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3 years ago

A 2.00 L bottle of Dr Pepper contains 225 g of sucrose, C12H22O11. What is the molality of that solution?

Chemistry

1 answer:

Firdavs [7]3 years ago

6 0

Answer:

Molality = 0.33 mol/Kg

Explanation:

Given data:

Volume of solvent = 2.00 L

Mass of solute = 225 g

Molality of solution = ?

Solution:

Formula:

Molality = Number of moles of solute / kg of solvent

Now we will calculate the moles of solute.

Number of moles = mass/ molar mass

Number of moles = 225 g/ 342.3 g/mol

Number of moles = 0.66 mol

Now we will calculate the Kg of solvent.

1 L = 1 Kg

2 L = 2 Kg

Molality = Number of moles of solute / kg of solvent

Molality = 0.66 mol / 2 kg

Molality = 0.33 mol/Kg

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maria [59]

Answer:

I think beaker three will take longer to boil since there is more water present,compared to the other beakers

3 0

3 years ago

The expression of the theoretical yield (TY) in function of limiting reagent (LR) of a reaction is as follows: TY = ideal mole r

spin [16.1K]

Answer: The theoretical yield of acetanilide is 6.5 grams.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For aniline:

Given mass of aniline = $4.50\times 10^0=4.50g$ (We know that: $10^0=1$ )

Molar mass of aniline = 93.13 g/mol

Putting values in equation 1, we get:

$\text{Moles of aniline}=\frac{4.50g}{93.13g/mol}=0.048mol$

For acetic anhydride:

To calculate the mass of acetic anhydride, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Volume of acetic anhydride = $(1.25\times \text{Mass of aniline})=1.25\times 4.50=5.625mL$

Density of acetic anhydride = 1.08 g/mL

Putting values in above equation:

$1.08g/mL=\frac{\text{Mass of acetic anhydride}}{5.625mL}\\\\\text{Mass of acetic anhydride}=(1.08g/mL\times 5.625mL)=6.08g$

Given mass of acetic anhydride = 6.08 g

Molar mass of acetic anhydride = 102.1 g/mol

Putting values in equation 1, we get:

$\text{Moles of acetic anhydride}=\frac{6.08g}{102.1g/mol}=0.06mol$

The chemical equation for the reaction of aniline and acetic anhydride follows:

$C_6H_5NH_2+CH_3COOCOCH_3\rightarrow C_6H_5NHCOCH_3+CH_3COOH$

By Stoichiometry of the reaction:

1 mole of aniline reacts with 1 mole of acetic anhydride

So, 0.048 moles of aniline will react with = $\frac{1}{1}\times 0.048=0.048mol$ of acetic anhydride

As, given amount of acetic anhydride is more than the required amount. So, it is considered as an excess reagent.

Thus, aniline is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of aniline produces 1 mole of acetanilide

So, 0.048 moles of aniline will produce = $\frac{1}{1}\times 0.048=0.048mol$ of acetanilide

Now, calculating the theoretical yield of acetanilide by using equation 1:

Moles of acetanilide = 0.048 moles

Molar mass of acetanilide = 135.17 g/mol

Putting values in equation 1, we get:

$0.048mol=\frac{\text{Mass of acetanilide}}{135.17g/mol}\\\\\text{Mass of acetanilide}=(0.048mol\times 135.17g/mol)=6.5g$

Hence, the theoretical yield of acetanilide is 6.5 grams.

3 0

3 years ago

Given the following heats of combustion. CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(l) ΔH°rxn = -726.4 kJ C(graphite) + O2(g) CO2(g) ΔH

sergeinik [125]

Answer:

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation:

The formation reaction of CH_3OH will be,

$C(s)+2H_2(g)+\frac{1}{2}O_2\rightarrow CH_3OH(g),\Delta H_{formation}=?$

The intermediate balanced chemical reaction will be,

$C(graphite)+O_2(g)\rightarrow CO_2(g), \Delta H_1=-393.5kJ/mole$ ..[1]

$H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l), \Delta H_2=-285.8kJ/mole$ ..[2]

$CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l) , \Delta H_3=-726.4kJ/mole$ ..[3]

Now we will reverse the reaction 3, multiply reaction 2 by 2 then adding all the equations, Using Hess's law:

We get :

$C(graphite)+O_2(g)\rightarrow CO_2(g) , \Delta H_1=-393.5kJ/mole$ ..[1]

$2H_2(g)+2O_2(g)\rightarrow 2H_2O(l) ,\Delta H_2=2\times (-285.8kJ/mole)=-571.6kJ/mol$ ..[2]

$CO_2(g)+2H_2O(l)\rightarrow CH_3OH(g)+\frac{3}{2}O_2(g) ,\Delta H_3=726.4kJ/mole$ [3]

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

$\Delta H_{formation}=\Delta H_1+2\times \Delta H_2+\Delta H_3$

$\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)$

$\Delta H=-238.7kJ/mole$

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

4 0

3 years ago

How are all cells similar?

Reika [66]

Answer:

All cells have structural and functional similarities. Structures shared by all cells include a cell membrane, an aqueous cytosol, ribosomes, and genetic material (DNA). All cells are composed of the same four types of organic molecules: carbohydrates, lipids, nucleic acids, and proteins.

Explanation:

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8 0

3 years ago

If 1.85 g of Mg(OH)2 reacts with 3.71 g of HCl,

Damm [24]

Both of them are a hope this helps

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3 years ago