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dangina [55]
2 years ago
15

A burning candle is covered by a jar as shown in the picture. The whole arrangement has a mass of 500 g. What will be the approx

imate mass of the arrangement when the candle is completely burnt, after four minutes?
A)
0 g



B)
50 g



C)
250 g



D)
500 g
Physics
2 answers:
topjm [15]2 years ago
7 0

Technically, I can't answer the question, because you won't
let me see the picture that goes along with it and is a part of it. 
But I'm familiar with the set-up, have dealt with the question before,
and I can answer it from my previous experience and general knowledge.

If there is 500g of mass inside the jar when you lower it over
the candle, then there will be 500g of mass at any time after that,
forever, or until you pick up the jar and take some mass out or put
some more in.  It doesn't matter how long you wait.  It also doesn't
matter whether or not the candle is burning, whether or not the sun
is shining on the jar, or whether somebody comes along and spray-paints
the outside of the jar with black paint.  Matter is not created or destroyed. 
Whatever mass was inside when the jar got closed stays in there.
 
weqwewe [10]2 years ago
7 0
The answer is D)<span>500 g  it will not change</span>
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Answer:

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Area = A = π r² = 0.0158 m²

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Charge = Q = C V = ( 1.11 x 10⁻¹⁰ F )(126) = 13.98 nC

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This exercise uses the radioactive decay model. After 3 days a sample of radon-222 has decayed to 58% of its original amount. (a
Vadim26 [7]

Answer: a) 3.85 days

b) 10.54 days

Explanation:-

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = ?

t = time taken for decomposition  = 3 days

a = let initial amount of the reactant  = 100 g

a - x = amount left after decay process  = \frac{58}{100}\times 100=58g

First we have to calculate the rate constant, we use the formula :

Now put all the given values in above equation, we get

k=\frac{2.303}{3}\log\frac{100}{58}

k=0.18days^{-1}

a) Half-life of radon-222:

t_{\frac{1}{2}}=\frac{0.693}{k}

t_{\frac{1}{2}}=\frac{0.693}{0.18}=3.85days

Thus half-life of radon-222 is 3.85 days.

b) Time taken for the sample to decay to 15% of its original amount:

where,

k = rate constant  = 0.18days^{-1}

t = time taken for decomposition  = ?

a = let initial amount of the reactant  = 100 g

a - x = amount left after decay process  = \frac{15}{100}\times 100=15g

t=\frac{2.303}{0.18}\log\frac{100}{15}

t=10.54days

Thus it will take 10.54 days for the sample to decay to 15% of its original amount.

3 0
3 years ago
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