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Aloiza [94]
3 years ago
14

a chemist dissolves 0.564 moles of manganese (IV) oxide (MnO2) in water, and adds enough water to make 0.510 L of solution. Calc

ulate the molarity of the solution.
Chemistry
1 answer:
ladessa [460]3 years ago
4 0

Answer:

The molarity of the solution is 1.1 \frac{moles}{liter}

Explanation:

Molarity is a measure of the concentration of that substance that is defined as the number of moles of solute divided by the volume of the solution.

The molarity of a solution is calculated by dividing the moles of the solute by the volume of the solution:

molarity=\frac{number of moles of solute}{volume}

Molarity is expressed in units \frac{moles}{liter}

In this case

  • number of moles of solute= 0.564 moles
  • volume= 0.510 L

Replacing:

molarity=\frac{0.564 moles}{0.510 L}

Solving:

molarity= 1.1 \frac{moles}{liter}

<u><em>The molarity of the solution is 1.1 </em></u>\frac{moles}{liter}<u><em></em></u>

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Black_prince [1.1K]

Answer:

breathing

Explanation:

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2 years ago
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If a gas at a pressure of 600 mmHg and a volume of 200 mL is changed to 780 mmHg, then what will be the new volume?
svlad2 [7]
P1V1=P2V2
(600/760)(.200)=(780/760)x
x=.1538L or 153.8mL
8 0
3 years ago
A container was found in the home of the victim that contained 130 g of ethylene glycol in 380 g of liquid. What was the percent
Kaylis [27]

Answer:

34.21%

Explanation:

Given parameters:

Mass of ethylene glycol in the container = 130g

Mass of the liquid in total = 380g

Unkown:

Percentage of ethylene glycol in the liquid =?

Solution

To find the percentage of the ethylene glycol in liquid, we use the expression below:

Percentage of ethylene glycol in liquid = \frac{mass of ethylene glycol}{mass of liquid} x 100

Percentage of ethylene glycol = \frac{130}{380} x 100

Percentage of ethylene glycol = 34.21%

7 0
3 years ago
If an element has the atomic number 20, what sub levels are present
katrin [286]
Atomic number = 20 -------> Ca (calcium)

Calcium has four shells (1, 2, 3, 4) and two sublevels ( s, p ).

1s² 2s² 2p⁶ 3s² 3p⁶ 4s²
8 0
3 years ago
The rate constants of some reactions double with every 10-degree rise in temperature. Assume that a reaction takes place at 295
LUCKY_DIMON [66]

Answer : The activation energy for the reaction is, 51.9 kJ

Explanation :

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

or,

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = rate constant at 295 K

K_2 = rate constant at 305 K = 2K_1

Ea = activation energy for the reaction = ?

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 295 K

T_2 = final temperature = 305 K

Now put all the given values in this formula, we get:

\log (\frac{2K_1}{K_1})=\frac{Ea}{2.303\times 8.314J/mole.K}[\frac{1}{295K}-\frac{1}{305K}]

Ea=51879.96J=51.9kJ

Therefore, the activation energy for the reaction is, 51.9 kJ

8 0
4 years ago
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