A student uses paper towels to clean up a small chemical spill in the lab.Where should the dirty paper towels be placed?
A in the lab sink
B in the wastebasket
C in the chemical waste container
D in the hazardous waste container

Final answer:

The number of grams in one mole of B2 can be calculated using the atomic mass of element B. This is found on the periodic table and then doubled for B2 since it's diatomic. If B is Oxygen for instance, 1 mole of B2 (O2) weighs 32 grams.

Explanation:

To find the number of grams in one mole of B2, we need to know the atomic mass of element B, which isn't provided in your question. However, you can find this information on the periodic table. Once you have the atomic mass of B, you can calculate the molar mass of B2 (which is two times the atomic mass of B) since 1 mole of a substance corresponds to its molar mass in grams.

For example, if element B is Oxygen (O), its atomic mass is approximately 16 g/mol. Therefore, the molar mass of B2 (O2 in this case) would be 32 g/mol. Hence, 1 mole of B2 (or O2) would weigh 32 grams.

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Answer:

28.2 g of NaOH

Explanation:

We need to calculate the grams of NaOH needed to react with 25.0 g of Cl₂ in the following reaction:

2 NaOH(aq) + Cl₂(g) → NaOCI(aq + H₂0(I) + NaCl(aq)

We are going to solve this by making use of the molar ratio between Cl₂ and NaOH given by the reaction equation where we see that every mol of Cl₂ will react with 2 moles of NaOH.

So first we need to convert the 25.0 g of Cl₂ to moles:

• Molar Mass of Cl₂ = 2 x 35.45 = 70.90 g/mol
• Moles of Cl₂ = 25.0 g / 70.90 g/mol = 0.3526 moles

Then we need to calculate the moles of NaOH needed to react with these moles of Cl₂ knowing that every mol of Cl₂ will react with 2 moles of NaOH:

• moles of NaOH = 2 x moles of Cl₂ = 2 x 0.3526 moles = 0.7052 moles

Next we must convert these moles to grams:

• Molar Mass of NaOH = 22.990 + 15.999 + 1.008 = 40.00 g/mol
• Mass of NaOH = 0.7052 moles x 40.00 g/mol = 28.2 g

28.2 g are needed to react with 25.0 g of Cl₂ in the production of NaOCl

Answer : The mass of sucrose added to 375 mL of water must be, 10.6 grams.

Explanation :

As we are given that 2.75 m/m percent solution of sucrose. That means, 2.75 grams of sucrose present in 100 grams of solution.

Mass of solution = 100 g

Mass of sucrose = 2.75 g

Mass of water = Mass of solution - Mass of sucrose

Mass of water = 100 g - 2.75 g

Mass of water = 97.25 g

First we have to calculate the mass of water.

Density of water = 1.00 g/mL

Volume of water = 375 mL

Now we have to calculate the mass of sucrose in 375 g of water.

As, 97.25 grams of water contain 2.75 grams of sucrose

So, 375 grams of water contain grams of sucrose

Therefore, the mass of sucrose added to 375 mL of water must be, 10.6 grams.

To make a 2.75% m/m sucrose solution, you need to add approximately 1062 grams of sucrose to 375 mL of water, considering the density of water as 1 g/mL.

To prepare a mass/mass (m/m) percent solution of sucrose, you need to calculate the mass of sucrose (in grams) that needs to be added to 375 mL of water to achieve a 2.75% concentration.

