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3 years ago

What is the new boiling point if 25 g of NaCl is dissolved in 1.0 kg of water

1 answer:

8 0

The correct answer for the question that is being presented above is this one:

Given that:
delta Tb = Kbm Kb H2O = 0.52 degrees C/m
delta Tf = Kfm Kf H2O = 1.86 degrees C/m

We need to know the formula for Molality.
molality = mol solute / kg solvent

We are given the amount of solute in grams
Since amount of solute is given in moles, we have to convert 25 g NaCl to moles. Divide by molar mass. 

25 g NaCl / 58.5 g/mol = 0.427 mol 

Then, use the formula for molality. 

molality = mol solute / kg solvent 
= 0.427 / 1 
= 0.427 m 

Use now the formula to get the boiling point.

delta Tb = Kbm 
= (0.52)(0.427) 
= 0.22C

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An inverted pyramid is being filled with water at a constant rate of 45 cubic centimeters per second. The pyramid, at the top, h

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Answer:

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We have:

Rate at which water is filled in cube = $\frac{dV}{dt}= 45 cm^3/s$

Square based pyramid:

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Volume of the square based pyramid = V

$V=\frac{1}{3}\times l^2\times h$

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$l=\frac{6h}{13}$

$V=\frac{1}{3}\times (\frac{6h}{13})^2\times h$

$V=\frac{12}{169}h^3$

Differentiating V with respect to dt:

$\frac{dV}{dt}=\frac{d(\frac{12}{169}h^3)}{dt}$

$\frac{dV}{dt}=3\times \frac{12}{169}h^2\times \frac{dh}{dt}$

$45 cm^3/s=3\times \frac{12}{169}h^2\times \frac{dh}{dt}$

$\frac{dh}{dt}=\frac{45 cm^3/s\times 169}{3\times 12\times h^2}$

Putting, h = 4 cm

$\frac{dh}{dt}=\frac{45 cm^3/s\times 169}{3\times 12\times (4 cm)^2}$

$=13.20 cm/s$

13.20 cm/s is the rate at which the water level is rising when the water level is 4 cm.

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Hence we add 2 as a coefficient of HCl to balance the equation.

To know more about balance equation, visit the below link:
brainly.com/question/26694427

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