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3 years ago

A mixture that results when substances dissolve to form a homogeneous mixture

Chemistry

1 answer:

MrRa [10]3 years ago

3 0

A mixture that results when substances dissolve to form a homogeneous mixture is a solution.

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Consider the following intermediate chemical equations.

QveST [7]

Answer: 250 kJ

Explanation: According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to Hess’s law, the chemical equation can be treated as algebraic expressions and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

$P_4(s)+6Cl_2\rightarrow 4PCl_3$ $\Delta H_1=-2439kJ$ (1)

$4PCl_5(g)\rightarrow P_4(s)+10Cl_2(g)$ $\Delta H_2=3438kJ$ (2)

Net chemical equation:

$PCl_5(g)\rightarrow PCl_3(g)+Cl_2(g)$ $\Delta H=?$ (3)

Adding 1 and 2 we get,

$4PCl_5(g)\rightarrow 4PCl_3(g)+4Cl_2$ $\Delta H_3=\Delta H_1+\Delta H_2=-2439+3438=1000kJ$ (4)

Now dividing equation (4) by 4, we get

$PCl_5(g)\rightarrow PCl_3(g)+Cl_2$

$\Delta H=\frac{\Delta H_3}{4}=\frac{1000kJ}{4}=250kJ$ (4)

8 0

2 years ago

Air at 25°C with a dew point of 10 °C enters a humidifier. The air leaving the unit has an absolute humidity of 0.07 kg of water

madreJ [45]

Explanation:

The given data is as follows.

Temperature of dry bulb of air = $25^{o}C$

Dew point = $10^{o}C$ = (10 + 273) K = 283 K

At the dew point temperature, the first drop of water condenses out of air and then,

Partial pressure of water vapor $(P_{a})$ = vapor pressure of water at a given temperature $(P^{s}_{a})$

Using Antoine's equation we get the following.

$ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{T(K) - 46.854}$

$ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{283 - 46.854}$

= 0.17079

$P^{s}_{a}$ = 1.18624 kPa

As total pressure $(P_{t})$ = atmospheric pressure = 760 mm Hg

= 101..325 kPa

The absolute humidity of inlet air = $\frac{P^{s}_{a}}{P_{t} - P^{s}_{a}} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}$

$\frac{1.18624}{101.325 - 1.18624} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}$

= 0.00735 kg $H_{2}O$ / kg dry air

Hence, air leaving the humidifier has a has an absolute humidity (%) of 0.07 kg $H_{2}O$ / kg dry air.

Therefore, amount of water evaporated for every 1 kg dry air entering the humidifier is as follows.

0.07 kg - 0.00735 kg

= 0.06265 kg $H_{2}O$ for every 1 kg dry air

Hence, calculate the amount of water evaporated for every 100 kg of dry air as follows.

$0.06265 kg \times 100$

= 6.265 kg

Thus, we can conclude that kg of water the must be evaporated into the air for every 100 kg of dry air entering the unit is 6.265 kg.

3 0

3 years ago

Help me with this work plz plz

storchak [24]

Answer:

a -4

b - 8

c - 5

d 2

Explanation:

Significant Figures: The number of digits used to express a measured or calculated quantity. By using significant figures, we can show how precise a number is. ... Accuracy: Refers to how closely individual measurements agree with the correct or true value.

7 0

2 years ago

David found that water can be created in a lab by burning hydrogen gas in air. He concluded that water is not a compound because

Virty [35]

Hi~~

H2O, commonly known as water, is definitely a compound.

David's conclusion is wrong because the hydrogen gas combined with oxygen from the air to form dihydrogen monoxide, also known as water.

7 0

3 years ago

Credible sources always identify the author. Please select the best answer from the choices provided T F

Lostsunrise [7]

This is a true statement because the author can then be further looked into and researched