How much energy in joules does it take to raise the temperature of 255g of water by 12.5C

Answer:

Explanation:

To solve this problem, we must use stoichiometry: the calculation of reactants and products in a chemical reaction using ratios.

1. Analyze the Reaction

Check the coefficients, which indicate the moles required for the reaction. 2 moles of sodium (Na) and 1 mole of chlorine (Cl₂) produces 2 moles of sodium chloride (NaCl).

2. Set Up Ratios.

We are asked to find the grams of NaCl produced when 0.548 moles of Na react with excess chlorine.

We need 2 ratios: 1 for finding the moles of NaCl produced and 1 for converting to grams.

The first ratio is found using the coefficients. Since there is excess chlorine, we only need to focus on the sodium and sodium chloride. According to their coefficients, 2 moles of Na produce 2 moles of NaCl. This is the first ratio.

The second ratio uses the molar mass. Since we are solving for the grams of NaCl, we have to find its molar mass.

First, locate these values on the Periodic Table for the individual elements.

• Na: 22.989769 g/mol
• Cl: 35.45 g/mol

There is 1 of each atom in 1 molecule, so we can add these values.

• NaCl: 58.439769 g/mol

Use this value as the second ratio.

3. Calculate

Make 1 expression using the 2 ratios and the initial value of moles.

Flip the ratios so the correct units cancel out.

Multiply. Note that the moles of Na (units) cancel and the moles of NaCl (units). cancel.

The original value of moles has 3 significant figures, so our answer must have the same. For the number we found, that is the tenth place.

• 32.02499341232

The 2 in the hundredth place tells us to leave the 0.

0.548 moles of sodium react with excess chlorine to produce 32.0 grams of sodium chloride.

Answer:

70.88 mL volume of 1.27 M of HCl is required.

Explanation:

Given data:

Initial volume = ?

Initial  molarity =  1.27 M

Final volume = 197.4 mL

Final molarity = 0.456 M

Solution:

Formula:

M₁V₁ = M₂V₂

Now we will put the values in formula.

1.27 M × V₁ =  0.456 M × 197.4 mL

V₁ = 0.456 M × 197.4 mL/1.27 M

V₁ = 90.014M.mL/1.27 M

V₁ = 70.88 mL

70.88 mL volume of 1.27 M of HCl is required.

Final answer:

To prepare 197.4 mL of 0.456 M HCl from 1.27 M HCl, you need 71.03 mL of 1.27 M HCl.

Explanation:

The subject of this problem involves using the concept of molarity in Chemistry. We can use a simple formula for dilution, M1V1 = M2V2, to find the volume. Here M1 (1.27 M) is the molarity of stock HCl, V1 is the required volume, M2 (0.456 M) is the desired molarity, and V2 (197.4 mL) is the volume of the solution. Solving for V1, we get V1 = M2V2 / M1 = (0.456 M * 197.4 mL) / 1.27 M = 71.03 mL. Therefore, 71.03 mL of 1.27 M HCl is needed to prepare 197.4 mL of 0.456 M HCl.

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Answer: Cis-1-bromo-4-tert-butylcyclohexane would undergo faster elimination reaction.

Explanation:

The two primary requirements for an E-2 elimination reaction are:

1.There must be availability of β-hydrogens that is presence of hydrogen on the carbon next to the leaving group.

2.The hydrogen and leaving group must have a anti-periplanar position .

Any substrate which would follow the above two requirements can give elimination reactions.

For the structure of trans-1-bromo-4-tert-butylcyclohexane and cis-1-bromo-4-tert-butylcyclohexane  to be stable it  must have the tert-butyl group in the equatorial position as it is a bulky group and at equatorial position it would not repel other groups. If it is kept on the axial position it would undergo 1,3-diaxial interaction and would destabilize the system and that structure would be unstable.

Kindly find the structures of trans-1-bromo-4-tert-butylcyclohexane and cis-1-bromo-4-tert-butylcyclohexane in attachment.

The cis- 1-bromo-4-tert-butylcyclohexane has the leaving group and β hydrogens in anti-periplanar position so they can give the E2 elimination reactions easily.

