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Ksenya-84 [330]
3 years ago
12

A 110.0-g sample of metal at 82.00°C is added to 110.0 g of H2O(l) at 27.00°C in an insulated container. The temperature rises t

o 30.56°C. Neglecting the heat capacity of the container, what is the specific heat of the metal? The specific heat of H2O(l) is 4.18 J/(g ∙ °C). Group of answer choices 4.18 J/(g ∙ °C) 60.4 J/(g ∙ °C) 0.289 J/(g ∙ °C) 0.289 J/(g ∙ °C) 14.4 J/(g ∙ °C)
Chemistry
2 answers:
erastovalidia [21]3 years ago
7 0

Answer:

The specific heat of the metal is 0.289 J/g°C

Explanation:

Step 1: Data given

Mass of metal = 110.0 grams

Temperature of the metal = 82.00 °C

MAss of water = 110.0 grams

Temperature of the water = 27.00 °C

The final temperature = 30.56 °C

The specific heat of H2O is 4.18 J/(g°C).

Step 2: Calculate the specific heat of the metal

Heat lost = heat gained

Qlost = - Qgained

Qmetal = - Qwater

Q =m*c*ΔT

m(metal)*c(metal)*ΔT(metal) = -m(water) * c(water) *ΔT(water)

⇒with m(metal) = the mass of the metal = 110.0 grams

⇒with c(metal) = the specific heat of the metal = TO BE DETERMINED

⇒with ΔT(metal) = the change of temperature of the metal = T2 - T1 = 30.56 - 82.00 °C= -51.44 °C

⇒with m(water) = the mass of water = 110.0 grams

⇒with c(water) = the specific heat of water = 4.18 J/g°C

⇒with ΔT(water) = the change of temperature of water = T2 - T1 = 30.56 °C - 27.00 °C = 3.56 °C

110.0 * c(metal) * -51.44 = -110.0 * 4.18 * 3.56

c(metal) = 0.289 J/g°C

The specific heat of the metal is 0.289 J/g°C

lina2011 [118]3 years ago
7 0

Answer:

Cp_{metal}=0.289\frac{J}{g^oC}

Explanation:

Hello,

In this case, as the water is cold (lower initial temperature) and the metal is hot (higher initial temperature), the heat lost by the metal is gained by the water to attain an equilibrium temperature of 30.56 °C, this in an equation turns out:

-\Delta H_{metal}=\Delta H_{water}

In such a way, in terms of masses, heat capacities and temperatures we have:

-m_{metal}Cp_{metal}(T_{eq}-T_{metal})=m_{water}Cp_{water}(T_{eq}-T_{water})

Hence, solving for the heat capacity of the metal:

Cp_{metal}=\frac{m_{water}Cp_{water}(T_{eq}-T_{water})}{-m_{metal}Cp_{metal}(T_{eq}-T_{metal})}

Thus, with the given data we obtain:

Cp_{metal}=\frac{110.0g*4.18\frac{J}{g^oC} (30.56^oC-27.00^oC)}{-110.0g(30.56^oC-82.00^oC)}\\\\Cp_{metal}=0.289\frac{J}{g^oC}

Best regards.

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A sample of solid sodium hydroxide, weighing 13.20 grams is dissolved in deionized water to make a solution. What volume in mL o
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<h3>Answer:</h3>

2.809 L of H₂SO₄

<h3>Explanation:</h3>

Concept tested: Moles and Molarity

In this case we are give;

Mass of solid sodium hydroxide as 13.20 g

Molarity of H₂SO₄ as 0.235 M

We are required to determine the volume of H₂SO₄ required

<h3>First: We need to write the balanced equation for the reaction.</h3>
  • The reaction between NaOH and H₂SO₄ is a neutralization reaction.
  • The balanced equation for the reaction is;

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

<h3>Second: We calculate the umber of moles of NaOH used </h3>
  • Number of moles = Mass ÷ Molar mass
  • Molar mass of NaOH is 40.0 g/mol
  • Therefore;

Moles of NaOH = 13.20 g ÷ 40.0 g/mol

                          = 0.33 moles

<h3>Third: Determine the number of moles of the acid, H₂SO₄</h3>
  • From the equation, 2 moles of NaOH reacts with 1 mole of H₂SO₄
  • Therefore, the mole ratio of NaOH: H₂SO₄ is 2 : 1.
  • Thus, Moles of H₂SO₄ = moles of NaOH × 2

                                    = 0.33 moles × 2

                                   = 0.66 moles of H₂SO₄

<h3>Fourth: Determine the Volume of the acid, H₂SO₄ used</h3>
  • When given the molarity of an acid and the number of moles we can calculate the volume of the acid.
  • That is; Volume = Number of moles ÷ Molarity

In this case;

Volume of the acid = 0.66 moles ÷ 0.235 M

                                = 2.809 L

Therefore, the volume of the acid required to neutralize the base,NaOH is 2.809 L.

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