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3 years ago

In a concentrated solution there is ____.a. no solvent c. a small amount of soluteb. a large amount of solute d. no solute

Chemistry

1 answer:

Gelneren [198K]3 years ago

6 0

Answer:a large amount of solute

Explanation:

A solution is composed of a solute and a solvent. If the amount of solute is greater than that of solvent, the solution is concentrated. A concentrated solution contains quite a large amount of solute while a dilute solution contains less amount of solute. This is the difference between diluted and concentrated solutions.

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Consider a bathtub filled with water, with a plug in the drain. How long would this be considered a closed system?

Monica [59]

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3 years ago

Read 2 more answers

A sample of gas in a cylinder as in the example in Part A has an initial volume of 48.0 L , and you have determined that it cont

masya89 [10]

Answer:

0.45 moles

Explanation:

The computation of the number of moles left in the cylinder is shown below:

As we know that

$\frac{n1}{V1} = \frac{n2}{V2}$

we can say that

$n2 = n1 \times \frac{V2}{V1}$

where,

n1 = 1.80 moles of gas

V2 = 12.0 L

And, the V1 = 48.0 L

Now placing these values to the above formula

So, the moles of gas in n2 left is

$= 1.80 \times \frac{12.0\ L}{48.0\ L}$

= 0.45 moles

We simply applied the above formulas so that the n2 moles of gas could arrive

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3 years ago

Calculate the energy that is required to change 50.0 g ice at -30.0°C to a liquid at 73.0°C. The heat of fusion = 333 J/g, the h

OverLord2011 [107]

Answer:

There is 3.5*10^4 J of energy needed.

Explanation:

Step 1: Data given

Mass of ice at -30.0 °C = 50.0 grams

Final temperature = 73.0 °C

The heat of fusion = 333 J/g

the heat of vaporization = 2256 J/g

the specific heat capacity of ice = 2.06 J/gK

the specific heat capacity of liquid water = 4.184 J/gK

Step 2: Calculate the heat absorbed by ice

q = m*c*(T2-T1)

⇒ m = the mass of ice = 50.0 grams

⇒ c = the heat capacity of ice = 2.06 J/gK = 2.06 J/g°C

⇒ T2 = the fina ltemperature of ice = 0°C

⇒ T1 = the initial temperature of ice = -30.0°C

q = 50.0 * 2.06 J/g°C * 30 °C

q = 3090 J

Step 3: Calculate heat required to melt the ice at 0°C:

q = m*(heat of fusion)

q = 50.0* 333J/g

q = 16650 J

Step 4: Calculate the heat required to raise the temperature of water from 0°C to 73.0°C

q = m*c*(T2-T1)

⇒ mass = 50.0 grams

⇒ c = the specific heat of water = 4.184 J/g°C

⇒ ΔT = T2-T1 = 73.0 - 0 = 73 °C

q = 50.0 * 4.184 * 73.0 = 15271.6 J

Step 5: Calculate the total energy

qtotal = 3090 + 16650 + 15271.6 = 35011.6 J = 3.5 * 10^4 J

There is 3.5*10^4 J of energy needed.

8 0

3 years ago