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grigory [225]
2 years ago
14

Two lamps (with a resistance of 10 Ohms each) are connected in a parallel circuit and are supplied with 100- volts of power. whi

ch of the following statements are true of the voltage that each lamp receives
a. Each lamp will recieve 50 volts

B. Each lamp will recieve 100 volts

C. The first lamp will recieve more voltage than the second lamp
Chemistry
1 answer:
adelina 88 [10]2 years ago
6 0

In the parallel circuit, each lamp will receive 100 volts.

<h3>Equivalent resistance of the parallel circuit</h3>

The equivalent resistance of the parallel circuit is the inverse sum of the resistance of the two resistors in the circuit. This is as a result of different current flowing in each resistor parallel circuit arrangement.

The equivalent resistance is calculated as follows;

1/R = 1/10 + 1/10

1/R = 2/10

R = 10/2

R = 5 ohms

<h3>Current in each resistor</h3>

I = V/R

I = 100/10 = 10 A

<h3>Voltage in each resistor</h3>

V = IR

V = 10 X 10 = 100 V

Generally, for a circuit in parallel connection, same voltage will flow in each resistor of the circuit while different current will flow in the resistors.

Thus, the correct statement is, "Each lamp will receive 100 volts".

Learn more about circuit in parallel arrangement here: brainly.com/question/19929102

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Identify structure(s) for all constitutional isomers with the molecular formula C4H8 that have one double bond. Select all that
Alchen [17]

Answer:

a) But-1-ene

b) E-But-2-ene

c) Z-But-2-ene

d) 2-Methylpropene

Explanation:

In this case, if we want to draw the <u>isomers</u>, we have to check the<u> formula </u>C_4_H8 in this formula we can start with a linear structure with 4 carbons. We also know that we have a double bond, so we can put this double bond between carbons 1 and 2 and we will obtain <u>But-1-ene.</u>

<u />

For the next isomer, we can move the double bond to carbons 2 and 3. When we do this can have two structures. When the methyl groups are placed on the same side we will obtain <u>Z-But-2-ene</u>. When the methyls groups are placed on opposite sides we will obtain <u>E-But-2-ene.</u>

<u />

Finally, we can use a linear structure of three carbons with a methyl group in the middle with a double bond, and we will obtain <u>2-Methylpropene.</u>

<u />

See figure 1 to further explanations.

I hope it helps!

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8 0
3 years ago
Consider the above unbalanced equation. What volume of CO2 is produced at 270. mm Hg and 38.5°C when 0.820 g of C4H8 reacts with
irinina [24]

Answer:

The volume of CO2 is 4.20 L

Explanation:

Step 1: Data given

Pressure = 270 mm Hg =  260 /760 = 0.355263 atm

Temperature : 38.5 °C = 311.65 K

Mass of C4H8 = 0.820 grams

Step 2: The balanced equation

C4H8 + 6O2 → 4CO2 + 4H2O

Step 3: Calculate moles C4H8

Moles C4H8 = mass C4H8 / molar mass C4H8

Moles C4H8 = 0.820 grams / 56.11 g/mol

Moles C4H8 = 0.0146 moles

Step 4: Caalculate moles CO2

For 1 mol C4H8 we need 6 moles O2 to produce 4 moles CO2 and 4 moles H2O

For 0.0146 moles we'll have 4*0.0146 = 0.0584 moles CO2

Step 6: Calculate Volume CO2

p*V = n*R*T

V = (n*R*T) /p

⇒ with V = the volume of CO2 = TO BE DETERMINED

⇒ with n = the moles of CO2 = 0.0584 moles

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

⇒ with T = The temperature = 311.65 K

⇒ with p = the pressure = 0.355263 atm

V = (0.0584 * 0.08206 * 311.65) / 0.355263

V = 4.20 L

The volume of CO2 is 4.20 L

8 0
3 years ago
Question 14(multiple choice worth 2 points) a student had a sample of pure water, and added an unknown substance to it. the stud
pantera1 [17]
The answer is A its basic

5 0
3 years ago
The procedure for testing your unknown solution in this week's lab is identical to the procedure which you conducted in Week 1.
lawyer [7]

Answer:

The correct answer is option C. The unknown solution definitely has Hg22+ present.

Explanation:

In the analysis of group 1 metal cation, the unknown solution is treated with sufficient quantity of 6 M HCl solution and if group 1 metal cations are present then white precipitate of Agcl, PbCl2 or Hg2Cl2 is formed. The precipitate of PbCl2 is soluble in hot water but the other two remains insoluble after treating with hot water. Precipitate of AgCl disappears upon treatment of NH3 solution but Hg2Cl2 becomes black in the reaction with NH3. The black Colour appears due to the formation of metallic Hg.

Balanced chemical equation of the reation is -

Hg2Cl2 + 2NH3  ---------> HgNH2Cl (white ppt.) + Hg (black ppt.) + NH4Cl

Therefore, from the given information the conclusion which can be drawn is that the unknown solution definitely has Hg22+ present.

8 0
3 years ago
Read 2 more answers
Determine the value of the equilibrium constant, Kgoal, for the reaction N2(g)+O2(g)+H2(g)⇌12N2H4(g)+NO2(g), Kgoal=? by making u
IgorC [24]

Answer : The value of the equilibrium constant is, K_{goal}=1.35\times 10^{-34}

Explanation :

The given main reaction is:

N_2(g)+O_2(g)+H_2(g)\rightleftharpoons \frac{1}{2}N_2H_4(g)+NO_2(g)

The intermediate reactions are:

N_2(g)+O_2(g)\rightleftharpoons 2NO(g);   K_1=4.10\times 10^{-31}

N_2(g)+2H_2(g)\rightleftharpoons N_2H_4(g);   K_2=7.40\times 10^{-26}

2NO(g)+O_2(g)\rightleftharpoons 2NO_2(g);   K_3=6.00\times 10^{-13}

Now we are adding all the equation, we get:

2N_2(g)+2O_2(g)+2NO(g)+2H_2(g)\rightleftharpoons 2NO(g)+N_2H_4(g)+2NO_2(g)

2N_2(g)+2O_2(g)+2H_2(g)\rightleftharpoons N_2H_4(g)+2NO_2(g)           ...........(1)

The equilibrium constant expression will be:

K_{goal}=K_1\times K_2\times K_3

Now we are dividing equation 1 by 2, we get:

N_2(g)+O_2(g)+H_2(g)\rightleftharpoons \frac{1}{2}N_2H_4(g)+NO_2(g)           ...........(1)

The equilibrium constant expression will be:

K_{goal}=(K_1\times K_2\times K_3)^{1/2}

Now put all the given values in this expression, we get:

K_{goal}=[(4.10\times 10^{-31})\times (7.40\times 10^{-26})\times (6.00\times 10^{-13})]^{1/2}

K_{goal}=1.35\times 10^{-34}

Thus, the value of the equilibrium constant is, K_{goal}=1.35\times 10^{-34}

5 0
3 years ago
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