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3 years ago

Boiling point of a solution formed when 15.2 grams of CaCl2 dissolves in 57.0 g of water. kB= 0.512 c/m.

1 answer:

Nookie1986 [14]3 years ago

4 0

100.133 degree celsius is the boiling point of the solution formed when 15.2 grams of CaCl2 dissolves in 57.0 g of water.

Explanation:

Balanced eaquation for the reaction

CaCl2 + 2H20 ⇒ Ca(OH)2 + HCl

given:

mass of CaCl2 = 15.2 grams

mass of the solution = 57 grams

Kb (molal elevation constant) = 0.512 c/m

i = vont hoff factor is 1 as 1 mole of the substance is given as product.

Molality is calculated as:

molality = $\frac{grams of solute}{grams of solution}$

= $\frac{15.2}{57}$

= 0.26 M

Boiling point is calculated as:

ΔT = i x Kb x M

= 1 x 0.512 x 0.26

= 0.133 degrees

The boiling point of the solution will be:

100 degrees + 0.133 degrees (100 degrees is the boiling point of water)

= 100.133 degree celcius is the boiling point of mixture formed.

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Explanation:

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3 years ago

You are given a solution of sugar and sand and water your task is to separate sugar from sand which of the following experimenta

Ostrovityanka [42]

Answer:

Explanation:

Filtration followed by evaporation:

To separate the mixture of sand and sugar, it is best to use the separation technique of filtration then evaporation.

Pour the water into the mixture. The sugar will dissolve with time in the water. Sand is made up of quartz and does not dissolve in water.

After the dissolution, filter the solution to separate the sand using a filter paper.

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3 years ago

Cans of regular soda have volumes with a mean of 14.5 oz and a standard deviation of 0.18 oz. What is the maximum "usual" value

tangare [24]

Answer:

14.68 oz

Explanation:

The mean volume is the average volume of all the regular soda cans put together; 14.5 oz. The standard deviation 0.18 oz is the maximum deviation from the mean. It can be 14.5-0.18= 14.32 oz which is the usual minimum or 14.5 + 0.18=14.68 oz which is the usual maximum

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3 years ago

As an athlete exercises, sweat is produced and evaporated to help maintain a proper body temperature. On average, an athlete los

dmitriy555 [2]

<u>Answer:</u> For the given amount of sweat lost, the amount of energy required will be 692,899 Joules.

<u>Explanation:</u>

We are given:

Heat of vaporization for water = 2257 J/g

Amount of sweat lost = 307 grams

Applying unitary method:

For 1 g of sweat lost, the energy required is 2257 Joules

So, for 307 grams of sweat lost, the energy required will be = $\frac{2257J}{1g}\times 307g=692,899J$

Hence, for the given amount of sweat lost, the amount of energy required will be 692,899 Joules.

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3 years ago

How many kJ to hear 0.700 pint of room temp water to boiling

irina1246 [14]

Use the equation Q = mCT.

Q is the amount of energy in joules.

m is the mass in grams.

C is the specific heat of water (4.184 J/g °C).

T is the change in temperature.

0.700 pints = 331.224 g of water

Q = (331.224 g)(4.184 J/g °C)(100° C - 30° C)

Q = 97,008.885 J

Divide by 1,000 to convert to kilojoules:

97.009 kJ

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3 years ago