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2 years ago

150.0 mL of a gas is collected at 50 kPa, What will the volume be at 125kpa?

Chemistry

1 answer:

slava [35]2 years ago

4 0

Answer:

C. 60 mL

Explanation:

Boyle's Law P₁V₁ = P₂V₂

(50kPa)(150.0mL) = (125kPa)(V₂)

V₂ = 7500/125

V₂ = 60 mL

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Most plants and all animals depend on _____ to meet their need for nitrogen.

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They depend on nitrogen-fixing bacteria, which convert atmospheric nitrogen into a usable form.

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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 acid dissociation constant $pK_a$ for acetic acid $\text{HAc}$ ; however, calculating the hydrolysis equilibrium taking place in this basic mixture requires the base 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 $pK_b$ :

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

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

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

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

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3 years ago

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Which of the following would release the most heat? Assume the same mass of in each case. Specific heats of ice, liquid water, a

lesya692 [45]

Answer:

The process which releases most heat is E)

Explanation:

As we know that water freezes at 0ºC and vaporizes at 100ºC, we calculate the heat as follows:

Processes with temperatures < 0ºC : by using specific heat of ice (Sh ice) multiplied by the change in temperature (ΔT= Final Temperature - Initial Temperature)⇒ Sh ice x ΔT

Processes of ice melting (at 0ºC): by using heat of fusion of ice (ΔH fus) multiplied by a conversor factor (1 mol H20= 18 g)⇒ ΔHfus x 1mol/18g

Processes between 0ºC and 100ºC: by using specific heat of liquid water (Sh liq) multiplied by change in temperature ⇒ Sh liq x ΔT

Processes of water evaporation (at 100ºC): by using heat of vaporization (ΔH vap) multiplied by the conversor factor ⇒ ΔH vap x 1mol/18 g

Processes at a temperature >100ºC: by using specific heat of water vapor (Sh vap) multiplied by the change in temperature ⇒ Sh vap x ΔT

A) Water at -25ºC is ice. Ice is heated from -25ºC to 0ºC, then it melts at 0ºC (ice became liquid water) and then liquid water is heated from 0ºC to 70ºC. T

This is the only process in with the heat is absorbed (not releases), so it cannot be the right answer, but we calculate the heat involved to practice:

Heat= (Sh ice x ΔT) + (ΔH fus x 1/18 g) + Sh liq x ΔT

Heat= (2.05 J/g ºC x (0ºC -(-25ºC) ) + (6.01 x 10³ J/mol x 1 mol/18 g) + (4.18 J/g ºC x (70ºC-0ºC)

Heat= 51.25 J + 333,8 J +292.6 J

Heat= 677.65 J (heat is absorbed)

B) Water is cooled from 13ºC to 0ºC, then it is freezed at 0ºC and then the ice is cooled from 0ºC to -2.6 ºC

Heat= (Sh liq x ΔT) + (-ΔH melt x 1/18 g) + (Sh ice x ΔT)

Heat= 4.18 J/ºC x (0ºC- 13ºC) + (-6.01 x 10³ J/mol x 1mol/18 g) + (2.05 J/ºC x (-2.5ºc - 0ºC)

Heat= -54.34 J - 333.8 J + 5.33 J

Heat= -393.47 J (heat is released)

C) Liquid water is cooled from 74ºC to 95ºC

Heat= Sh liq x ΔT

Heat= 4.18 J/ºC x (74ºC - 95ºC)

Heat = -87.78 J (heat is released)

D) Water at 140ºC is in vapor state. Vapor at 140ºC is cooled to 110ºC (still vapor).

Heat = Sh vap x ΔT

Heat= 2.01 J/ºC x (110ºC - 140ºC)

Heat= -60.3 J (heat is released)

E) Vapor at 106ºC is cooled to 100ºC, then it condenses at 100ºC (convertion from gas to liquid), and liquid water is cooled from 100ºC to 88ºC.

Heat= (Sh vap x ΔT) + (-ΔHvap x 1mol/18g) + (Sh liq x ΔT)

Heat= (2.01 J/ºC x (100ºC-106ºC)) - (40.7 x 10³ J/mol x 1mol/18 g) + (4.18 J/ºC x (88ºC -100ºC)

Heat= -2323.32 J (heat is released) THIS IS THE RIGHT ANSWER (the more negative= the more released)

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3 years ago

8. How many grams are there in 5 moles of lithium?

nikitadnepr [17]

Answer: m = n·M = 34.7 g

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