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3 years ago

Can some1 please answer these questions

Chemistry

1 answer:

Yakvenalex [24]3 years ago

4 0

Answer:

the answer of your question is yes

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If 125 ml of o2 gas exerts a pressure of 1.0 atm inside a cylinder, what will the pressure be if the cylinder compressed the vol

Lelu [443]

<h2>Hello!</h2>

The answer is: The new pressure is 1.67 atm.

From the statement, we know that the temperature remains constant and the gas volume is changing, meaning that the new pressure will be different than the first pressure.

Since the temperature remains constant, we can calculate the new pressure using the Boyle's Law.

The Boyle's Law states that:

$P_{1}V_{1}=P_{2}V_{2}$

Where,

P is the pressure of the gas.

V is the volume of the gas.

Then, the given information is:

$V_{1}=125ml=0.125L\\P_{1}=1atm\\V_{2}=75ml=0.075L$

Remember, 1 L is equal to 1000 mL.

So,

$125mL*\frac{1L}{1000mL}=\frac{125mL*1L}{1000mL}=0.125L\\\\75mL*\frac{1L}{1000mL}=\frac{75mL*1L}{1000mL}=0.075L$

So, calculating the new volume, we have:

$P_{1}V_{1}=P_{2}V_{2}\\\\1atm*0.125L=P_{2}*0.075L\\\\P_{2}=\frac{1atm*0.125L}{0.075l}=1.67atm$

Hence, the new pressure is 1.67 atm.

Have a nice day!

7 0

3 years ago

6 State Your Claim Make a claim about how you<br> could separate a mixture of rocks and sand.

Sloan [31]

Answer: Sand can be separated from the sand by the process of Sieving. sorry if I'm wrong please give me brainlest

Explanation:

4 0

2 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 $pK_b$ :