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inn [45]
3 years ago
14

A gas has a volume of 25.6 L at a temperature of 278 K. What will be the new temperature, in C, if

Chemistry
1 answer:
Inessa05 [86]3 years ago
7 0
<h2>Hello!</h2>

The answer is: -97.37° C

<h2>Why?</h2>

According to the Charles and Gay-Lussac's Law we have that:

\frac{V}{T}=k

Where:

<em>v</em> is the volume of the gas

<em>t</em> is the temperature of the gas

<em>k </em>is the proportionality constant

From the Gay-Lussac's Law we also have the following relation:

\frac{V1}{T1}=\frac{V2}{T2}

Since we need to find the new temperature (T2) we can use the last equation:

\frac{V1}{T1}=\frac{V2}{T2}\\\\T2=\frac{V2*T1}{V1}=\frac{16.2L*278K}{25.6L}=175.92K

We are asked to find the temperature in Celsius degrees, so, we must convert the result (in K) to Celsius degrees:

Celsius=K-273.15=175.92-273.15=-97.37

So, the temperature is -97.37° C

Have a nice day!

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A 100-g piece of metal initially at T = 75C is submerged in 100 g of water initially at T = 25C. The specific heat capacity of i
san4es73 [151]
I’m not so sure but I think it’s b
6 0
3 years ago
A compound is found to contain 50. 05% sulfur and 49. 95% oxygen by mass. What is the empirical formula for this compound? SO S2
jekas [21]

The empirical formula for the given compound has been \rm SO_2. Thus, option C is correct.

The empirical formula has been the whole unit ratio of the elements in the formula unit.

<h3>Computation for the Empirical formula</h3>

The given mass of Sulfur has been, 50.05 g

The given mass of oxygen has been 49.95 g.

The moles of elements in the sample has been given by:

\rm Moles=\dfrac{Mass}{Molar\;mass}

  • Moles of Sulfur:

\rm Moles\;S=\dfrac{50.05}{32}\\&#10; Moles\;S=1.56\;mol

The moles of sulfur in the unit has been 1.56 mol.

  • Moles of Oxygen:

\rm Moles\;O=\dfrac{49.95}{16} \\&#10;Moles\;O=3.12\;mol

The moles of oxygen in the unit has been 3.12 mol.

The empirical formula unit has been given as:

\rm S_{1.56}O_{3.12}=SO_2

Thus, the empirical formula for the given compound has been \rm SO_2. Thus, option C is correct.

Learn more about empirical formula, here:

brainly.com/question/11588623

8 0
2 years ago
Where the oxygen comes from the air (21% O2 and 79% N2). If oxygen is fed from air in excess of the stoichiometric amount requir
guajiro [1.7K]

Answer:

y_{O2} =4.3%

Explanation:

The ethanol combustion reaction is:

C_{2}H_{5} OH+3O_{2}→2CO_{2}+3H_{2}O

If we had the amount (x moles) of ethanol, we would calculate the oxygen moles required:

x*1.10(excess)*\frac{3 O_{2}moles }{etOHmole}

Dividing the previous equation by x:

1.10(excess)*\frac{3 O_{2}moles}{etOHmole}=3.30\frac{O_{2}moles}{etOHmole}

We would need 3.30 oxygen moles per ethanol mole.

Then we apply the composition relation between O2 and N2 in the feed air:

3.30(O_{2} moles)*\frac{0.79(N_{2} moles)}{0.21(O_{2} moles)}=121.414 (N_{2} moles )

Then calculate the oxygen moles number leaving the reactor, considering that 0.85 ethanol moles react and the stoichiometry of the reaction:

3.30(O_{2} moles)-0.85(etOHmoles)*\frac{3(O_{2} moles)}{1(etOHmoles)} =0.75O_{2} moles

Calculate the number of moles of CO2 and water considering the same:

0.85(etOHmoles)*\frac{3(H_{2}Omoles)}{1(etOHmoles)}=2.55(H_{2}Omoles)

0.85(etOHmoles)*\frac{2(CO_{2}moles)}{1(etOHmoles)}=1.7(CO_{2}moles)

The total number of moles at the reactor output would be:

N=1.7(CO2)+12.414(N2)+2.55(H2O)+0.75(O2)\\ N=17.414(Dry-air-moles)

So, the oxygen mole fraction would be:

y_{O_{2}}=\frac{0.75}{17.414}=0.0430=4.3%

6 0
3 years ago
Can someone tell me plz :P
NemiM [27]

Answer:

There's nothing there

Explanation:

8 0
3 years ago
Which group on the periodic table contains only metals?
Natalija [7]

Answer:

D. Along the bottom

5 0
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