All categories

2 years ago

What is the partial pressure of 0.50 mol Ne gas combined with 1.20 mol Kr gas at a final pressure of 730 torr?

2 answers:

8 0

The partial pressure of 0.50 Ne gas is 214.71 torr

calculation

the partial pressure of Ne = moles of Ne/total moles x final pressure

find the total moles of the air mixture

that is moles of Ne + moles of K= 0.50 + 1.20 = 1.70 moles

The partial pressure is therefore = 0.50 /1.70 x 730 = 214.71 torr

Scorpion4ik [409]2 years ago

3 0

it would be 215 torr for the answer for edg

You might be interested in

PLEASE HELP: You have been given 0.507 moles of cesium fluoride (CsF), determine the mass in grams of cesium fluoride that you h

Degger [83]

Answer:

C because 77.0 CsF

Explanation:

That is the correct answer because I have that homework and i got it right

5 0

2 years ago

How do you calculate mass using density and volume?

skad [1K]

Example:

Mass = ?

Density = 25 g/mL

Volume = 5 mL

therefore:

d = m / V

25 = m / 5

m = 25 x 5

m = 125 g

hope this helps!

7 0

3 years ago

Suppose you have just added 100 ml of a solution containing 0.5 mol of acetic acid per liter to 400 ml of 0.5 m naoh. what is th

Tpy6a [65]

$pH = 13.5$

Explanation:

Sodium hydroxide completely ionizes in water to produce sodium ions and hydroxide ions. Hydroxide ions are in excess and neutralize all acetic acid added by the following ionic equation:

$\text{HAc} + \text{OH}^{-} \to \text{Ac}^{-} + \text{H}_2\text{O}$

The mixture would contain

$0.4 \times 0.5 - 0.1 \times 0.5 = 0.15 \; \text{mol}$ of $\text{OH}^{-}$ and
$0.1 \times 0.5 = 0.05 \; \text{mol}$ of $\text{Ac}^{-}$

if $\text{Ac}^{-}$ undergoes no hydrolysis; the solution is of volume $0.1 + 0.4 = 0.5 \; \text{L}$ after the mixing. The two species would thus be of concentration $0.30 \; \text{mol} \cdot \text{L}^{-1}$ and $0.10 \; \text{mol} \cdot \text{L}^{-1}$ , respectively.

Construct a RICE table for the hydrolysis of $\text{Ac}^{-}$ under a basic aqueous environment (with a negligible hydronium concentration.)

$\begin{array}{cccccccc} \text{R} & \text{Ac}^{-}(aq) &+ & \text{H}_2\text{O}(aq) & \leftrightharpoons & \text{HAc}(aq) & + & \text{OH}^{-} (aq)\\ \text{I} & 0.10 \; \text{M} & & & & & &0.30 \; \text{M}\\ \text{C} & -x \; \text{M}& & & & +x \; \text{M}& & +x \; \text{M} \\ \text{E} & (0.10 - x) \; \text{M} & & & & x \; \text{M} & & (0.30 +x) \; \text{M} \end{array}$

The question supplied the <em>acid</em> dissociation constant $pK_a$ for acetic acid $\text{HAc}$ ; however, calculating the hydrolysis equilibrium taking place in this basic mixture requires the <em>base</em> dissociation constant $pK_b$ for its conjugate base, $\text{Ac}^{-}$ . The following relationship relates the two quantities:

$pK_{b} (\text{Ac}^{-}) = pK_{w} - pK_{a}( \text{HAc})$

... where the water self-ionization constant $pK_w \approx 14$ under standard conditions. Thus $pK_{b} (\text{Ac}^{-}) = 14 - 4.7 = 9.3$ . By the definition of $pK_b$ :

$[\text{HAc} (aq)] \cdot [\text{OH}^{-} (aq)] / [\text{Ac}^{-} (aq) ] = K_b = 10^{-pK_{b}}$

$x \cdot (0.3 + x) / (0.1 - x) = 10^{-9.3}$

$x = 1.67 \times 10^{-10} \; \text{M} \approx 0 \; \text{M}$

$[\text{OH}^{-}] = 0.30 +x \approx 0.30 \; \text{M}$

$pH = pK_{w} - pOH = 14 + \text{log}_{10}[\text{OH}^{-}] = 14 + \text{log}_{10}{0.30} = 13.5$

6 0

3 years ago

2. What do acidic solutions have high concentrations of?

Xelga [282]

Hydrogen ions
mark me brainlist

7 0

2 years ago

Read 2 more answers

What would happen to a diver who does no exhale while surfacing from a 30 m dive?

sertanlavr [38]

Their lungs would try to expand to about 4 timed the normal volume which would force air into the various body tissues. this can cause a lung expansion injury and it could case air embolism. Air embolism is when air bubbles get trapped in blood vessels. This can lead to a blockage which will could be fatal.

6 0

3 years ago