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LekaFEV [45]
3 years ago
5

Please help me thanks so much!?!?!?

Chemistry
2 answers:
Marina86 [1]3 years ago
8 0

Answer:

Lighting the stove and cooking the meat are both chemical changes.

Explanation:

Both fire and cooking are chemical changes.

TiliK225 [7]3 years ago
5 0

Answer:

the second one

Explanation:

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I need some chem help :(
alexandr402 [8]
Hi, can you post what you need help with??
7 0
3 years ago
A fossilized leaf contains 24% of its normal amount of carbon 14. how old is the fossil (to the nearest year)? use 5600 years as
sergij07 [2.7K]
The Half life is the time taken for a radioisotope or a radioactive substance to decay by half its original amount. The half life of carbon 14 is 5600 years. 
Original mass is 100%
Remaining amount is 24% 
Therefore; 0.24 = 1 × (1/2)^n
                     n = log 0.24/log 0.5
                        = 2.06
therefore, the age of the fossil is 5600×2.06
 = 11529.8 
 ≈ 11529 years

7 0
4 years ago
SHOW WORK
Mademuasel [1]

Answer: Molarity of the solution: 0.12

Molarity of sulfate ion: 0.360

Molarity of aluminum ion: 0.240

Explanation:

I’m not sure if that’s what you’re looking for

6 0
3 years ago
Pollution comes from<br> many small sources.
Digiron [165]

Answer: True

Hope this helps

5 0
4 years ago
ubstance A undergoes a first order reaction A®B with a half-life, t½, of 20 min at 25 °C. If the initial concentration of A in a
Stells [14]

Answer : The concentration of A after 80 min is, 0.100 M

Explanation :

Half-life = 20 min

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

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

k=\frac{0.693}{20\text{ min}}

k=3.465\times 10^{-2}\text{ min}^{-1}

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  = 3.465\times 10^{-2}\text{ min}^{-1}

t = time passed by the sample  = 80 min

a = initial amount of the reactant  = 1.6 M

a - x = amount left after decay process = ?

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

80=\frac{2.303}{3.465\times 10^{-2}}\log\frac{1.6}{a-x}

a-x=0.100M

Therefore, the concentration of A after 80 min is, 0.100 M

3 0
3 years ago
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