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11111nata11111 [884]
3 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]3 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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<u>Answer:</u> The moles of oxygen and carbon dioxide in air is 3.63\times 10^{19}mol and 7.18\times 10^{16}mol respectively

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of atmosphere = 5.00\times 10^{18}kg=5.00\times 10^{21}g

Average molar mass of atmosphere = 28.96 g/mol

Putting values in above equation, we get:

\text{Moles of atmosphere}=\frac{5.00\times 10^{21}g}{28.96g/mol}=1.73\times 10^{20}mol

We know that:

Percent of oxygen in air = 21 %

Percent of carbon dioxide in air = 0.0415 %

Moles of oxygen in air = \frac{21}{100}\times 1.73\times 10^{20}=3.63\times 10^{19}mol

Moles of carbon dioxide in air = \frac{0.0415}{100}\times 1.73\times 10^{20}=7.18\times 10^{16}mol

Hence, the moles of oxygen and carbon dioxide in air is 3.63\times 10^{19}mol and 7.18\times 10^{16}mol respectively

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What is the pH of a solution that is 0.40 M NaBrO and 0.50 M HBrO (hypobromous acid) (Ka for HBrO = 2.3 x 10^-9)
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Answer

pH=8.5414

Procedure

The Henderson–Hasselbalch equation relates the pH of a chemical solution of a weak acid to the numerical value of the acid dissociation constant, Kₐ. In this equation, [HA] and [A⁻] refer to the equilibrium concentrations of the conjugate acid-base pair used to create the buffer solution.

pH = pKa + log₁₀ ([A⁻] / [HA])

Where

pH = acidity of a buffer solution

pKa = negative logarithm of Ka

Ka =acid disassociation constant

[HA]= concentration of an acid

[A⁻]= concentration of conjugate base

First, calculate the pKa

pKa=-log₁₀(Ka)= 8.6383

Then use the equation to get the pH (in this case the acid is HBrO)

pH=8.6383+\log_{10}(\frac{0.40\text{ M}}{0.50\text{ M}})=8.5414

8 0
1 year ago
What is the wavelength of light that has a frequency of 3.7 x 1014 Hz? Show all work!!!
kramer

Answer:

Explanation:

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