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VikaD [51]
3 years ago
10

Use the worked example above to help you solve this problem. An electrical heater is operated by applying a potential difference

of 49.4 V to nichrome wire of total resistance 8.04 Ω. (a) Find the current carried by the wire and the power rating of the heater. I = 6.14 Correct: Your answer is correct. A P = 303.5 Correct: Your answer is correct. W (b) Using this heater, how long would it take to heat 2.10 103 moles of diatomic gas (e.g., a mixture of oxygen and nitrogen, or air) from a chilly 10°C to 25°C? Take the molar specific heat at constant volume of air to be 5 2 R. s (c) How many kilowatt-hours of electricity are used during the time calculated in part (b) and at what cost, at $0.14 per kilowatt-hour? U = kWh cost $ EXERCISEHINTS: GETTING STARTED | I'M STUCK! (a) A hot-water heater is rated at 4.69 103 W and operates at 2.40 102 V. Find the resistance in the heating element and the current. R = Ω I = A (b) How long does it take to heat 125 L of water from 25.6°C to 54.0°C, neglecting conduction and other losses? (The specific heat of water is 4.186 J/g · K.) s (c) How much does it cost at $0.14/kWh? $

Chemistry
1 answer:
ioda3 years ago
5 0

Answer:

Part A:

a) Current = 6.14 A; Power = 303.5 W

b) time = 0.6 hours

c) cost = $ 0.025

Part 2:

a) Resistance = 12.28 ohms; Current = 19.54 A

b) time = 0.88 hours

c) cost = $0.578

Explanation:

The explanation is found in the attachments below:

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In a aqueous solution of 4-chlorobutanoic acid , what is the percentage of -chlorobutanoic acid that is dissociated
lord [1]

Answer:

Explanation:

Let assume that the missing aqueous solution of 4-chlorobutanoic acid = 0.76 M

Then, the dissociation of 4-chlorobutanoic acid can be expressed as:

\mathsf{C_3H_6ClCO_2H }          ⇄      \mathsf{C_3H_6ClCO_2^-}      +      \mathsf{H^+}

The ICE table can be computed as:

                   \mathsf{C_3H_6ClCO_2H }          ⇄      \mathsf{C_3H_6ClCO_2^-}      +      \mathsf{H^+}

Initial              0.76                                 -                           -

Change            -x                                  +x                         +x

Equilibrium   0.76 - x                              x                          x  

K_a = \dfrac{[\mathsf{C_3H_6ClCO_2^-}] [\mathsf{H^+}]}{\mathsf{[C_3H_6ClCO_2H ]}}

K_a = \dfrac{[x] [x]}{ [0.76-x]}

where:

K_a = 3.02*10^{-5}

3.02*10^{-5} = \dfrac{x^2}{ [0.76-x]}

however, the value of x is so negligible:

0.76 -x = 0.76

Then:

3.02*10^{-5}*0.76 = x^2

x=\sqrt{3.02*10^{-5}*0.76 }

x = 0.00479 M

∴

x = \mathsf{[C_3H_6ClCO_2^-] = [H^+]=} 0.00479 M

\mathsf{C_3H_6ClCO_2H }  = (0.76 - 0.00479) M

= 0.75521 M

Finally, the percentage of the acid dissociated is;

= ( 0.00479 / 0.76) × 100

= 0.630 M

7 0
3 years ago
What is the molar concentration of H2SO4 in a solution made by reacting 188.9 mL of 2.086 M H2SO4 with 269.3 mL of 0.4607 M NaOH
MrRissso [65]

Answer:

0.7246 M

Explanation:

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

For H_2SO_4 :

Molarity = 2.086 M

Volume = 188.9 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 188.9×10⁻³ L

Thus, moles of H_2SO_4 :

Moles=2.086 \times {188.9\times 10^{-3}}\ moles

Moles of H_2SO_4  = 0.39405 moles

For NaOH :

Molarity = 0.4607 M

Volume = 269.3 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 269.3×10⁻³ L

Thus, moles of NaOH :

Moles=0.4607 \times {269.3\times 10^{-3}}\ moles

Moles of NaOH  = 0.1241 moles

According to the given reaction:

H_2SO_4_{(aq)}+2NaOH_{(aq)}\rightarrow Na_2SO_4_{(aq)}+2H_2O_{(aq)}

1 moles of H_2SO_4 react with 2 moles of NaOH to form 1 mole of sodium sulfate.

Thus,

2 moles of NaOH react with 1 mole of H_2SO_4

1 mole of NaOH react with 1/2 mole of H_2SO_4

0.1241 moles of NaOH react with (1/2)×0.1241 mole of H_2SO_4

Moles of H_2SO_4 that got reacted = 0.06205 moles

Unreacted moles = Total moles - Moles that got reacted = 0.39405 - 0.06205 moles = 0.332 moles

Total volume = 188.9×10⁻³ L + 269.3×10⁻³ L = 458.2×10⁻³ L

Concentration of H_2SO_4 :

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Molarity_{H_2SO_4}=\frac{0.332}{458.2\times 10^{-3}}

<u>Concentration of H_2SO_4 = 0.7246 M</u>

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4 years ago
In the compound KHSO4, there is an ionic bond between the
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the ionic bond would be between the potassium, and hydrogen sulfate.
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Which of the following appliances increase CFC's in the atmosphere?
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Answer:

Television cause it has a satellite

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3 years ago
PLEASE HELP ME WITH THIS QUESTION!!!!
elena-s [515]
I believe that would be D] CuSO4(s). 
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