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3 years ago

What is the new boiling point if 25 g of NaCl is dissolved in 1.0 kg of water

1 answer:

8 0

The correct answer for the question that is being presented above is this one:

Given that:
delta Tb = Kbm Kb H2O = 0.52 degrees C/m
<span>delta Tf = Kfm Kf H2O = 1.86 degrees C/m
</span>
We need to know the formula for Molality.
molality = mol solute / kg solvent

<span>We are given the amount of solute in grams
Since amount of solute is given in moles, we have to convert 25 g NaCl to moles. Divide by molar mass. </span>

<span>25 g NaCl / 58.5 g/mol = 0.427 mol </span>

<span>Then, use the formula for molality. </span>

<span>molality = mol solute / kg solvent </span>
<span>= 0.427 / 1 </span>
<span>= 0.427 m </span>

<span>Use now the formula to get the boiling point.</span>

<span>delta Tb = Kbm </span>
<span>= (0.52)(0.427) </span>
<span>= 0.22C </span>

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Dmitrij [34]

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2 years ago

A student has a sample of isopropanol (C3H,OH) that has a mass of 78.6 g. The molar mass of isopropanol is 60.1 g/mol. How

Sauron [17]

I think it goes like this.

mass = 78.6 g
molar mass = 60.1 g/mol
amount (in mols) = mass/mol
78.6 g / 60.1 g/mol = 1.30782 mol

^^rounded to 3 sig figs, final answer = 1.31mol

4 0

4 years ago

Read 2 more answers

How can we specify / identify a certain electron?<br> (Quantum forces)

Serga [27]

Answer:

electrons are less than 1 amu, they are also negative charged. they are very small too

3 0

3 years ago

A crystal of potassium permanganate is placed in a beaker of water. The purple colour of potassium permanganate gradually spread

cricket20 [7]

When the crystals of potassium permanganate are preserved in water, the purple-coloured crystals of potassium permanganate break further into smaller particles that populate the distance between the molecules of water imparting a purple colour to the water. This is an example of diffusion.

<h3>What are the two conclusions given out in the method of diffusion?</h3>

Diffusion is the process of movement of solvent from higher concentration to lower concentration through a semipermeable membrane. So, we can form the decision that it cannot occur through a thick membrane from which small molecules cannot pass through.

<h3>What is difference between osmosis and diffusion?</h3>

Osmosis is the direction of solvent particles from a solution that is diluted to a more concentrated one. In contrast, diffusion is the movement of particles from a higher concentration region to a part of lower concentration.

To learn more about potassium permanganate, refer

brainly.com/question/24204185

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6 0

2 years ago

Calculate Delta H in KJ for the following reactions using heats of formation:

lozanna [386]

Answer:

$\Delta H\textdegree = -2856.8\;\text{kJ}$ per mole reaction.

$\Delta H\textdegree = -22.3\;\text{kJ}$ per mole reaction.

Explanation:

What is the standard enthalpy of formation $\Delta H_f\textdegree{}$ of a substance? $\Delta H_f\textdegree{}$ the enthalpy change when one mole of the substance is formed from the most stable allotrope of its elements under standard conditions.

Naturally, $\Delta H_f\textdegree{} = 0$ for the most stable allotrope of each element under standard conditions. For example, oxygen $\text{O}_2$ (not ozone $\text{O}_3$ ) is the most stable allotrope of oxygen. Also, under STP $\text{O}_2$ is a gas. Forming $\text{O}_2\;(g)$ from itself does not involve any chemical or physical change. As a result, $\Delta H_f\textdegree{} = 0$ for $\text{O}_2\;(g)$ .

Look up standard enthalpy of formation $\Delta H_f\textdegree{}$ data for the rest of the species. In case one or more values are not available from your school, here are the published ones. Note the state symbols of the compounds (water/steam $\text{H}_2\text{O}$ in particular) and the sign of the enthalpy changes.

$\text{C}_2\text{H}_6\;(g)$ : $-84.0\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{CO}_2\;(g)$ : $-393.5\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{H}_2\text{O}\;{\bf (g)}$ : $-241.8\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{PbO}\;(s)$ : $-217.9\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{PbO}_2\;(s)$ : $-276.6\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{Pb}_3\text{O}_4\;(s)$ : $-734.7\;\text{kJ}\cdot\text{mol}^{-1}$

How to calculate the enthalpy change of a reaction $\Delta H_\text{rxn}$ (or simply $\Delta H$ from enthalpies of formation?

Multiply the enthalpy of formation of each product by its coefficient in the equation.
Find the sum of these values. Label the sum $\Sigma (n\cdot \Delta_f(\text{Reactants}))$ to show that this value takes the coefficients into account.
Multiply the enthalpy of formation of each reactant by its coefficient in the equation.
Find the sum of these values. Label the sum $\Sigma (n\cdot \Delta_f(\text{Products}))$ to show that this value takes the coefficient into account.
Change = Final - Initial. So is the case with enthalpy changes. $\Delta H_\text{rxn} = \Sigma (n\cdot \Delta_f(\textbf{Products})) - \Sigma (n\cdot \Delta_f(\textbf{Reactants}))$ .

For the first reaction:

$\Sigma (n\cdot \Delta_f(\text{Reactants})) = 4\times (-393.5) + 6\times (-241.8) = -3024.8\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\Sigma (n\cdot \Delta_f(\text{Products})) = 2\times (-84.0) + 7\times 0 = -168.0\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\begin{aligned}\Delta H_\text{rxn} &= \Sigma (n\cdot \Delta_f(\textbf{Products})) - \Sigma (n\cdot \Delta_f(\textbf{Reactants}))\\ &= (-3024.8\;\text{kJ}\cdot\text{mol}^{-1}) - (-168.0\;\text{kJ}\cdot\text{mol}^{-1})\\ &= -2856.8\;\text{kJ}\cdot\text{mol}^{-1} \end{aligned}$ .

Try these steps for the second reaction:

$\Delta H_\text{rxn} = -22.3\;\text{kJ}\cdot\text{mol}^{-1}$ .

6 0

3 years ago