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babunello [35]
3 years ago
7

Why can more sugar be dissolved in hot water than cold water (more solute be dissolved in hot solvent then cold solvent)

Chemistry
1 answer:
lesantik [10]3 years ago
4 0
Sugar dissolves faster in hot water than it does in cold water because hot water has more energy than cold water. when water is heated, the molecules gain energy and because of that, move faster. as they move faster, they come in contact with the sugar more often, causing it to dissolve faster.
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) Given the following balanced equation, determine the rate of reaction with respect to [O2]. If the rate of formation of O2is 7
Travka [436]

Answer:

Rate of the reaction is 0.2593 M/s

-0.5186 M/s is the rate of the loss of ozone.

Explanation:

The rate of the reaction is defined as change in any one of the concentration of reactant or product per unit time.

2O_3\rightleftharpoons 3O_2

Rate of formation of oxygen : 7.78\times 10^{-1} M/s

Rate of the reaction(R) =\frac{-1}{2}\frac{d[O_3]}{dt}=\frac{1}{3}\frac{d[O_2]}{dt}

R=\frac{1}{3}\frac{d[O_2]}{dt}

Rate of formation of oxygen=3 × (R)

7.78\times 10^{-1} M/s=3\times R

Rate of the reaction(R): 0.2593 M/s

Rate of the reaction is 0.2593 M/s

Rate of disappearance of the ozone:

R=-\frac{1}{2}\frac{d[O_3]}{dt}

\frac{d[O_3]}{dt}=-2\times R=-2\times 0.2593\times M/s=-0.5186M/s

-0.5186 M/s is the rate of the loss of ozone.

6 0
3 years ago
At standard temperature and pressure. 0.500 mole of xenon gas occupies
ANEK [815]

Answer:

0.500 mole of Xe (g) occupies 11.2 L at STP.

General Formulas and Concepts:

<u>Gas Laws</u>

  • STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

<u>Stoichiometry</u>

  • Mole ratio
  • Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

<em>Identify.</em>

0.500 mole Xe (g)

<u>Step 2: Convert</u>

  1. [DA] Set up:                                                                                                  \displaystyle 0.500 \ \text{mole Xe} \bigg( \frac{22.4 \ \text{L Xe}}{1 \ \text{mole Xe}} \bigg)
  2. [DA] Evaluate:                                                                                               \displaystyle 0.500 \ \text{mole Xe} \bigg( \frac{22.4 \ \text{L Xe}}{1 \ \text{mole Xe}} \bigg) = 11.2 \ \text{L Xe}

Topic: AP Chemistry

Unit: Stoichiometry

3 0
2 years ago
Nitrogen forms a surprising number of compounds with oxygen. A number of these, often given the collective symbol NOx (for "nitr
kvv77 [185]

Answer:

9.2

Explanation:

Let's do an equilibrium chart of this reaction:

2NO(g) + O₂(g) ⇄ 2NO₂(g)

4.9 atm    5.1 atm    0       Initial

-2x             -x           +2x    Reacts (stoichiometry is 2:1:2)

4.9-2x      5.1-x        2x      Equilibrium

The mole fraction of NO₂ (y) can be calculated by the Raoult's law, that states that the mole fraction is the partial pressure divided by the total pressure:

y = 2x/(4.9 - 2x + 5.1 -x + 2x)

0.52 = 2x/(10 - x)

2x = 5.2 -0.52x

2.52x = 5.2

x = 2.06 atm

Thus, the partial pressure at equilibrium are:

pNO = 4.9 -2*2.06 = 0.78 atm

pO₂ = 5.1 - 2.06 = 3.04 atm

pNO₂ = 2*2.06 = 4.12 atm

Thus, the pressure equilibrium constant Kp is:

Kp = [(pNO₂)²]/[(pNO)²*(pO₂)]

Kp = [(4.12)²]/[(0.78)²*3.04]

Kp = [16.9744]/[1.849536]

Kp = 9.2

4 0
3 years ago
Select the situation below that would produce a total displacement of zero
likoan [24]

Answer:

uhhhhhjjj??????????

6 0
3 years ago
Read 2 more answers
23.5 L of h2 is stored at a pressure of 58.7 Kpa what volume would the gas take up at stp
stepan [7]

Answer:-  13.6 L

Solution:- Volume of hydrogen gas at 58.7 Kpa is given as 23.5 L. It asks to calculate the volume of hydrogen gas at STP that is standard temperature and pressure. Since the problem does not talk about the original temperature so we would assume the constant temperature. So, it is Boyle's law.

Standard pressure is 1 atm that is 101.325 Kpa.

Boyle's law equation is:

P_1V_1=P_2V_2

From given information:-

P_1 = 58.7 Kpa

V_1 = 23.5 L

P_2 = 101.325 Kpa

V_2 = ?

Let's plug in the values and solve it for final volume.

58.7Kpa*23.5L=101.325Kpa*V_2

On rearranging the equation for V_2

V_2=\frac{58.7Kpa*23.5L}{101.325Kpa}

V_2 = 13.6 L

So, the volume of hydrogen gas at STP for the given information is 13.6 L.

3 0
3 years ago
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