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3 years ago

Calculate the heat released when 454 grams of steam at 100 degrees C condenses.

Chemistry

1 answer:

sasho [114]3 years ago

5 0

<h3>Answer:</h3>

1026.05 kJ

<h3>Explanation:</h3>

The Quantity of heat is calculated by multiplying the mass of a substance by specific heat capacity and the change in temperature.
However, condensation takes place at a constant temperature.
Therefore, Quantity of heat is calculated by multiplying mass by the latent heat of vaporization.

That is; Q = m × Lv

In this case we are given;

Mass of steam as 454 g

To calculate the amount of heat released we must know the value of latent heat of vaporization;

The latent heat of vaporization of steam is 2260 J/g

Therefore;

Heat released, Q = 454 g × 2260 J/g

=1026040 Joules

But, 1000 joules = 1 kilo-joules

Thus, Q = 1026.04 kJ

Hence, the amount of heat released is 1026.05 kJ

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Read 2 more answers

A 1.10 g sample contains only glucose and sucrose. When the sample is dissolved in water to a total solution volume of 25.0L, th

olga_2 [115]

Answer:

$\large \boxed{79 \, \%}$

Explanation:

I assume the volume is 2.50 L. A volume of 25.0 L gives an impossible answer.

We have two conditions:

(1) Mass of glucose + mass of sucrose = 1.10 g

(2) Osmotic pressure of glucose + osmotic pressure of sucrose = 3.78 atm

Let g = mass of glucose

and s = mass of sucrose. Then

g/180.16 = moles of glucose, and

s/342.30 = moles of sucrose. Also,

g/(180.16×2.50) = g/450.4 = molar concentration of glucose. and

s/(342.30×2.50) = s/855.8 = molar concentration of sucrose.

1. Set up the osmotic pressure condition

Π = cRT, so

$\begin{array}{rcl}\Pi_{\text{g}} +\Pi_{\text{s}}&=&\Pi_{\text{tot}}\\\dfrac{g}{450.4}\times8.314\times298 + \dfrac{s}{855.8}\times8.314\times298 & = & 3.78\\\\5.501g + 2.895s & = & 3.78\\\end{array}$

Now we can write the two simultaneous equations and solve for the masses.

2. Calculate the masses

$\begin{array}{lrcl}(1)& g + m & = & 1.10\\(2) &5.501g +2.895s & = & 3.78\\(3) & m & = &1.10 - g\\&5.501g + 2.895(1.10 - g) & = & 3.78\\&2.606g + 3.185 & = & 3.78\\ &2.606g & = & 0.595\\(4) & g & = & \mathbf{0.229}\\&0.229 + s & = & 1.10\\& s & = & \mathbf{0.871}\\\end{array}$

We have 0.229 g of glucose and 0.871 g of sucrose.

3. Calculate the mass percent of sucrose

$\text{Mass percent} = \dfrac{\text{Mass of component}}{\text{Total mass}} \times \, 100\%\\\\\text{Percent sucrose} = \dfrac{\text{0.871 g}}{\text{1.10 g}} \times \, 100\% = 79 \, \%\\\\\text{The mixture is $\large \boxed{\mathbf{79 \, \%}}$ sucrose}$

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3 years ago