All categories

2 years ago

How does the number of osmoles factor in when trying to calculate the molar mass of a substance?

Chemistry

1 answer:

Tom [10]2 years ago

7 0

To calculate the molar mass of substance, divide the number of moles by the given mass of substance.

The quantity of a substance is known as a mole. You can measure it in terms of grams, liters, atoms, molecules, or particles.

Use the substance's molar mass to convert between grams and moles. Divide the given mass by the number of moles to convert from moles to molar mass.

Using the formula, Number of moles = Given mass/ molar mass

The molar mass of a substance can be calculated as-

Molar mass = Given mass/number of moles

Learn more about mole -

brainly.com/question/26416088

#SPJ4

You might be interested in

Find the moles in 0.0023 of Co, at STP:

Alexus [3.1K]

What is does that mean 0.0023 cooooo

7 0

3 years ago

You are given a solution of sugar and sand and water your task is to separate sugar from sand which of the following experimenta

Ostrovityanka [42]

Answer:

Explanation:

Filtration followed by evaporation:

To separate the mixture of sand and sugar, it is best to use the separation technique of filtration then evaporation.

Pour the water into the mixture. The sugar will dissolve with time in the water. Sand is made up of quartz and does not dissolve in water.

After the dissolution, filter the solution to separate the sand using a filter paper.

Dry the sand thereafter then proceed to evaporate the sugar with water solution. Evaporation will turn water into vapor and the sugar crystals will be left behind.

8 0

3 years ago

What is the total energy change for the following reaction:CO+H2O-CO2+H2

Alekssandra [29.7K]

Answer:

$\large \boxed{\text{-41.2 kJ/mol}}$

Explanation:

Balanced equation: CO(g) + H₂O(g) ⟶ CO₂(g) + H₂(g)

We can calculate the enthalpy change of a reaction by using the enthalpies of formation of reactants and products

$\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)$

(a) Enthalpies of formation of reactants and products

$\begin{array}{cc}\textbf{Substance} & \textbf{$\Delta_{\text{f}}$H/(kJ/mol}) \\\text{CO(g)} & -110.5 \\\text{H$_{2}$O} & -241.8\\\text{CO$_{2}$(g)} & -393.5 \\\text{H$_{2}$(g)} & 0 \\\end{array}$

(b) Total enthalpies of reactants and products

$\begin{array}{ccr}\textbf{Substance} & \textbf{Contribution)/(kJ/mol})&\textbf{Sum} \\\text{CO(g)} & -110.5& -110.5 \\\text{H$_{2}$O(g)} &-241.8& -241.8\\\textbf{Total}&\textbf{for reactants} &\mathbf{ -352.3}\\&&\\\text{CO}_{2}(g) & -393.5&-393.5 \\\text{H}_{2} & 0 & 0\\\textbf{Total}&\textbf{for products} & \mathbf{-393.5}\end{array}$

(c) Enthalpy of reaction $\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)= \text{-393.5 kJ/mol - (-352.3 kJ/mol}\\= \text{-393.5 kJ/mol + 352.3 kJ/mol} = \textbf{-41.2 kJ/mol}\\ \text{The total enthalpy change is $\large \boxed{\textbf{-41.2 kJ/mol}}$}$

4 0

2 years ago

I’ll give brainliest!!!

vodka [1.7K]

Is this prier to a lab you've done?

4 0

3 years ago

Empirical formula of C6H10O6

KiRa [710]

Answer:

I don't know.

Explanation:

I actually don't know the answer so I wrote the answer is "I don't know" or simply "I dunno".

7 0

2 years ago