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shepuryov [24]
2 years ago
6

A 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C. If the final temperature of water is 42.0 °C, what

was the initial temperature of the copper piece? (5 points)
Specific heat of copper = 0.39 J/g °C

Group of answer choices

322 °C

345 °C

356 °C

364 °C
Chemistry
1 answer:
ludmilkaskok [199]2 years ago
8 0

The initial temperature of the copper piece if a 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C is 345.5°C

<h3>How to calculate temperature?</h3>

The initial temperature of the copper metal can be calculated using the following formula on calorimetry:

Q = mc∆T

mc∆T (water) = - mc∆T (metal)

Where;

  • m = mass
  • c = specific heat capacity
  • ∆T = change in temperature

According to this question, a 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C. If the final temperature of water is 42.0 °C, the initial temperature of the copper is as follows:

400 × 4.18 × (42°C - 24°C) = 240 × 0.39 × (T - 24°C)

30,096 = 93.6T - 2246.4

93.6T = 32342.4

T = 345.5°C

Therefore, the initial temperature of the copper piece if a 240.0 gram piece of copper is dropped into 400.0 grams of water at 24.0 °C is 345.5°C.

Learn more about temperature at: brainly.com/question/15267055

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When 2.69 g 2.69 g of a nonelectrolyte solute is dissolved in water to make 345 mL 345 mL of solution at 26 °C, 26 °C, the solut
Gre4nikov [31]

Answer:

The molar concentration of this solution is 0.0463 mol/L

Explanation:

Step 1 : Data given

Mass of a nonelectrolyte solute = 2.69 grams

Volume of water = 345 mL = 0.345 L

Temperature = 26.0°CC = 273 + 26 = 299 K

The osmotic pressure = 863 torr

⇒ 863torr /760 = 1.13553 atm

Step 2: Calculate the molar concentration of this solution

Π = i*M*R*T

⇒with Π = the osmotic pressure = 1.13553 atm

⇒with i = the van't Hoff factor of the nonelectrolyte solute = 1

⇒with M = the molar concentration = TO BE DETERMINED

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temperature = 299 K

1.13553 atm = 1 * M * 0.08206 L*atm/mol*K * 299 K

M = 1.13553 / (0.08206*299)

M = 0.0463 mol/L

The molar concentration of this solution is 0.0463 mol/L

5 0
3 years ago
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postnew [5]
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3 years ago
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A gas is contained in a thick-walled balloon. When the pressure changes from 45.0 bar to 96.0 bar, the volume changes from 1.20
notsponge [240]

P1 * V1 ÷ T1 = P2 * V2 ÷ T2  

45 * 1.20 ÷ 314 = 96 * V2 ÷ 420  

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Lead(II) nitrate is added slowly to a solution that is 0.0800 M in Cl− ions. Calculate the concentration of Pb2+ ions (in mol /
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Answer:

[Pb^{2+}]=3.9 \times 10^{-2}M

this is the concentration required to initiate precipitation

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PbCl_2  ⇄ Pb^{2+}+2Cl^-

Precipitation starts when ionic product is greater than solubility product.

Ip>Ksp

Precipitation starts only when solution is supersaturated because solution become supersaturated then it does not stay in this form and precipitation starts itself only solution become saturated.

This usually happens when two solutions containing separate sources of cation and anion are mixed together and here also we are mixing lead (||)nitrate solution(source of lead(||)) into the Cl- solution.

Ip=[Pb^{2}][2Cl^-]^2=Ksp

Ksp=2.4\times 10^{-4}

lets solubility=S

[Pb^{2+}] = S

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Ksp=[Pb^{2+}]\times [Cl^-]^2

Ksp=S \times (2S)^2

Ksp=4S^3

S=\sqrt[3]{\frac{Ksp}{4} }

S=3.9\times 10^{-2}

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Answer: 1.) His action is not justified because he could have done something else like talking with someone.

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