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2 years ago

Solve for b if R = x(A + B)

Chemistry

1 answer:

Tatiana [17]2 years ago

7 0

This is what I got. Hope it helps :)

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Answer:

ebcuase

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2 years ago

To titrate 45.00 milliliters of an unknown HCl sample, use 25.00 milliliters of a standard 0.2000 M NaOH titrant. What is the mo

gtnhenbr [62]

V ( HCl ) = 45.00 mL in liters : 45.00 / 1000 => 0.045 L

M ( HCl ) = ?

V ( NaOH ) = 25.00 / 1000 => 0.025 L

M ( NaOH) = 0.2000 M

number of moles NaOH :

n = M x V = 0.2000 x 0.025 => 0.005 moles of NaOH

Mole ratio:

HCl + NaOH = NaCl + H2O

1 mole HCl ---------- 1 mole NaOH
? mole HCl ---------- 0.005 moles NaOH

moles HCl = 0.005 x 1 / 1

= 0.005 moles of HCl :

M ( HCl ) = n / V

M ( HCl ) = 0.005 / 0.045

= 0.1111 M

hope this helps!

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3 years ago

In the front of the room, there is a bottle that contains a 32.00 g sample of sulfur. This is

Vera_Pavlovna [14]

6.02x10^23 atoms is the answer

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2 years ago

Use the big bang theory to explain why scientists do not expect to find galaxies that have large blue shifts.

STALIN [3.7K]

Answer:

becoz they want so

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2 years ago

Argon (Ar) and helium (He) are initially in separate compartments of a container at 25°C. The

love history [14]

Answer:

(a) $V_B=11.68L$

(b) $x_{He}=0.533$

Explanation:

Hello,

In this case, since the both gases behave ideally, with the given information we can compute the moles of He in A:

$n_A=\frac{0.082\frac{atm*L}{mol*K}*298K}{1.974 atm*6.00L}=2.063mol$

Thus, since the final pressure is 3.60 bar, we can write:

$P=x_{Ar}P_A+x_{He}P_B\\\\P=\frac{n_{Ar}}{n_{Ar}+n_{He}} P_A+\frac{n_{He}}{n_{Ar}+n_{He}} P_B\\\\3.60bar=\frac{2.063mol}{2.063mol+n_{He}} *2.00bar+\frac{n_{He}}{2.063mol+n_{He}} *5.00bar$

The moles of helium could be computed via solver as:

$n_{He}=2.358mol$

Or algebraically:

$3.60bar=\frac{1}{2.063mol+n_{He}} *(4.0126+5.00*n_{He})\\\\7.314+3.60n_{He}=4.013+5.00*n_{He}\\\\7.314-4.013=5.00*n_{He}-3.60n_{He}\\\\n_{He}=\frac{3.3}{1.4}=2.358mol$

In such a way, the volume of the compartment B is:

$V_B=\frac{n_{He}RT}{P_B}=\frac{2.358mol*0.082\frac{atm*L}{mol*K}*298.15K}{4.935atm}\\ \\V_B=11.68L$

Finally, he mole fraction of He is:

$x_{He}=\frac{2.358}{2.358+2.063}\\ \\x_{He}=0.533$

Regards.

8 0

2 years ago