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2 years ago

What items could scientists use to date a layer of rock

Chemistry

1 answer:

Charra [1.4K]2 years ago

4 0

By Using relative and radiometric dating methods hope this helps!!

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Diagonal lines in the graph represent the temperature of the substance is ___.

labwork [276]

Gdnndjfndmnxndndndjdjdjxncncncnnc

5 0

3 years ago

Name the homologous series for this compound and show its structure<br>ch3ch2cc(ch2)2ch3

kotykmax [81]

Answer:

Alkyne

Explanation:

The homologous series for this compound is the Alkyne because of the presence of the triple bond present between the third and fourth carbon atoms as seen in the picture attached.

5 0

2 years ago

A 1.25 g gas sample occupies 663 ml at 25∘ c and 1.00 atm. what is the molar mass of the gas?

lakkis [162]

Using ideal gas equation,

$P\times V=n\times R\times T$

Here,

P denotes pressure

V denotes volume

n denotes number of moles of gas

R denotes gas constant

T denotes temperature

The values at STP will be:

P=1 atm

T=25 C+273 K =298.15K

V=663 ml=0.663L

R=0.0821 atm L mol ⁻¹

Mass of gas given=1.25 g g

Molar mass of gas given=?

$Number of moles of gas, n= \frac{Given mass of the gas}{Molar mass of the gas}$

$Number of moles of gas, n= \frac{1.25}{Molar mass of the gas}$

Putting all the values in the above equation,

$1\times 0.663=\frac{1.25}{Molar mass of the gas}\times 0.0821\times 298.15$

Molar mass of the gas=46.15

3 0

2 years ago

The half-life of 131I is 80 days. How much of a 500.0 mg sample remains after 3 half lives?

I am Lyosha [343]

Answer:

62.5 mg

Explanation:

Just multiply the original amount by 1/2 three times:

500 mg x 1/2 x 1/2 x 1/2 = 62.5 mg

6 0

1 year ago

If a radioactive isotope has a half-life of 26.5 days, how many days does it take for a sample of the isotope to decrease by 35.

dangina [55]

Answer:

16.5 days

Explanation:

Given that:

Half life = 26.5 days

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,

$k=\frac {ln\ 2}{t_{1/2}}$

$k=\frac {ln\ 2}{26.5}\ days^{-1}$

The rate constant, k = 0.02616 days⁻¹

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given:

35.0 % is decomposed which means that 0.35 of $[A_0]$ is decomposed. So,

$\frac {[A_t]}{[A_0]}$ = 1 - 0.35 = 0.65

t = 7.8 min

$\frac {[A_t]}{[A_0]}=e^{-k\times t}$

$0.65=e^{-0.02616\times t}$

7 0

2 years ago