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2 years ago

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A 55-g sample of a gaseous fuel mixture contains 0.43 mole fraction propane; the remainder of the mixture is butane. What are th

e masses of propane and butane in the sample?

1 answer:

Andru [333]2 years ago

7 0

I know if you swallow gasoline you will probably die.

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If a particular ore contains 55.4 % calcium phosphate, what minimum mass of the ore must be processed to obtain 1.00 kg of phosp

gizmo_the_mogwai [7]

The ore contains 55.4% calcium phosphate (related to the mineral apatite) so the amount of Ca3(PO4)2 is 55.4%x=1000g so x=1000/0.554= 1.805kg. Now for the % of P in this amount of calcium phosphate, use all the masses of the elements in Ca3PO4= Ca=40.078 x 3= 120.23 and (PO4)2= (30.974+64)2=189.95 (NB oxygen is 16 mass x 4 =64) so the total mass is 310.2 and we have 61.95 of P (Pmass x 2) so 61.95/3102.= 0.19 or 19% P. So of the 1.805 x 0.19= 0.34kg of phosphorus.

8 0

2 years ago

Which is the correct Lewis structure for fluorine, which is a group 7A element?

Yuliya22 [10]

The correct Lewis structure for Fluorine is A.

4 0

2 years ago

Read 2 more answers

A sample consisting of 1.0 mol of perfect gas molecules with CV = 20.8 J K−1 is initially at 4.25 atm and 300 K. It undergoes re

Marat540 [252]

Answer : The value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

Explanation :

First we have to calculate the value of $\gamma$ .

$\gamma=\frac{C_p}{C_v}$

As, $C_p=R+C_v$

So, $\gamma=\frac{R+C_v}{C_v}$

Given :

$C_v=20.8J/K\\\\R=8.314J/K$

$\gamma=\frac{8.314+20.8}{20.8}=1.4$

Now we have to calculate the initial volume of gas.

Formula used :

$P_1V_1=nRT_1$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$V_1$ = initial volume of gas = ?

$T_1$ = initial temperature of gas = 300 K

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

$(4.25atm)\times V_1=(1.0mol)\times (0.0821L.atm/mol.K)\times (300K)$

$V_1=5.80L$

Now we have to calculate the final volume of gas by using reversible adiabatic expansion.

$P_1V_1^{\gamma}=P_2V_2^{\gamma}$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$P_2$ = final pressure of gas = 2.50 atm

$V_1$ = initial volume of gas = 5.80 L

$V_2$ = final volume of gas = ?

$\gamma$ = 1.4

Now put all the given values in above formula, we get:

$(4.25atm)\times (5.80L)^{1.4}=(2.50atm)\times V_2^{1.4}$

$V_2=8.47L$

Now we have to calculate the final temperature of gas.

Formula used :

$P_2V_2=nRT_2$

where,

$P_2$ = final pressure of gas = 2.50 atm

$V_2$ = final volume of gas = 8.47 L

$T_2$ = final temperature of gas = ?

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in above formula, we get:

$(2.50atm)\times (8.47L)=(1.0mol)\times (0.0821L.atm/mol.K)\times T_2$

$T_2=257.9K\approx 258K$

Now we have to calculate the work done.

$w=nC_v(T_2-T_1)$

where,

w = work done = ?

n = number of moles of gas =1.0 mol

$T_1$ = initial temperature of gas = 300 K

$T_2$ = final temperature of gas = 258 K

$C_v=20.8J/K$

Now put all the given values in above formula, we get:

$w=(1.0mol)\times (20.8J/K)\times (258-300)K$

$w=-873.6J$

Therefore, the value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

8 0

2 years ago

For which of the following properties does sodium have a larger value than rubidium? Select all that apply.

Doss [256]

Answer:

Ionization energy

Electronegativity

Explanation:

-due to its smaller ionic radius....the electron in the outter most shell tends to expierence a stronger nuclear attraction...which makes it harder to remove the electron from the sodium atom

-Rubidium has lesser ionization energy because its (i) affected by its larger ionic radius which tends to lessen the nuclear attraction ...hence making it easier to remove the electron...(ii)and also by the screening effect done by the inner shells, which also tends to lessen the nuclear attraction.

Sodium has a higher electronegativity than rubidium;

Electronegativity is the charge density of electrons in an atom...in which its high when the atomic radius is smaller...

So hence due to the sodium atomic radius being smaller...it tends to have a higher charge density than rubidium....which then gives it a higher electronegativity value

4 0

2 years ago

A buffer is a solution that is a mixture of either a weak acid and its conjugate base or a weak base and its conjugate acid. Whe

babymother [125]

Answer:

b. The weak base of an alkaline buffer will accept hydrogen protons when a strong acid is added to the solution

d.The conjugate acid of an alkaline buffer will donate hydrogen protons when a strong base is added to the solution.

Explanation:

A buffer is a solution that resist pH change, it shows minimal change upon addition of small amount of strong acid or strong base. An alkaline buffer will have a basic pH, above 7. It is made by mixing a weak base and its salt with a strong acid. An example of an alkaline buffer is carbonate-bicarbonate buffer that is prepared using varying amount of anhydrous sodium carbonate and volume of solution of sodium bicarbonate to get pH range between 9.2 to 10.7

Within the buffer,the salt is completely ionized while the weak base is partly ionized. on addition of acid, the released protons will be removed by the bicarbonate ion to form sodium carbonate; on addition of base, the hydroxide ion released by the base will be removed by the hydrogen ions to form water and the pH remains relatively the same

6 0

2 years ago