Here's how you can calculate it:

1. Convert the volume of water to grams, considering the density of water:

  Density of water ≈ 1 g/mL

  Mass of water = Volume of water × Density of water

  Mass of water = 375 mL × 1 g/mL = 375 g

2. Determine the desired mass of sucrose as a percentage of the total mass:

  Desired m/m percent = 2.75%

3. Calculate the mass of sucrose needed:

  Mass of sucrose = (Desired m/m percent / 100) × Total mass

  Mass of sucrose = (2.75 / 100) × (375 g + Mass of sucrose)

4. Rearrange the equation to solve for the mass of sucrose:

  Mass of sucrose = (2.75 / 100) × (375 g) / (1 - (2.75 / 100))

Now, calculate:

Mass of sucrose = (2.75 / 100) × (375 g) / (1 - 0.0275)

Mass of sucrose ≈ (2.75 / 100) × (375 g) / 0.9725

Mass of sucrose ≈ (2.75 × 375 g) / 100 / 0.9725

Mass of sucrose ≈ (1031.25 g) / 0.9725

Mass of sucrose ≈ 1061.98 g

So, approximately 1062 grams of sucrose must be added to 375 mL of water to prepare a 2.75 m/m percent solution of sucrose.

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Answer:

Tinitial = 78.6°C

Explanation:

In a calorimetry experiment, the flow heat is measured for a system that has a state change. In this case, there isn't happening a physical change, so the heat is called sensitive heat, and it's calculated by:

q = mxCpxΔT

Where q is the heat, m is the mass, Cp is the specific heat and ΔT is the difference of final and initial temperature (Tfinal - Tinitial).

Copper is losing heat, so q is negative, then:

-1860 = 92x0.377x(25 - Ti)

34.684(25 - Ti) = -1860

25 - Ti = -53.63

-Ti = -78.63

Ti = 78.6ºC

Answer:

ΔH°rxn = 54.08 kJ

Explanation:

Let's consider the following equations.

a) ClO(g) + O₃(g) ⇄ Cl(g) + 2 O₂(g)                     ΔH°rxn = –29.90 kJ

b) 2 O₃(g) ⇄ 3 O₂(g)                                            ΔH°rxn = 24.18 kJ

We have to determine the value of heat of reaction for the following reaction: Cl(g) + O₃(g) ⇄ ClO(g) + O₂(g)

According to Hess's law, the enthalpy change in a chemical reaction is the same whether the reaction takes place in one or in several steps. That means that we can find the enthalpy of a reaction by adding the corresponding steps and adding their enthalpies. According to Lavoisier-Laplace's law, if we reverse a reaction, we also have to reverse the sign of its enthalpy.

Let's reverse equation a) and add it to equation b).

-a) Cl(g) + 2 O₂(g) ⇄ ClO(g) + O₃(g)                    ΔH°rxn = 29.90 kJ

b) 2 O₃(g) ⇄ 3 O₂(g)                                            ΔH°rxn = 24.18 kJ

-------------------------------------------------------------------------------------------------

Cl(g) + 2 O₂(g) + 2 O₃(g) ⇄ ClO(g) + O₃(g) + 3 O₂(g)

Cl(g) + O₃(g) ⇄ ClO(g) +O₂(g)

ΔH°rxn = 29.90 kJ + 24.18 kJ = 54.08 kJ

Final answer:

The heat of the reaction (ΔH°rxn) for the reaction Cl(g) + O3(g) ? ClO(g) + O2(g) is calculated using Hess's Law. The sum of the heat of reversed first reaction and the second reaction provided is 54.08 kJ.

Explanation:

The chemistry question asks to determine the heat of the reaction for the reaction Cl(g) + O3(g) ? ClO(g) + O2(g). In Hess's Law, the heat of the reaction or ΔH for a reaction can be calculated from the sum of the heats of other reactions that sum to the desired reaction. In this case, we want to reverse the first reaction provided (which changes the sign of ΔH) and add it to the second reaction provided.

So, reversing the first reaction we get: Cl(g) + 2 O2(g) ? ClO(g) + O3(g) ?H°rxn = 29.90 kJ

Adding this to the second reaction: 2 O3(g) ? 3 O2(g), ?H°rxn = 24.18 kJ, gives the reaction Cl(g) + O3(g) ? ClO(g) + O2(g). Adding the ΔH values gives the ΔH for this reaction: 29.90 kJ + 24.18 kJ = 54.08 kJ. So, ?H°rxn for the reaction Cl(g) + O3(g) ? ClO(g) + O2(g) is 54.08 kJ.

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