The trans-1-bromo-4-tert-butylcyclohexane  does not have the leaving group and βhydrogen in anti periplanar position so they would not give elimination reaction easily.

so only the cis-1-bromo-4-tert butyl cyclohexane would give elimination reaction.

Final answer:

Trans-1-bromo-4-tert-butylcyclohexane is expected to undergo E2 elimination faster than cis-1-bromo-4-tert-butylcyclohexane due to less steric hindrance.

Explanation:

In determining the rate of E2 elimination, the trans-1-bromo-4-tert-butylcyclohexane would undergo E2 elimination faster than the cis-1-bromo-4-tert-butylcyclohexane. This is due to the larger degree of steric hindrance in the case of the cis isomer.

In trans-1-bromo-4-tert-butylcyclohexane, the bromine is at the equatorial position while the tert-butyl group is axial. It forms a structure that allows the compound to experience less steric hindrance with bromine in a more favorable position for leaving.

In comparison, cis-1-bromo-4-tert-butylcyclohexane has a bromine and tert-butyl group both at equatorial positions. This causes steric hindrance, and in turn, slows down the E2 elimination rate. Despite the more stable conformation, the bromine is not well-oriented for a leaving group in E2 elimination.

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Answer:

To 318.18 mL would you need to dilute 20.0 mL of a 1.40 M solution of LiCN to make a 0.0880 M solution of LiCN

Explanation:

Dilution is the reduction of the concentration of a chemical in a solution and consists simply of adding more solvent.

In a dilution the amount of solute does not vary. But as more solvent is added, the concentration of the solute decreases, as the volume (and weight) of the solution increases.

In a solution it is fulfilled:

Ci* Vi = Cf* Vf

where:

• Ci: initial concentration
• Vi: initial volume
• Cf: final concentration
• Vf: final volume

In this case:

• Ci= 1.40 M
• Vi= 20 mL
• Cf= 0.088 M
• Vf= ?

Replacing:

1.40 M* 20 mL= 0.088 M* Vf

Solving:

Vf= 318.18 mL

To 318.18 mL would you need to dilute 20.0 mL of a 1.40 M solution of LiCN to make a 0.0880 M solution of LiCN

Final answer:

To make a 0.0880 M solution of LiCN, you would need to dilute 20.0 mL of the 1.40 M solution to a final volume of 318.18 mL.

Explanation:

To dilute a solution, you can use the formula:

M1V1 = M2V2

where M1 and V1 are the initial molarity and volume, and M2 and V2 are the final molarity and volume. Rearranging the formula, we can solve for V2:

V2 = (M1 · V1) / M2

Plugging in the values given:

V2 = (1.40 M · 20.0 mL) / 0.0880 M = 318.18 mL

To make a 0.0880 M solution of LiCN, you would need to dilute 20.0 mL of the 1.40 M solution to a final volume of 318.18 mL.

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Answer:

8.0 mol O₂

Explanation:

Let's consider the complete combustion reaction of C₉H₁₂.

C₉H₁₂ + 12 O₂ → 9 CO₂ + 6 H₂O

The molar ratio of C₉H₁₂ to O₂ is 1:12. The moles of O₂ required to react with 0.67 moles of C₉H₁₂ are:

0.67 mol C₉H₁₂ × (12 mol O₂/1 mol C₉H₁₂) = 8.0 mol O₂

8.0 moles of O₂ are required to completely react with 0.67 moles of C₉H₁₂.

Answer:

To react with 0.67 moles C9H12 we need 8.04 moles of O2

Explanation:

Step 1: Data given

Number of moles C9H12 = 0.67 moles

Step 2: The balanced equation

C9H12 + 12O2 → 9CO2 + 6H2O

Step 3: Calculate moles of O2 required

For 1 mol C9H12 we need 12 moles of O2 to produce 9 moles of CO2 and 6 moles of H2O

For 0.67 moles of C9H12 we need 12 *0.67 = 8.04 moles of O2

To produce 9*0.67 = 6.03 moles of CO2 and 6*0.67 = 4.02 moles H2O

To react with 0.67 moles C9H12 we need 8.04 moles of O2

Answer:

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