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jeyben [28]
3 years ago
7

If a 12-volt battery produces a current of 20 amps, what is the resistance?

Chemistry
1 answer:
mixas84 [53]3 years ago
4 0

Answer:

0.6 Ω

Explanation:

From the question given above, the following data were obtained:

Voltage (V) = 12 V

Current (I) = 20 A

Resistance (R) =?

From Ohm's law,

V = IR

Where:

V => is the voltage

I => is the current

R => R is the resistance

With the above formula, we can obtain the resistance as follow:

Voltage (V) = 12 V

Current (I) = 20 A

Resistance (R) =?

V = IR

12 = 20 × R

Divide both side by 20

R = 12 / 20

R = 0.6 Ω

Thus the resistance is 0.6 Ω

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If the density of an object is 5.2g/cm3, and it’s volume is 3.7 cm3, what is it’s mass?
lord [1]
Here's the equation you use: Density = mass/volume 

1) 5.2g/cm^3 = m/3.7cm^3 

2) m = 5.2g/cm^3 x 3.7cm^3 

3) m = 19.24g 

You can check the answer by plugging it in 

19.24g/3.7cm^3 
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5 0
3 years ago
Gravity is matter because it does not have mass and volume.
aksik [14]
False, gravity is a force, matter has mass and volume.
6 0
3 years ago
What would a scientist’s next steps be if his/her data failed to support their hypothesis?
Paul [167]
To change only one variable which is very important than to test the experiment to match the hypothesis again, I think. It’s been a while since I was on that lesson‍♀️
6 0
3 years ago
Someone please help! this is the last question<br>I only need help with B.<br><br>​
ludmilkaskok [199]

1. mol ratio of Al(NO₃)₃ : Na₂CO₃ = 2 : 3

2. Na₂CO₃ as a limiting reactant

<h3>Further explanation</h3>

Given

Reaction

2 Al(NO₃)₃ + 3 Na₂CO₃ → Al₂(CO₃)₃ + 6 NaNO₃

Required

mol ratio

Limiting reactant

Solution

The reaction coefficient in the chemical equation shows the mole ratio of the components of the compound involved in the reaction (reactants and products)

1. From the equation mol ratio of Al(NO₃)₃ : Na₂CO₃ = 2 : 3

2. mol : coefficient of Al(NO₃)₃ : Na₂CO₃ = 2 mole/2 : 2 mole/3 = 1 : 0.67

Na₂CO₃ as a limiting reactant (smaller)

6 0
3 years ago
At a certain temperature, the vapor pressure of pure benzene ‍ is atm. A solution was prepared by dissolving g of a nondissociat
erastova [34]

Answer:

Molar mass of solute: 300g/mol

Explanation:

<em>Vapor pressure of pure benzene: 0.930 atm</em>

<em>Assuming you dissolve 10.0 g of the non-volatile solute in 78.11g of benzene and vapour pressure of solution was found to be 0.900atm</em>

<em />

It is possible to answer this question based on Raoult's law that states vapor pressure of an ideal solution is equal to mole fraction of the solvent multiplied to pressure of pure solvent:

P_{sln} = X_{solvent}P_{solvent}^0

Moles in 78.11g of benzene are:

78.11g benzene × (1mol / 78.11g) = <em>1 mol benzene</em>

Now, mole fraction replacing in Raoult's law is:

0.900atm / 0.930atm = <em>0.9677 = moles solvent / total moles</em>.

As mole of solvent is 1:

0.9677× total moles = 1 mole benzene.

Total moles:

1.033 total moles. Moles of solute are:

1.033 moles - 1.000 moles = <em>0.0333 moles</em>.

As molar mass is the mass of a substance in 1 mole. Molar mass of the solute is:

10.0g / 0.033moles = <em>300g/mol</em>

8 0
3 years ago
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