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4 years ago

How much heat is lost when 0.440 mol of steam condenses at 100 °C?

Chemistry

1 answer:

dusya [7]4 years ago

8 0

Answer:

17,890 J

Explanation:

The amount of heat released by a gaseous substance when it condenses is given by the formula

$Q=n\lambda_v$

where

n is the number of moles of the substance

$\lambda_v$ is the latent heat of vaporization

The formula can be applied if the substance is at its vaporization temperature.

In this problem, we have:

n = 0.440 mol is the number of moles of steam

$\lambda_v=40,660 J/mol$ is the latent heat of vaporization of water

And the steam is already at 100C, so we can apply the formula:

$Q=(0.440)(40660)=17,890 J$

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Look at the diagram below.

damaskus [11]

Answer:

it will gain electrons to fill its outer shell

Explanation:

This element is boron which has 5 electrons.

6 0

3 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

Given 50.0 grams of Strontium, find grams of Phosphorus required for complete reaction

Sveta_85 [38]

Answer:

Mass = 11.78 g of P₄

Explanation:

The balance chemical equation is as follow:

6 Sr + P4 → 2 Sr₃P₂

Step 1: Calculate moles of Sr as;

Moles = Mass / M/Mass

Moles = 50.0 g / 87.62 g/mol

Moles = 0.570 moles

Step 2: Find moles of P₄ as;

According to equation,

6 moles of Sr reacted with = 1 mole of P₄

So,

0.570 moles of Sr will react with = X moles of P₄

Solving for X,

X = 1 mol × 0.570 mol / 6 mol

X = 0.0952 mol of P₄

Step 3: Calculate mass of P₄ as,

Mass = Moles × M.Mass

Mass = 0.0952 mol × 123.89 g/mol

Mass = 11.78 g of P₄

8 0

3 years ago

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lions [1.4K]

Answer:

Explanation:

Definitely not in the cell theory

3 0

3 years ago

The reducing agent in this catalytic hydrogenation reaction was molecular hydrogen (H2), which was produced in situ (in the reac

creativ13 [48]

Answer:

a) After the balloon inflated after 440 uL of dropwise due to the reaction of 1-Decene and the solution in the conical vial. b) $4NaBH_{4} + 2HCl + 7H_{2}O$ ⇒ 16 $H_{2(g)}+ 2NaCl + Na_{2}B_{4}O_{7}$ c) No $H_{2}$ was not the limiting reactant.

Explanation:

Generally, hydrogenation is the chemical reaction between a compound or element and molecular hydrogen in the presence of catalysts such as platinum.

a) After the balloon inflated after 440 uL of dropwise 1-Decene solution was added due to the reaction between 1-Decene and the solution in the conical vial.

b) $4NaBH_{4} + 2HCl + 7H_{2}O$ ⇒ 16 $H_{2(g)}+ 2NaCl + Na_{2}B_{4}O_{7}$

c) $H_{2}$ was not the limiting reactant based on the mol to mol ratio of $H_{2}$ and decane which is 1:1. Therefore, if 0.8 mol of decane was produced then 0.8 mol of $H_{2}$ would also be produced.

4 0

3 years ago