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Lady bird [3.3K]
2 years ago
6

A balloon containing helium gas expands from 230

Chemistry
1 answer:
Anit [1.1K]2 years ago
7 0

The answer for the following problem is mentioned below.

  • <u><em>Therefore the final  moles of the gas is 14.2 × </em></u>10^{-4}<u><em> moles.</em></u>

Explanation:

Given:

Initial volume (V_{1}) = 230 ml

Final volume (V_{2}) = 860 ml

Initial moles (n_{1}) = 3.8 ×10^{-4} moles

To find:

Final moles (n_{2})

We know;

According to the ideal gas equation;

    P × V = n × R × T

where;

P represents the pressure of the gas

V represents the volume of the gas

n represents the no of the moles of the gas

R represents the universal gas constant

T represents the temperature of the gas

So;

    V ∝ n

\frac{V_{1} }{V_{2} } = \frac{n_{1} }{n_{2} }

where,

(V_{1}) represents the initial volume of the gas

(V_{2}) represents the final volume of the gas

(n_{1}) represents the initial  moles of the gas

(n_{2}) represents the final moles of the gas

Substituting the above values;

   \frac{230}{860} = \frac{3.8 * 10^-4}{n_{2} }

  n_{2} = 14.2 × 10^{-4} moles

<u><em>Therefore the final  moles of the gas is 14.2 × </em></u>10^{-4}<u><em> moles.</em></u>

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calculate the specific heat capacity for gold n 105 joules are required to heat 30.0 grams of gold from 27.7c to 54.9c
Pepsi [2]

<u>Answer:</u>

<em>The specific heat capacity for gold in 105 joules which are required to heat 30.0 grams of gold is 0.129 J/(g℃)</em>

<u>Explanation:</u>

We make use of the formula

Q=m \times c \times \Delta T

where

∆T = final T - initial T

= 54.9℃ - 27.7℃ = 27.2℃

Q is the heat energy in Joules = 105J

c is the specific heat capacity = ?

m is the mass of Gold = 30.0g

Q=m \times c \times \Delta T

Rearranging the formula

c= \frac {Q}{(m\times \Delta T)}

= \frac {105J}{(30.0g \times 27.2 ^\circ{C})}\\\\= \frac {105J}{(816g^\circ{C})}

So,

c = 0.129 J/(g℃)

(Answer)

7 0
2 years ago
Which contain covalent bonds?
Ilia_Sergeevich [38]
The only compound that contains covalent bonds would be A. BCl4-.
3 0
2 years ago
Zinc metal is added to hydrochloric acid to generate hydrogen gas and is collected over a liquid whose vapor pressure is the sam
mafiozo [28]

Question:

Zinc metal is added to hydrochloric acid to generate hydrogen gas and is collected over a liquid whose vapor pressure is the same as pure water at 20.0 degrees C (18 torr). The volume of the mixture is 1.7 L and its total pressure is 0.987 atm. Determine the number of moles of hydrogen gas present in the sample.

A. 0.272 mol

B. 0.04 mol

C. 0.997 mol

D. 0.139 mol

E. 0.0681 mol

Answer:

The correct option is;

E. 0.0681 mol

Explanation:

The equation for the reaction is

Zn + HCl = H₂ + ZnCl₂

Vapor pressure of the liquid = 18 torr = 2399.803 Pa

Total pressure of gas mixture H₂ + liquid vapor = 0.987 atm  

= 100007.775 Pa

Therefore, by Avogadro's law, pressure of the hydrogen gas is given by the following equation

Pressure of H₂ = 100007.775 Pa - 2399.803 Pa = 97607.972 Pa

Volume of H₂ = 1.7 L = 0.0017 m³

Temperature = 20 °C = 293.15 K

Therefore,

n = \frac{PV}{RT} =  \frac{100007.775 \times 0.0017 }{8.3145 \times 293.15} = 0.068078 \ moles

Therefore, the number of moles of hydrogen gas present in the sample is n ≈ 0.0681 moles.

7 0
2 years ago
6.0 g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of 13
vodomira [7]

<u>Answer:</u>

<em>The molecular formula of X is given as C_7 H_6 O_3</em>

<em></em>

<u>Explanation:</u>

Moles $C O_{2}=\frac{\text { mass }}{\text { molar mass }}=\frac{13.39 \mathrm{g}}{44.01 \mathrm{g} \text { per mole }}=0.304 \mathrm{mol}$\\\\moles $\mathrm{C}=$ moles $\mathrm{CO}_{2}=0.304 \mathrm{mol}$

mass $C=$ moles $\times$ molar mass $=0.304 \mathrm{mol} \times 12 \frac{g}{m o l}=3.65g$\\\\moles $\mathrm{H}_{2} \mathrm{O}=\frac{2.35 \mathrm{g}}{18.02 \mathrm{g} \text { permole }}=0.130 \mathrm{mol}$\\\\moles $\mathrm{H}=2 \times$ moles $\mathrm{H}_{2} \mathrm{O}=0.130 \times 2=0.260 \mathrm{mol}$\\\\Mass $\mathrm{H}=0.260 \mathrm{mol} \times 1.008 \frac{g}{\mathrm{mol}}=0.262 \mathrm{g}$

mass O = Total mass of the compound - (mass of C + mass of H)

=6.0 g - ( 3.65 + 0.262 ) g

=2.09 g

moles $O=\frac{2.09 g}{16 g \text { per mole }}=0.131 \mathrm{mol}$

Least moles is for O that is 0.131mol and dividing all by the least we get

$\begin{aligned} C &=\frac{0.304}{0.131}=2.3 \\\\ H &=\frac{0.260}{0.131}=2 \\\\ O &=\frac{0.131}{0.131}=1 \end{aligned}$

Since 2.3 is a fraction it has to be converted to a whole number so we multiply all the answers by 3

\\$C 2.3 \times 3=7$\\\\$H 2 \times 3=6$\\\\$O 1 \times 3=3$

So the empirical formula is C_7 H_6 O_3

Empirical formula mass

=(7 \times 12) +(6\times1.008)+(3\times16)=138.048g

$n=\frac{\text { molar mass }}{\text { empirical formula mass }}=\frac{138}{138.048}=1$

Molecular formula =n × empirical formula

=1 \times C_7 H_6 O_3

Compound X  = C_7 H_6 O_3  is the Answer

8 0
2 years ago
at what temperature (inc) would the volume a gas be equal to 45.7L if the volume of gas was 33.9L at 12.4c
vesna_86 [32]

Answer:

The answer to your question is  T1 = 384.7 °K

Explanation:

Data

Volume 1 = V1 = 45.7 l

Temperature 1 = T1 = ?

Volume 2 = V2 = 33.9 l

Temperature 2 = T2 = 12.4°C

To solve this problem use Charles' law

              V1/T1 = V2/T2

                    T1 = V1T2/V2

-Convert temperature to °K

T2 = 12.4 + 273 = 285.4°K

-Substitution

                    T1 = (45.7 x 285.4) / 33.9

-Simplification

                    T1 = 13042.8 / 33.9

-Result

                    T1 = 384.7 °K

7 0
2 years ago
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