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OleMash [197]
3 years ago
11

Calculate the approximate enthalpy of the reaction in joules. Estimate that 1.0 mL of vinegar has the same thermal mass as 1.0 m

L of water. It takes 4.2 joules to change 1.0 grm (1.0 mL) of water 1.0 Celcius. Is it endotherrmic or exothermic? Volume of vinegar 25 mL, Initial temp of vinegar 17 Celcius and final temp is 14 Celcius.
Chemistry
1 answer:
Nesterboy [21]3 years ago
5 0

Explanation:

The given data is as follows.

   Density of vinegar = 1.0 g/ml

   Specific heat capacity = 4.25 J/g ^{o}C

   T_{1} = 17 ^{o}C,  and   T_{2} = 14 ^{o}C

Relation between enthalpy and specific heat is as follows.

                   \Delta H = mC \Delta T

Hence, putting the values into the above formula as follows.

          \Delta H = mC \Delta T

                  = 25 \times 1.0 \times 4.25 J/g ^{o}C \times -3^{o}C          (as density = \frac{mass}{volume})

                               = - 315 J

Thus, we can conclude that the enthalpy of reaction is -315 J.

As the value is negative so, it means that heat is releasing. Hence, the reaction is exothermic in nature.

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What is the precipitation reaction of sodium nitrate and lead (II)?
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2 years ago
In one step in the synthesis of the insecticide Sevin, naphthol reacts with phosgene as shown.
Serhud [2]

Answer:

Explanation:

Chemical equation:

C₁₀H₈O + COCl₂  → C₁₁H₇O₂Cl + HCl

A. How many kilograms of C₁₁H₇O₂Cl form from 2.5×10*2 kg of naphthol?

Given data:

Mass of naphthol = 2.5 ×10² kg ( 250×1000 = 250000 g)

Mass of C₁₁H₇O₂Cl = ?

Solution:

Number of moles of naphthol = mass/ molar mass

Number of moles of naphthol = 250000 g/ 144.17 g/mol

Number of moles of naphthol = 1734.1 mol

Now we will compare the moles of naphthol with C₁₁H₇O₂Cl.

                     C₁₀H₈O       :         C₁₁H₇O₂Cl

                        1               :                1

                       1734.1         :             1734.1

Mass of C₁₁H₇O₂Cl:

Mass = number of moles × molar mass

Mass = 1734.1 mol × 206.5 g/mol

Mass = 358091.65 g

Gram to kilogram:

1 kg×358091.65 g/ 1000 g  = 358.1 kg

B. If 100. g of naphthol and 100. g of phosgene react, what is the theoretical yield of C11H7O2Cl?

Given data:

Mass of naphthol = 100 g

Mass of COCl₂ = 100 g

Theoretical yield of C₁₁H₇O₂Cl = ?

Solution:

Number of moles of naphthol:

Number of moles of naphthol = mass/ molar mass

Number of moles of naphthol = 100 g/ 144.17 g/mol

Number of moles of naphthol = 0.694 mol

Number of moles of phosgene:

Number of moles  = mass/ molar mass

Number of moles =  100 g/ 99 g/mol

Number of moles = 1.0 mol

Now we will compare the moles of naphthol and phosgene with C₁₁H₇O₂Cl.

                     C₁₀H₈O        :         C₁₁H₇O₂Cl

                        1                :                1

                       0.694        :              0.694

                    COCl₂          :             C₁₁H₇O₂Cl

                        1                :                1

                       1.0              :              1.0

The number of moles of C₁₁H₇O₂Cl produced by C₁₀H₈O are less so it will limiting reactant and limit the yield of  C₁₁H₇O₂Cl.

Mass of C₁₁H₇O₂Cl:

Mass = number of moles × molar mass

Mass =  0.694 mol × 206.5 g/mol

Mass = 143.3 g

Theoretical yield  =  143.3 g

C. If the actual yield of C11H7O2Cl in part b is 118 g, what is the percent yield?

Given data:

Actual yield of C₁₁H₇O₂Cl = 118 g

Theoretical yield = 143.3 g

Percent yield = ?

Solution:

Formula :

Percent yield = actual yield / theoretical yield × 100

Now we will put the values in formula.

Percent yield = 118 g/ 143.3 g × 100

Percent yield = 0.82 × 100

Percent yield = 82%

5 0
3 years ago
An unknown aqueous metal analysis yielded a detector response of 0.255. When 1.00 mL of a solution containing 100.0 ppm of the m
AVprozaik [17]

Answer:

1.022ppm is the unknown concentration of the metal

Explanation:

Based on Lambert-Beer law, the increasing in signal of a detector is directly proportional to its concentration.

The unknown concentration (X) produces a signal of 0.255

99mL * X + 1mL * 100ppm / 100mL produces a signal of 0.502

0.99X + 1ppm produce 0.502, thus, X is:

0.255 * (0.99X + 1 / 0.502) =

X = 0.503X + 0.508

0.497X = 0.508

X =

1.022ppm is the unknown concentration of the metal

3 0
2 years ago
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