Would you expect hydrogen chloride to be a gas, a liquid or a solid, at room temperature and pressure? Explain your answer

1 answer:

6 0

At room temperature hydrogen chloride is a colorless gas with a sharp or pungent odor. Under pressure or at temperatures below –85°C (-121°F), it is a clear liquid.

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hi, can you check if my answers are correct? I'm not so confident in this topic (topic: Classifying Types of Chemical Reactions)

NikAS [45]

I am pretty sure your answer are correct , from what I know. Good job!

7 0

3 years ago

How many solutions are there in the following equation.

givi [52]

When they ask you for the solution they are usually asking for x. So solve for X.
X=-12 so you only have one solution. so the answer is B

6 0

3 years ago

If 97.7 ml of silver nitrate solution reacts with excess potassium chloride solution to yield 0.326 g of precipitate, what is th

BlackZzzverrR [31]

The complete balanced chemical reaction is written as:

AgNO3 + KCl ---> AgCl + KNO3

where AgCl is our precipitate

So calculating for moles of AgCl produced: MM AgCl = 143.5 g/mol

moles AgCl = 0.326 g / (143.5 g/mol) = 2.27 x 10^-3 mol

we see that there is 1 mole of Ag per 1 mole of AgCl so:

moles Ag = 2.27 x 10^-3 mol

The molarity is simply the ratio of number of moles over volume in Liters, therefore:

Molarity = 2.27 x 10^-3 mol / 0.0977 L

<span>Molarity = 0.0233 M</span>

4 0

3 years ago

The question is in the picture below

Rus_ich [418]

Answer:

$\Delta\text{H}_1+2\Delta\text{H}_2-\Delta\text{H}_3$

Explanation:

Hess's Law of Constant Heat Summation states that if a chemical equation can be written as the sum of several other chemical equations, the enthalpy change of the first chemical equation is equal to the sum of the enthalpy changes of the other chemical equations. Thus, the reaction that involves the conversion of reactant A to B, for example, has the same enthalpy change even if you convert A to C, before converting it to B. Regardless of how many steps it takes for the reactant to be converted to the product, the enthalpy change of the overall reaction is constant.

With Hess's Law in mind, let's see how A can be converted to 2C +E.

$\bf{\text{A} \rightarrow 2\text{B}}$ (Δ $\text{H}_1$ ) -----(1)

Since we have 2B, multiply the whole of II. by 2:

$\bf{2\text{B} \rightarrow 2\text{C} +2\text{D}}$ (2Δ $\text{H}_2$ ) -----(2)

This step converts all the B intermediates to 2C +2D. This means that the overall reaction at this stage is $\text{A} \rightarrow 2\text{C} +2\text{D}$ .

Reversing III. gives us a negative enthalpy change as such:

$\bf{2\text{D} \rightarrow \text{E}}$ (-Δ $\text{H}_3$ ) -----(3)

This step converts all the D intermediates formed from step (2) to E. This results in the overall equation of $\text{A} \rightarrow 2\text{C} +\text{E}$ , which is also the equation of interest.