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Andreyy89
3 years ago
8

A generator does work on an electric heater by forcing an electric current through it. Suppose 1 kJ of work is done on the heate

r and in turn 1 kJ of energy as heat is transferred to its surroundings. What is the change in internal energy of the heater?
Chemistry
2 answers:
alekssr [168]3 years ago
8 0

Answer:

-2 kJ.

Explanation:

Given:

∆Q = 1 kJ

W = 1 kJ

Using the first law of thermodynamics in application to conservation of energy,

∆U = ∆Q - W

Where,

∆U = change in internal energy

∆Q = heat supplied to the system

W = workdone by the system

∆U = -1 - 1

= -2 kJ.

azamat3 years ago
3 0

Answer:

The change in internal energy of the heater is 0 kJ

Explanation:

∆U = Q - W

Q is quantity of heat transferred = 1 kJ

W is work done on the heater = 1 kJ

Change in internal energy (∆U) = 1 - 1 = 0 kJ

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A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli
vladimir1956 [14]

Answer:

109.09°C

Explanation:

Given that:

the capacity of the cooling car system = 5.6 gal

volume of solute = volume of the water; since a 50/50 blend of engine coolant and water (by volume) is used.

∴ \frac{5.60}{2}gallons = 2.80 gallons

Afterwards, the mass of the solute and the mass of the water can be determined as shown below:

mass of solute = (M__1}) = Density*Volume

                          = 1.1g/mL *2.80*\frac{3785.41mL}{1gallon}

                         = 11659.06grams

On the other hand; the mass of water = (M__2})= Density*Volume

                         = 0.998g/mL *2.80*\frac{3785.41mL}{1gallon}

                        = 10577.95 grams

Molarity = \frac{massof solute*1000}{molarmassof solute*massofwater}

              =  \frac{11659.06*1000}{62.07*10577.95}

              = 17.757 m

              ≅ 17.76 m

∴  the boiling point of the solution is calculated using the  boiling‑point elevation constant for water and the Molarity.

\Delta T_{boiling} = k_{boiling}M

where,

k_{boiling} = 0.512 °C/m

\Delta T_{boiling} =  100°C + 17.56 × 0.512

              = 109.09 °C

6 0
4 years ago
Which of the following is an ion with a correct charge ?<br><br> O 4-<br> Ne8+<br> Ca2+<br> Na1-
KengaRu [80]

Answer:

Ca2+ is an ion with a correct charge.

8 0
3 years ago
Omg pls help i dunno what the frick frack this is
snow_lady [41]

Answer:

1. Mass of KCl produced = 774.8 g of KCl

2. Mass of KNO₃ produced = 13.837g

3. Moles of NaOH made = 0.846 moles

4. Moles of LiCl produced = 0.846 moles

5. Moles of CO₂ produced = 207.6 moles

Explanation:

1. From the equation of reaction, 1 mole of ZnCl₂ produces, 2 moles of KCl.

5.02 moles of ZnCl₂ will produce, 2 × 5.02 moles of KCl = 10.4 moles of KCl

Molar mass of KCl = (39 + 35.5) g/mol = 74.5 g/mol

10.4 moles of KCl = 10.4 × 74.5 g

Mass of KCl produced = 774.8 g of KCl

2. Mole ratio of KNO₃ and KOH = 1:1

O.137 moles of KOH will produce 0.137 moles of KNO₃

Molar mass of KNO₃ = 101 g/mol

Mass of KNO₃ produced = 0.137 × 101 g = 13.837g

3. Molar mas of Ca(OH)₂ = 74.0 g

Moles of Ca(OH)₂ in 31.3 g = 31.3/74.0 = 0.423 moles of Ca(OH)₂

Mole ratio of NaOH and Ca(OH)₂ in the reaction = 2 : 1

Moles of NaOH made = 2 × 0.423 = 0.846 moles

4. Molar mass of MgCl₂ = 95.0 g

Moles of MgCl₂ in 40.2 g = 40.2/95.0 = 0.423 moles

From the reaction equation, mole ratio of MgCl₂ and LiCl = 1:2

Moles of LiCl produced = 2 × 0.423 = 0.846 moles

5. From the equation of reaction, 1 mole of C₆H₁₀O₅ produces 6 moles of cO₂

34.6 moles of C₆H₁₀O₅ will produce 34.6 × 6 moles of CO₂

Moles of CO₂ produced = 207.6 moles

4 0
3 years ago
What do you use to mesure wind
mezya [45]

Answer:

Either a windmill or an anemometer

Explanation:

A windmill is usually used as a harvester for wind energy. Anemometers measure wind speed and determine wind direction. Using these sets of data, meteorologists can calculate wind pressure.

7 0
3 years ago
Octane has a density of 0.692 g/ml at 20oc. how many grams of o2 are required to burn 15.0 gal of c8h18 (the average capacity of
Hatshy [7]

Octane has a density of 0.692 g/ml at 20°C. Grams of O₂ required to burn 15.0 gal of C₈H₁₈ is 1.37868×10⁵g.

Octane is a hydrocarbon having eight carbon atoms and have single bonds only.

Given,

Density of octane = 0.692g/ml

Temperature = 20°C = 293K

Amount of Octane present = 15gal

Molar mass of octane = 114g

We know that,

1 gal = 3.78541L = 3785.4ml

Hence, 15 gal = 56781 ml

Now let's calculate the Mass of octane required:

Mass of octane = 0.692 x 56781

Mass of octane = 39292.45 g

According to the given equation,

C₈H₁₈ + 12.5O₂ ---------> 8CO₂ + 9H₂O

Also, 114g of octane needs-------> 400g of oxygen

39292.45g of octane needs ---------> 137868.24g of oxygen

Hence, oxygen needed to burn octane is 1.37868×10⁵g

Learn more about Octane here, brainly.com/question/21268869

#SPJ4

4 0
2 years ago
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