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3 years ago

Tulle

Chemistry

1 answer:

Andre45 [30]3 years ago

5 0

Answer:

RH = 20%

Explanation:

Data given:

water pressure at 0°C = 4.85 g/m³

Actual water vapor density = 0.97 g/m³

Relative humidity (RH) = ?

Solution:

Formula used

Relative humidity (RH) = $\frac{actual vapor density}{saturated vapor density of water}$ x 100%

As we know that at 0°C water vapor have 4.85 g/m³ saturated vapor density.

So,

put values in the above formula

Relative humidity (RH) = 0.97 g/m³ / 4.85 g/m³ x 100%

Relative humidity (RH) = 0.2 x 100%

Relative humidity (RH) = 20%

So,

Relative humidity (RH) = 20%

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A gaseous fuel mixture stored at 747 mmHg and 298 K contains only methane (CH4) and propane (C3H8). When 11.1 L of this fuel mix

Alisiya [41]

Answer:

$M_f=38.8\%$

Explanation:

From the question we are told that:

Pressure $P=747mmHg$

Temperature $T=298K$

Volume $V=11.1$

Heat Produced $Q=780kJ$

Generally the equation for ideal gas is mathematically given by

$PV=nRT$

$n= (747/760) *11.1/ (0.0821*298)$

$n=0.446mol$

Therefore

$x+y=0.446$

$x=0.446-y .....1$

Since

Heat of combustion of Methane=889 kJ/mol

Heat of combustion of Propane=2220 kJ/mol

Therefore

$x(889) + y(2220) = 760 ...... 2$

Comparing Equation 1 and 2 and solving simultaneously

$x=0.446-y .....1$

$x(889) + y(2220) = 760 ...... 2$

$x=0.173$

$y=0.273$

Therefore

Mole fraction 0f Methane is mathematically given as

$M_f=\frac{x}{n}*100\%$

$M_f=\frac{1.173}{0.446}*100\%$

$M_f=38.8\%$

7 0

3 years ago

3. How much energy is needed to raise 45 grams of water from 40°C to 115 °C?

Dafna1 [17]

Answer:

Q = 114349.5 J

Explanation:

Hello there!

In this case, since this a problem in which we need to calculate the total heat of the described process, it turns out convenient to calculate it in three steps; the first one, associated to the heating of the liquid water from 40 °C to 100 °C, next the vaporization of liquid water to steam at constant 100 °C and finally the heating of steam from 100 °C to 115 °C. In such a way, we calculate each heat as shown below:

$Q_1=45g*4.18\frac{J}{g\°C}*(100\°C-40\°C)=11286J\\\\Q_2=45g* 2260 \frac{J}{g} =101700J\\\\Q_3=45*2.02\frac{J}{g\°C}*(115\°C-100\°C)=1363.5J$

Thus, the total energy turns out to be:

$Q_T=11286J+101700J+1363.5J\\\\Q_T=114349.5J$

Best regards!

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3 years ago

Which statement best explains what would happen if a reactant were added to a system in equilibrium?

expeople1 [14]

system will shift the equilibrium to the right, towards the side of the products

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2 years ago

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Pure chlorobenzene (C6H5Cl) has a normal boiling point of 131.00 °C. A solution of 32.5 g of 2,8-dibromodibenzofuran (C12H6Br2O)

vichka [17]

Answer:

Kb → 1.56 °C / m

Explanation:

This is all about boiling point elevation, the colligative property that shows that boiling point for a solution is higher than boiling point of pure solvent.

This is the formula: ΔT = Kb . m . i

where i is the Van't Hoff factor (ions dissolved in solution). As these are organic compounds, we assume they are non electrolytic,

m is molality (mol of solute / 1kg of solvent)

Kb is our unknown. The value for ebulloscopic constant, it is specific for each solvent.

ΔT = T° boiling from solution - T° boiling from solute

First of all, let's determine the moles of solute.

Mass / Molar mass → 32.5 g/ 113.45 g/mol = 0.286 mol

Molality is mol of solute/ 1 kg of solvent

We must convert the mass from g to kg

195g . 1kg /1000 = 0.195 kg

Molality = 0.286 mol / 0.195 kg = 1.47 m

Let's replace the values in the formula

133.30 °C - 131°C = Kb . 1.47m .1

2.30°C / 1.47 m = Kb → 1.56 °C / m

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