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3 years ago

2C4H10+13o2= 8CO2+10H2O

Chemistry

1 answer:

Gennadij [26K]3 years ago

5 0

Answer:

your answer will be between 16-17 moles

Explanation:

The combustion of butane is 2 C4H10 + 13 O2 = 8 CO2 + 10 H2O. Every two moles of C4H10 can produce 10 moles of water,

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Question 2 The metal molybdenum becomes superconducting at temperatures below 0.90K. Calculate the temperature at which molybden

LenKa [72]

Answer:

Temperature at which molybdenum becomes superconducting is-272.25°C

Explanation:

Conductor are those hard substances which allows path of electric current through them. And super conductors are those hard substances which have resistance against the flow of electric current through them.

As given, molybdenum becomes superconducting at temperatures below 0.90 K.

Temperature in Kelvins can be converted in °C by relation:

T(°C)=273.15-T(K)

Molybdenum becomes superconducting in degrees Celsius.

T(°C)=273.15-0.90= -272.25 °C

Temperature at which molybdenum becomes superconducting is -272.25 °C

5 0

3 years ago

A mixture containing 20 mole % butane, 35 mole % pentane and rest

notka56 [123]

Answer:

2.5 % butane, 42.2 % pentane and 55.3 % hexane

Explanation:

Hello,

In this case, the mass balance for each substance is given by:

$Butane:z_bF=y_bD+x_bB\\\\Pentane: z_pF=y_pD+x_pB\\\\Hexane: z_hF=y_hD+x_hB$

Whereas $y$ accounts for the fractions at the outlet distillate and $x$ for the fractions at the outlet bottoms. Moreover, with the 90 % recovery of butane, we can write:

$0.9=\frac{y_bD}{z_bF}$

So we can compute the product of the molar fraction of butane at the distillate by total distillate flow by assuming a 100-mol feed:

$y_bD=0.9*z_bF=0.9*0.2*100mol=18mol$

The total distillate flow:

$y_bD=18mol\\\\D=\frac{18mol}{0.95} =18.95mol$

And the total bottoms flow:

$F=D+B\\\\B=F-D=100mol-18.95mol=81.05mol$

Next, by using the mass balance of butane, we compute the molar fraction of butane at the bottoms:

$x_b=\frac{z_bF-y_bD}{B} =\frac{0.2*100mol-18mol}{81.05} =0.025$

Then, the molar fraction of pentane and hexane:

$x_p=\frac{z_pF-y_pD}{B} =\frac{0.35*100mol-0.04*18.95mol}{81.05} =0.422$

$x_h=\frac{z_hF-y_hD}{B} =\frac{(1-0.2-0.35)*100mol-(1-0.95-0.04)*18.95mol}{81.05} =0.553$

Therefore, the molar composition of the bottom product is 2.5 % butane, 42.2 % pentane and 55.3 % hexane.

NOTE: notice the result is independent of the value of the assumed feed, it means that no matter the basis, the compositions will be the same for the same recovery of butane at the feed, only the flows will change.

Regards.

8 0

3 years ago

Read 2 more answers

10 pts) A student titrates a 20.00 mL sample of an aqueous borax solution with 1.03 M H2SO4. If 2.07 mL of acid are needed to re

Natasha2012 [34]

Answer: The molarity of the borax solution is 0.107 M

Explanation:

The neutralization reaction is:

$Na_2B_4O_7.10H_2O+H_2SO_4(aq)\rightarrow Na_2SO_4+4H_3BO_3+5H_2O$

According to neutralization law:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1$ = basicity of $H_2SO_4$ = 2

$n_2$ = acidity of borax = 2

$M_1$ = concentration of $H_2SO_4$ = 1.03 M

$M_2$ = concentration of borax =?

$V_1$ = volume of $H_2SO_4$ = 2.07ml

$V_2$ = volume of borax = 20.0 ml

Now put all the given values in the above law, we get the molarity of borax:

$(2\times 1.03\times 2.07)=(2\times M_2\times 20.0)$

By solving the terms, we get :

$M_2=0.107M$

Thus the molarity of the borax solution is 0.107 M

7 0

3 years ago

A Greek philosopher who was first to use the word atoms.

12345 [234]

Answer:

I think it's Democritus

Explanation:

5 0

3 years ago

Weather balloon is filled with 62.3 L of helium gas at 760 mm of Hg after being released the volume increases to 130 L as the ba

enot [183]

Answer : The pressure at this new altitude is, 364.2 mmHg

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

$P\propto \frac{1}{V}$