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nadya68 [22]
3 years ago
5

A 6.22-kg piece of copper metal is heated from 20.5 °C to 324.3 °C. The specific heat of Cu is 0.385 Jg-1°C-1.a.Calculate the he

at absorbed (in kJ) by the metal.b.How close is this heat capacity to the expected heat capacity at the classical limit?
Chemistry
1 answer:
iren2701 [21]3 years ago
8 0

Answer:

a) 727.5 kJ

Explanation:

Step 1: Data given

Mass of the piece of copper = 6.22 kg

Initial temperature of the copper = 20.5 °C

Final temperature of the copper = 324.3 °C

Specific heat of copper = 0.385 J/g°C

Step 2:

Q = m*c*ΔT

⇒ with Q = heat transfer (in J)

⇒ with m = the mass of the object (in grams) = 6220 grams

⇒ with c = the specific heat capacity = 0.385 J/g°C

⇒ with ΔT = T2 -T1 = 324.3 -  20.5 = 303.8

Q = 6220 grams * 0.385 J/g°C * 303.8 °C

Q = 727509.9 J = 727.5 kJ

b) This heat capacity is the heat capacity given for a copper at a temperature of 25°C

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For the reaction 2Co3+(aq)+2Cl−(aq)→2Co2+(aq)+Cl2(g). E∘=0.483 V what is the cell potential at 25 ∘C if the concentrations are [
sveta [45]

Explanation:

The given data is as follows.

     E^{o} = 0.483,     [Co^{3+}] = 0.173 M,

     [Co^{2+}] = 0.433 M,     [Cl^{-}] = 0.306 M,

     P_{Cl_{2}} = 9.0 atm

According to the ideal gas equation, PV = nRT

or,             P = \frac{n}{V}RT    

Also, we know that

                Density = \frac{mass}{volume}

So,         P = MRT

and,          M = \frac{P}{RT}

                    = \frac{9.0 atm}{0.0820 L atm/mol K \times 298 K}

                    = \frac{9.0}{24.436}

                    = 0.368 mol/L

Now, we will calculate the cell potential as follows.

          E = E^{o} - \frac{0.0591}{n} log \frac{[Co^{2+}]^{2}[Cl_{2}]}{[Co^{3+}][Cl^{-}]^{2}}

             = 0.483 - \frac{0.0591}{2} log \frac{(0.433)^{2}(0.368)}{(0.173)(0.306)^{2}}

             = 0.483 - 0.02955 log \frac{0.0689}{0.0162}

             = 0.483 - 0.02955 \times 0.628

             =  0.483 - 0.0185

             = 0.4645 V

Thus, we can conclude that the cell potential of given cell at 25^{o}C is 0.4645 V.

4 0
4 years ago
You are asked to write observations about a 100g sample of Nitrogen-16 before it decays, but you're running late. In order to ma
san4es73 [151]

Answer:

You cannot make observations if you are 57 seconds late into the lab.

Explanation:

The atomic nucleus can split by decay into 2 or more particles as a result of the instability of its atomic nucleus due to the fact that radioactive elements possess an unstable atomic nucleus.

Now, the primary particles which are emitted by radioactive elements in order to make them decay are alpha, beta & gamma particles.

The half life equation is;

N_t = N₀(½)^(t/t_½)

Where:

t = duration of decay

t_½ = half-life

N₀ = number of radioactive atoms initially

N_t = number of radioactive atoms remaining after decay over time t

We are given;

t = 57 secs

N₀ = 100 g

Now, half life of Nitrogen-16 from online sources is 7.2 seconds. t_½ = 7.2

Thus;

N_t = 100(1/2)^(57/7.2)

N_t = 0.4139g

We are told that In order to make observations, you require at least .5g of material.

The value of N_t you got is less than 0.5g, therefore you cannot make observations if you are 57 seconds late.

5 0
3 years ago
What quantity of sodium azide in grams is required to fill a 56.0 liters air bag with nitrogen gas at 1.00 atm and exactly 0 °C:
Margarita [4]

Answer:

108.6 g

Explanation:

  • 2NaN₃(s) → 2Na(s) + 3N₂(g)

First we use the <em>PV=nRT formula</em> to <u>calculate the number of nitrogen moles</u>:

  • P = 1.00 atm
  • V = 56.0 L
  • n = ?
  • R = 0.082 atm·L·mol⁻¹·K⁻¹
  • T = 0 °C ⇒ 0 + 273.2 = 273.2 K

<u>Inputting the data</u>:

  • 1.00 atm * 56.0 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 273.2 K
  • n = 2.5 mol

Then we <u>convert 2.5 moles of N₂ into moles of NaN₃</u>, using the <em>stoichiometric coefficients of the balanced reaction</em>:

  • 2.5 mol N₂ * \frac{2molNaN_3}{3molN_2} = 1.67 mol NaN₃

Finally we <u>convert 1.67 moles of NaN₃ into grams</u>, using its <em>molar mass</em>:

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6 0
3 years ago
How many mL of a .132 M aqueous solution of sodium chloride, NaCl, must be taken to obtain 3.59 grams of the salt?
Sever21 [200]

Answer:

465mL

Explanation:

Volume of a solution, V =Mass of substance, m/(Molarity of the solution of the substance, M × molar mass of the substance, M.m)

Given in the question,

M=.132M

M.m=23+35.5 = 58.5g/mol

m=3.59g

V= 3.59/(.132×58.5)

V = 0.465L

Volume in mL = volume in L × 1000

= 0.465 × 1000 = 465mL

Therefore, 465mL of a .132M aqueous solution of sodium chloride, NaCl, must be taken to obtain 3.59 grams of the salt

4 0
4 years ago
Assignment Directions:
Alja [10]

Answer:

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I was not surprised to see that, but I thought it was going to be injuries as well.

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Explanation:

This is not 300 words, it is 244. You can it. Make sure to change it so it sounds like you wrote it.  

8 0
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