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BARSIC [14]
3 years ago
12

What is the simplest form of the expression (14.2a + 9.8b) – (13.1b – 0.2a) – (3.7a + 4.8b)?

Mathematics
2 answers:
Sati [7]3 years ago
8 0

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<em>Look</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>attached</em><em> </em><em>picture</em><em>⤴</em>

<em>Hope</em><em> </em><em>it</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>u</em><em>.</em><em>.</em><em>.</em>

marshall27 [118]3 years ago
5 0

Answer:

a

Step-by-step explanation:

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Travka [436]

Answer :

The amount after 1000 years will be, 5.19 grams.

The amount after 10000 years will be, 0.105 grams.

Step-by-step explanation :

Half-life = 1599 years

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

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

k=\frac{0.693}{1599\text{ years}}

k=4.33\times 10^{-4}\text{ years}^{-1}

Now we have to calculate the amount after 1000 years.

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  = 4.33\times 10^{-4}\text{ years}^{-1}

t = time passed by the sample  = 1000 years

a = initial amount of the reactant  = 8 g

a - x = amount left after decay process = ?

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

1000=\frac{2.303}{4.33\times 10^{-4}}\log\frac{8}{a-x}

a-x=5.19g

Thus, the amount after 1000 years will be, 5.19 grams.

Now we have to calculate the amount after 10000 years.

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  = 4.33\times 10^{-4}\text{ years}^{-1}

t = time passed by the sample  = 10000 years

a = initial amount of the reactant  = 8 g

a - x = amount left after decay process = ?

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

10000=\frac{2.303}{4.33\times 10^{-4}}\log\frac{8}{a-x}

a-x=0.105g

Thus, the amount after 10000 years will be, 0.105 grams.

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Answer:

so all x's have to be added together.

8x+5x+10x+7x=30x

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