All categories

3 years ago

If an unknown quantity of gas is held at a temperature of 789 K in a container with a volume of

Chemistry

1 answer:

Orlov [11]3 years ago

8 0

Answer:

n = 0.34 mol

Explanation:

Given data:

Given temperature = 789 K

Pressure = 765 torr ( 765/760 = 1.0 atm

Volume of container = 22.0 L

Number of moles of gas = ?

Solution:

Formula:

PV = nRT

n = PV/RT

n = 1.0 atm × 22.0 L / 0.0821 atm.L/mol.K ×789 K

n = 22.0 L.atm /64.8 atm.L/mol

n = 0.34 mol

You might be interested in

Melted rock can ooze out from below earths surface through a crack in the crust called a(n)?

Katen [24]

Melted rock can ooze out from below the earth's surface through a crack called a fault.

5 0

4 years ago

25 g of a compound is added to 500 mL of water if the freezing point of the resulting solution is

vlabodo [156]

Answer:

a) 119 g/mol

Explanation:

-We apply the formula for freezing point depression to obtain the molality of the solution:

$\bigtriangleup T_f=K_fm, \ \ K_f=1.36\textdegree C/m\\\\\therefore m=\frac{\bigtriangleup T_f}{K_f}\\\\=\frac{0.57\textdegree C}{1.36\textdegree C}\\\\=0.4191\ mol/Kg\\\\$

#We use the molality above to calculate the molar mass:

$m=\frac{0.4191\ mol}{1\ Kg}=\frac{25\ g}{0.5\ Kg}\\\\\therefore 1 \ mol=\frac{25\ g}{0.5\ Kg}\times\frac{1\ Kg}{ 0.4191}\\\\=119.3033\approx 119\ g/mol$

Hence, the molar mass of the compound is 119 g/mol

5 0

4 years ago

7.<br> How many grams are contained in 3.9 x 1023 sulfur atoms?<br> atoms → moles<br> grams<br> I

Anni [7]

Answer: 20.775 g S

Explanation: 3.9x10^23 atoms = 0.648 mol

Atomic mass S = 32.08

S in grams = 20.775

4 0

3 years ago

What does air have to do with light refraction?

-BARSIC- [3]

When light travels from air into water,it slows down,causing it to change directions slightly

4 0

3 years ago

Read 2 more answers

If acetic acid is the only acid that vinegar contains (ka=1.8×10−5), calculate the concentration of acetic acid in the vinegar.

kicyunya [14]

Ethanoic (Acetic) acid is a weak acid and do not dissociate fully. Therefore its equilibrium state has to be considered here.

$CH_{3}COOH \ \textless \ ---\ \textgreater \ H^{+} + CH_{3}COO^{-}$

In this case pH value of the solution is necessary to calculate the concentration but it's not given here so pH = 2.88 (looked it up)

pH = 2.88 ==> $[H^{+}]$ = $10^{-2.88}$ = 0.001 $moldm^{-3}$

The change in Concentration Δ $[CH_{3}COOH]$ = 0.001 $moldm^{-3}$

  CH3COOH H+ CH3COOH
Initial   $moldm^{-3}$      $x$ 0 0

Change $moldm^{-3}$ -0.001 +0.001 +0.001

Equilibrium $moldm^{-3}$ $x$ - 0.001 0.001 0.001


Since the $k_{a}$ value is so small, the assumption
$[CH_{3}COOH]_{initial} = [CH_{3}COOH]_{equilibrium}$ can be made.

$k_{a} = [tex]= 1.8*10^{-5} = \frac{[H^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH]} = \frac{0.001^{2}}{x}$

Solve for x to get the required concentration.

note: 1.)Since you need the answer in 2SF don&t round up values in the middle of the calculation like I've done here.

2.) The ICE (Initial, Change, Equilibrium) table may come in handy if you are new to problems of this kind

Hope this helps!

8 0

3 years ago