years ago
If the density of carbon tetrachloride is 1.59 g/ml, what is the volume in l, of 4.21 kg of carbon tetrachloride
makvit [3.9K]

Density is the ratio of mass to the volume. The formula of density is:

density = \frac{mass}{volume}    -(1)

Density of carbon tetrachloride = 1.59 g/ml   (given)

Mass of carbon tetrachloride = 4.21 kg   (given)

Since, 1 kg = 1000 g

So, 4.21 kg = 4210 g

Substituting the values in formula (1):

1.59 g/mL = \frac{4210 g}{volume}

volume = \frac{4210 g}{1.59 g/mL}

volume = 2647.799 mL

Since, 1 mL = 0.001 L

So, 2647.799 mL = 2.65 L

Hence, the volume of carbon tetrachloride is 2.65 L.


6 0
2 years ago
If the standard solutions had unknowingly been made up to be 0.0024 m agno3 and 0.0040 m k2cro4, would this have affected your r
Mnenie [13.5K]
2AgNO3+K2CrO4⇒Ag2CrO4(s)+2KNO3
Hence by mixing 0.0024M AgNO3 and 0.004M
K2CrO4, we will have Ag2CrO4 which is precipitated out and leave us with 
0.0024M KN03 which is mixed with (0.004-0.0024/2)M, it can be 0.0028M, of K2Cr04
6 0
2 years ago
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