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Ksivusya [100]
3 years ago
9

A typical lead acid storage battery like the one found in your car is 12 V and is rated at 70 amp-hr (70 amp of current drawn fo

r 1 hour). Volts x Amps gives you Watts.
So the energy stored in the fully charged battery is 840 Watt-hours (Wh). Remember that a kWh = 3600 kJ.

If a Snicker's bar contains 250 Cal and 1 Cal = 4.184 kJ, how many Snicker's bars worth of energy is stored in your battery?

a. Fewer than 5

b. 5-20

c. 20-50

d. 50-100

e. More than 100
Chemistry
1 answer:
xeze [42]3 years ago
3 0

Answer:

Almost 3 snicker's bar will contain the worth of energy stored in the battery which makes option A correct (Fewer than 5)

Explanation:

Step 1

The energy stored in a charged battery is 840 Watt- hours

Recall from the question that 1 KWh = 3600 KJ

So how many Kilo-joules will be in 840 Watt- hours?

I will convert 840 Watt- hours to KWh = 840 / 1000

                                                              = 0.840 KWh

Since I KWh gives 3600 KJ

then, 0.840 KWh will give x

                                          x = 3024 KJ

3024 KJ of energy is stored in the battery

Step 2

A snicker bar contains 250 cal and 1 cal = 4.184 KJ

Convert 250 cal to KJ?

250 cal = 1046 KJ

Each snicker bar contains 1046 KJ

To calculate how many snicker's bar worth energy stored in the battery will be (y) ?

y = 3024 / 1046

y = 2.9 snicker's bar approximately 3

The correct option is Fewer than 5

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lana [24]

Answer:

The new equilibrium concentration of HI: <u>[HI] = 3.589 M</u>          

Explanation:

Given: Initial concentrations at original equilibrium- [H₂] = 0.106 M; [I₂] = 0.022 M; [HI] = 1.29 M        

Final concentrations at new equilibrium- [H₂] = 0.95 M; [I₂] = 0.019 M; [HI] = ? M

<em>Given chemical reaction:</em> H₂(g) + I₂(g) → 2 HI(g)

The equilibrium constant (K_{c}) for the given chemical reaction, is given by the equation:

K_{c} = \frac {[HI]^{2}}{[H_{2}]\: [I_{2}]}

<u><em>At the original equilibrium state:</em></u>

K_{c} = \frac {(1.29\: M)^{2}}{(0.106\: M) \times (0.022\: M)}

K_{c} = \frac {1.6641}{0.002332} = 713.59

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K_{c} = \frac {[HI]^{2}}{(0.95\: M) \times (0.019\: M)}

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<h3>Further explanation </h3>

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