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soldier1979 [14.2K]
3 years ago
8

A small toy car draws a 0.50-mA current from a 3.0-V NiCd (nickel-cadmium) battery. In 10 min of operation, (a) how much charge

flows through the toy car, and (b) how much energy is lost by the battery? 4. (Resistance and Ohm’s law, Prob. 17.16, 1.0 point) How
Chemistry
1 answer:
adelina 88 [10]3 years ago
4 0

Answer:

0.3 Coulomb charge flows through the toy car.

0.9 Joules of energy is lost by the battery

Explanation:

Current(I)=\frac{charge(Q)}{Time (T)}

a)

Current drawn from the battery = 0.50 mA = (0.50 × 0.001 A)

milli Ampere = 0.001 Ampere

Duration of time current drawn = T = 10 min = 10 × 60 s = 600 s

1 min = 60 seconds

Charge flows through the toy car be Q

0.50\times 0.001 A=\frac{Q}{600 s}

Q=0.50 \times 0.001 A\times 600 s=0.3 C

0.3 Coulomb charge flows through the toy car.

b)

Heat(H)=Voltage(V)\times Current(I)\times Time(T)

Voltage of the battery = V = 3.0 V

Current drawn from the battery = 0.50 mA = (0.50 × 0.001 A)

Duration of time current drawn = T = 10 min = 10 × 60 s = 600 s

H=V\times I\time T=3.0 V\times 0.50 \times 0.001 A\times 600 s

H = 0.9 Joules

0.9 Joules of energy is lost by the battery

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Answer:

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Part B = The mass of 1 silver atom is 1.79 * 10^-22 grams

Explanation:

Part A

Step 1: Data given

A mixture of carbon and sulfur has a mass of 9.0 g

Mass of the product = 27.1 grams

X = mass carbon

Y = mass sulfur

x + y = 9.0  grams

x = 9.0 - y

x(molar mass CO2/atomic mass C) + y(molar mass SO2/atomic mass S) = 22.6

(9 - y)*(44.01/12.01) + y(64.07/32.07)

(9-y)(3.664) + y(1.998)

32.976 - 3.664y + 1.998y = 22.6

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y = 6.228 = mass sulfur

x = 9.0 - 6.228 = 2.772 grams = mass C

The mass of sulfur is 6.228 grams

Part B

Calculate the mass, in grams, of a single silver atom (mAg = 107.87 amu ).

Calculate moles of 1 silver atom

Moles = 1/ 6.022*10^23

Moles = 1.66*10^-24 moles

Mass = moles * molar mass

Mass = 1.66*10 ^-24 moles *107.87

Mass = 1.79 * 10^-22 grams

The mass of 1 silver atom is 1.79 * 10^-22 grams

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(I)how many atoms are present in 7g of lithium?
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Answer :

(i) The number of atoms present in 7 g of lithium are, 6.07\times 10^{23}

(ii) The number of atoms present in 7 g of lithium are, 1.204\times 10^{24}

(iii) The number of moles of F_2 is, 1 mole

The number of moles of CO_2 is, 0.5 mole

The number of moles of OH^- is, 1 mole

Explanation :

<u>Part (i) :</u>

First we have to calculate the moles of lithium.

\text{Moles of }Li=\frac{\text{Mass of }Li}{\text{Molar mass of }Li}

Molar mass of Li = 6.94 g/mole

\text{Moles of }Li=\frac{7g}{6.94g/mol}=1.008mole

Now we have to calculate the number of atoms present.

As, 1 mole of lithium contains 6.022\times 10^{23} number of atoms

So, 1.008 mole of lithium contains 1.008\times 6.022\times 10^{23}=6.07\times 10^{23} number of atoms

Thus, the number of atoms present in 7 g of lithium are, 6.07\times 10^{23}

<u>Part (ii) :</u>

First we have to calculate the moles of carbon.

\text{Moles of }C=\frac{\text{Mass of }C}{\text{Molar mass of }C}

Molar mass of C = 12 g/mole

\text{Moles of }C=\frac{24g}{12g/mol}=2mole

Now we have to calculate the number of atoms present.

As, 1 mole of carbon contains 6.022\times 10^{23} number of atoms

So, 2 mole of carbon contains 2\times 6.022\times 10^{23}=1.204\times 10^{24} number of atoms

Thus, the number of atoms present in 7 g of lithium are, 1.204\times 10^{24}

<u>Part (iii) :</u>

<u>To calculate the moles of </u>F_2<u> :</u>

\text{Moles of }F_2=\frac{\text{Mass of }F_2}{\text{Molar mass of }F_2}

Molar mass of F_2 = 38 g/mole

\text{Moles of }F_2=\frac{19g}{19g/mol}=1mole

Thus, the number of moles of F_2 is, 1 mole

<u>To calculate the moles of </u>CO_2<u> :</u>

\text{Moles of }CO_2=\frac{\text{Mass of }CO_2}{\text{Molar mass of }CO_2}

Molar mass of CO_2 = 44 g/mole

\text{Moles of }CO_2=\frac{22g}{44g/mol}=0.5mole

Thus, the number of moles of CO_2 is, 0.5 mole

<u>To calculate the moles of </u>OH^-<u> ions :</u>

\text{Moles of }OH^-=\frac{\text{Mass of }OH^-}{\text{Molar mass of }OH^-}

Molar mass of OH^- = 17 g/mole

\text{Moles of }OH^-=\frac{17g}{17g/mol}=1mole

Thus, the number of moles of OH^- is, 1 mole

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3 years ago
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