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3 years ago

. Which of the following is NOT a common state of matter under room temperature conditions?

Chemistry

2 answers:

Musya8 [376]3 years ago

6 0

Answer:

plasma

Explanation:

aka firey substance

storchak [24]3 years ago

4 0

Answer: plasma

Explanation:

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What is CdF^4 compound name

topjm [15]

Answer:

well it might be 1.)Cadmium Fluoride CdF2 150.4078

2.)Cadmium(II) Perfluorate Cd(FO4) 2278.403

3.)Cadmium Ferrocyanide Cd2Fe(CN) 6436.7714

Explanation:

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2 years ago

Which of the following compounds contains the tin (IV) ion? A. SnO2 B. SnCl2 C. Sn2O D. Sn4O3

Zolol [24]

A. SnO2 is the correct answer. I hope this helps

3 0

2 years ago

Read 2 more answers

Select the correct answer.

Nata [24]

C. Ninhydrin is used to detect prints on a non-porous surface

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2 years ago

Find the standard enthalpy of formation of ethylene, C2H4(g), given the following data: C2H4(g) + 3 O2(g) --> 2CO2(g) + 2 H2O

Brums [2.3K]

Answer : The standard enthalpy of formation of ethylene is, 52.4 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 this law, the chemical equation can be treated as ordinary algebraic expression 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.

The formation reaction of $C_2H_4$ will be,

$2C(s)+2H_2(g)\rightarrow C_2H_4(g)$ $\Delta H_{formation}=?$

The intermediate balanced chemical reaction will be,

(1) $C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)$ $\Delta H_1=-1411kJ$

(2) $C(s)+O_2(g)\rightarrow CO_2(g)$ $\Delta H_2=-393.5kJ$

(3) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)$ $\Delta H_3=-285.8kJ$

Now we will reverse the reaction 1, multiply reaction 2 and 3 by 2 then adding all the equation, we get :

(1) $2CO_2(g)+2H_2O(l)\rightarrow C_2H_4(g)+3O_2(g)$ $\Delta H_1=+1411kJ$

(2) $2C(s)+2O_2(g)\rightarrow 2CO_2(g)$ $\Delta H_2=2\times (-393.5kJ)=-787kJ$

(3) $2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)$ $\Delta H_3=2\times (-285.8kJ)=-571.6kJ$

The expression for enthalpy of formation of $C_2H_4$ will be,

$\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H=(+1411kJ)+(-787kJ)+(-571.6kJ)$

$\Delta H=52.4kJ$

Therefore, the standard enthalpy of formation of ethylene is, 52.4 kJ

7 0

2 years ago

2. Which of the following expressions is generally used for solubility? (1 point)

miskamm [114]

Answer:

For 2: The correct answer is grams of solute per 100 grams of solvent.

For 3: The correct answer is supersaturated.

For 4: the correct answer is the solubility decreases.

Explanation:

For 2:

Solubility is defined as the property which refers to the ability of the solute that can be dissolved in a solvent. It is defined as the number of grams of solute per 100 grams of solvent.

For 3:

Unsaturated solution is defined as the solution in which amount of solute that is dissolved in the solvent is less.

Saturated solution is defined as the solution in which no more solute can be dissolved in the given amount of solvent.

Emulsion is defined as the dispersion of one liquid in another liquid in which it is not soluble.

Supersaturated solution is defined as the solution in which solvent contains more amount of solute than the required amount. These solutions help in the process of crystallization.

When a crystal is added to a supersaturated solution, more and more particles come out of the solution and this process is known as crystallization.

For 4:

According to the Henry's Law

The solubility of the gas in a liquid is directly proportional to the partial pressure of the gas.

$\text{Solubility of a gas in a liquid}\propto \text{Partial pressure of the gas}$

With increase in the partial pressure, the solubility of the gas in liquid also increases and vice-versa.

Hence, the correct answer is the solubility decreases.

7 0

3 years ago

Read 2 more answers