A student dissolves 12.g of sucrose C12H22O11 in 300.mL of a solvent with a density of 1.01/gmL . The student notices that the volume of the solvent does not change when the sucrose dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to 2 significant digits. molarity = molality =

Answers

Answer 1

Answer:

Answer:

Molarity → 0.17 M

Molality → 0.11 m

Explanation:

The student notices that the volume of the solvent does not change when the sucrose dissolves in it; therefore we assume the volume of solvent as solution.

Molarity = Mol of solute/L

Let's calculate the mol of solute (mass / molar mass)

12 g / 342 g/mol = 0.0351 moles

Let's conver the volume (mL) to L

300 mL / 1000 = 0.3 L

Molarity (mol/L) = 0.0351 mol / 0.3L → 0.17 M

Molality = mol of solute / 1kg of solvent.

Let's find out the mass of solvent with the density

Solvent density = Solvent mass / Solvent volume

1.01 g/mL = Solvent mass / 300 mL

1.01 g/mL . 300 mL = Solvent mass →303 g

Let's convert the mass to kg

303 g / 1000 = 0.303 kg

Molality (mol/kg) → 0.0351 mol / 0.303kg = 0.11 m

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What happens if more product is added to a system at equilibrium

Answers

Answer:

More reactants will be produced

Explanation:

Le Chatelier's principle; adding additional product or reactant will move the equilibrium left or right to compensate and come back to equilibrium

By adding more product to your system at equilibrium, the equilibrium will shift towards reactants, more reactants will be produced

Answer: Liquid molecules forming a gas and gas molecules forming a liquid are equal in number

Explanation: :/

A certain first-order reaction has a rate constant of 2.75 10-2 s−1 at 20.°c. what is the value of k at 45°c if ea = 75.5 kj/mol? webassign will check your answer for the correct number of significant figures. 0.0352 incorrect: your answer is incorrect.

Answers

With an activation energy $($E_a$)$ of 75.5 kJ/mol, the rate constant k for a first-order reaction at 20°C is 2.75 × 10⁻² s⁻¹. At 45°C, k is approximately 0.095 s⁻¹, determined using the Arrhenius equation.

The Arrhenius equation relates the rate constant k, temperature T, activation energy $($E_a$)$ , and the gas constant R:

$k = Ae^{-(E_a)/(RT)}$

Given that $k_1 = 2.75 * 10^(-2) \, \text{s}^(-1)$ at $T_1 = 20^\circ \text{C} = 293.15 \, \text{K}$ and $E_a = 75.5 \, \text{kJ/mol}$ , we want to find $k_2$ at $T_2 = 45^\circ \text{C} = 318.15 \, \text{K}$ .

First, let's find the value of A using the Arrhenius equation at T_1:

$2.75 * 10^(-2) = A e^{-((75.5 * 10^3))/((8.314)(293.15))}$

Solving for A:

$A \approx 3.65 \, \text{s}^(-1)$

Now, use the Arrhenius equation at $T_2$ to find $k_2$ :

$k_2 = (3.65) e^{-((75.5 * 10^3))/((8.314)(318.15))}$

Calculate $k_2$ .

$k_2 \approx 0.095 \, \text{s}^(-1)$

Therefore, the value of k at $45^\circ \text{C}$ is approximately $0.095 \, \text{s}^(-1)$ .

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Final answer:

To find the new rate constant of a first-order reaction under different temperature conditions, we can use the Arrhenius equation, which relates the rate constant, activation energy, and temperature of a reaction.

Explanation:

The student is interested in finding the value of the rate constant (k) at a different temperature for a first-order reaction. The answer can be found using the Arrhenius equation, which defines the relationship between the rate constant (k) of a reaction and the temperature at which the reaction occurs. The activation energy (Ea) is also necessary.

The Arrhenius equation is: k = A * exp(-Ea/(R*T)), where A is the pre-exponential factor, R is the universal gas constant (the value of R should be 8.314 J/mol.K to match the Ea units), and T is the temperature in Kelvin.

At the first condition, you have the value of k and the corresponding T (convert Celsius to Kelvin by adding 273.15). With these values and the known Ea, you can solve for A. Then, using the value of A, Ea, and the second T (also converted to Kelvin), you can solve for the new k.

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Write the number 4.92 x 10-3 from scientific notation to regular.
Please help?????

Answers

Explanation:

4.92*10-3=4.92/10^3=

4.92/1000=0.00492

How many moles of H2O are produced from 6.23g of C2H6 in the following reaction? 2C2H6+7O2 -> 4CO2+6H2O

Answers

Answer:

0.623 moles of H₂O.

Explanation:

balanced equation: 2C₂H₆ + 7O₂ -> 4CO₂+ 6H₂O

Given:

6.23g of C₂H₆

C₂H₆ molar mass is 30

solve for moles of C₂H₆

moles = mass/mr

moles = 6.23/30

moles = 0.2077

solve for moles of H₂O using molar ratio

2C₂H₆ : 6H₂O

2 : 6

0.2077 : 0.623

Therefore, found that 0.623 moles of H₂O is produced.

The figure below represents a reaction.What type of reaction is shown?SO3+ H2SO4 →→ HSO4
synthesis
decomposition
single displacement
double displacement

Answers

SO₃+ H₂SO₄ → H₂SO₄ This reaction is synthesis type of reaction, because that would be combination (synthesis) A + B → AB. Therefore, option A is correct.

What are the types of reaction ?

There are five basic types of chemical reactions are combination, decomposition, single-replacement, double-replacement, and combustion.

Biochemical reactions are chemical reactions that occur within living things. Metabolism refers to the sum of all biochemical reactions in an organism. Exothermic and endothermic chemical reactions are both part of metabolism.

Hydration is the process of combining water molecules with another substance to form a single, new compound. SO3 is an acidic oxide that reacts with and dissolves in water to form sulfuric acid, H2SO4.

Thus, option A is correct.

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Are you sure it isn’t SO3+H2O = H2SO4 because that would be combination (synthesis) A+ B=AB

Or SO3 + H2SO4 = H2S2O7
Because that would also be synthesis

Draw 1,2,3,4,5,6-hexachlorocyclohexane with :a. all the chloro groups in axial positions.
b. all the chloro groups in equatorial positions.

Answers

Answer:

This is required answer.

Explanation:

Given that,

1,2,3,4,5,6-hexachlorocyclohexane

(a). We need to draw 1,2,3,4,5,6-hexachlorocyclohexane with all the chloro groups in axial positions

Using given data

We draw 1,2,3,4,5,6-hexachlorocyclohexane with all the chloro groups in axial positions.

When we say that all the chloro groups in axial position that means axial bonds are vertical.

(b). We need to draw 1,2,3,4,5,6-hexachlorocyclohexane with all the chloro groups in equatorial positions

Using given data

We draw 1,2,3,4,5,6-hexachlorocyclohexane with all the chloro groups in equatorial positions.

When we say that all the chloro groups in equatorial position that means axial bonds are horizontal.

Hence, This is required answer.

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