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soldier1979 [14.2K]
1 year ago
7

A sample of 5.72 g of liquid 1-propanol, C, H, O, is combustod with 43.4 g of oxygen gas, Carbon dioxide and water are

Chemistry
1 answer:
bekas [8.4K]1 year ago
6 0

Answer:

2C3H8O + 9O2 ==> 6CO2 + 8H2O ... balanced equation

moles propanol = 5.26 g x 1 mol/60.1 g = 0.0875 moles

moles O2 = 31.8 g x 1 mol/31.9 g = 0.997 moles O2

Propanol is limiting based on the mol ratio in balance equation of 2 : 9

To find mass of O2 (excess reagent) left over, we will first find moles O2 used up.

moles O2 used = 0.0875 mol propanol x 9 mol O2/2 mol propanol = 0.394 moles O2 used

moles O2 left over = 0.997 mol - 0.394 mol = 0.603 mol O2 left

mass O2 left = 0.603 mol O2 x 32 g/mol = 19.3 g O2 left over

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A gas occupies 200ml at a temperature of 26 degrees Celsius and 76mmHg pressure. Find the volume at -3degree Celsius with the pr
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Answer:

184.62 ml

Explanation:

Let p_1, v_1, and T_1 be the initial and p_2, v_2, and T_2 be the final pressure, volume, and temperature of the gas respectively.

Given that the pressure remains constant, so

p_1=p_2 ...(i)

v_1 = 200 ml

T_1= 26 ^{\circ}C = 273+26 =299 K

T_2= 3 ^{\circ}C = 273+3 =276 K

From the ideal gas equation, pv=mRT

Where p is the pressure, v is the volume, T is the temperature in Kelvin, m is the mass of air in kg, R is the specific gas constant.

For the initial condition,

p_1v_1=mRT_1 \\\\mR= \frac{p_1v_1}{T_1}\cdots(ii)

For the final condition,

p_2v_2=mRT_2 \\\\mR= \frac{p_2v_2}{T_2}\cdots(iii)

Equating equation (i), and (ii)

\frac{p_1v_1}{T_1}=\frac{p_2v_2}{T_2}

\frac{v_1}{T_1}=\frac{v_2}{T_2}  [from equation (i)]

v_2=\frac{T_2}{T_1} \times v_1

Putting all the given values, we have

v_2=\frac{276}{299} \times 200 = 184.62 \; ml

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3.05 = 2.21 + log [neutral form] / [Protonated form]

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<h2> The correct option is (a).</h2>

Explanation:

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Hence, option (a) is correct.

Rest of the options (b), (c), (d), (e) is not correct because of the following reasons:

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