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Eva8 [605]
3 years ago
13

what is the amount of electrical energy utilized in moving the electrons from one terminal point in a battery called

Chemistry
1 answer:
lilavasa [31]3 years ago
4 0

Answer:

Explanation:

In a similar way charged particles such as electrons need to have work done to move them from one point to another. The amount of work per unit of charge is called is called the electric potential difference between the two points. The unit of potential difference is called the volt.

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A balloon filled with 1.22 L of gas at 286 K is heated until the
TiliK225 [7]

Answer: 670K

Explanation:

Given that,

Original volume of gas V1 = 1.22 L

Original temperature T1 = 286 K

New volume V2 = 2.86 L

New temperature T2 = ?

Since volume and temperature are involved while pressure is constant, apply the formula for Charles law

V1/T1 = V2/T2

1.22 L/286 K = 2.86 L/ T2

Cross multiply

1.22 L x T2 = 286 K x 2.86 L

1.22T2 = 817.96

Divide both sides by 1.22

1.22T2/1.22 = 817.96/1.22

T2 = 670.459 K (Round to the nearest whole number as 670 K)

Thus, the temperature of the gas is 670 Kelvin

4 0
3 years ago
Who was the first to state the concept of an atom
marin [14]
Democritus, a Greek philosopher, first developed the idea of atoms (around 460 B.C., I believe).
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6 0
3 years ago
8. What volume (ml) of a 5.45 M lead nitrate solution must be diluted to 820.7 ml to make a
Ganezh [65]

212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.

Answer:

Option A.

Explanation:

Similar to Avagadro's law, there is another law termed as dilution law. As the product of volume and normality of the reactant is equal to the product of volume and normality of the product from the Avagadro's law. In dilution law, it will be as product of volume and concentration of the solute of the reactant is equal to the product of volume and concentration of solution.

C_{1} V_{1} =C_{2} V_{2}

So, as per the given question C1 = 5.45 M of lead nitrate and V1 has to be found. While C2 is 1.41 M of lead nitrate and V2 is 820.7 ml.

Then, (5.45*V_{1} ) = (820.7*1.41)

V_{1}=\frac{820.7*1.41}{5.45}=  212.33 ml

So nearly 212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.

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3 years ago
What is the mass of 25 Liters of nitrogen dioxide gas?
ValentinkaMS [17]

Answer:

cacfat lebron

Explanation:

mruno bars is the wae

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coileach

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7 0
3 years ago
Which of the following interactions can contribute to the intrinsic binding energy during enzymatic catalysis? choose all that a
Lady_Fox [76]
1. electrostatic interactions 
<span>3. van de waals interactions </span>
<span>4. hydrogen bonding </span>
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3 years ago
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