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3 years ago

I need a answer ASAP

Chemistry

1 answer:

svet-max [94.6K]3 years ago

4 0

Answer:

<h3>3.2 millions years old</h3>

Explanation:

i hope it helps :)

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Astatine-210 has a half-life of 8.08 days. What fraction of a sample of astatine-210 is left unchanged after 16.16 days?

Flauer [41]

Answer:

Explanation:

Given parameters:

Half-life = 8.08days

Unknown:

What fraction is left unchanged after 16.16days = ?

Solution:

The half - life of a substance is the time taken for the half of a radioactive material to decay to half.

Day 0 Day 8.08 Day 16.16

100% 50% 0% Parent

0% 50% 100% Daughter

After 16.16 days, non of the original sample will remain unchanged.

6 0

3 years ago

In glycolysis, what starts the process of glucose breakdown?

Lostsunrise [7]

Answer:

That would be ATP, hope this helps! :)

6 0

3 years ago

The half-life of a positron is very short. It reacts with an electron, and the masses of both are converted to two gamma-ray pho

sp2606 [1]

Explanation:

(a) It is known that relation between energy and mass is as follows.

$E = 2 \times mc^{2}$

where, E = energy

m = mass

c = speed of light = $3 \times 10^{8}$ m/s

As it is given that mass is $9.109 \times 10^{-31}$ kg. So, putting the given values into the above formula as follows.

$E = 2 \times mc^{2}$

= $2 \times 9.109 \times 10^{-31} kg \times 3 \times 10^{8}m/s$

= $1.638 \times 10^{-13} J$

Therefore, we can conclude that the energy produced by the reaction between one electron and one positron is $1.638 \times 10^{-13} J$ .

(b) When gamma ray photons are produced then they will have the same frequency. Relation between energy and frequency is as follows.

E = $h \times \nu$ ..... (1)

where, h = plank's constant = $6.626 \times 10^{-34} J.s$

$\nu$ = frequency

Also, $E = 2 \times mc^{2}$ ........ (2)

Hence, equating equations (1) and (2) as follows.

$h \times \nu$ = $2 \times mc^{2}$

So,

$6.626 \times 10^{-34} Js \times \nu$ = $1.638 \times 10^{-13} J$

$\nu$ = $1.236 \times 10^{20} Hz$

Thus, we can conclude that the frequency is $1.236 \times 10^{20} Hz$ .

5 0

3 years ago

I get 2 marks for this so plz answer. How many particles are in a 53.2 L volume of gas at STP?

exis [7]

The number of particles= 1.43 x 10²⁴ particles

<h3>Further explanation</h3>

Given

53.2 L volume of gas at STP

Required

The number of particles

Solution

Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters/mol.

So mol for 53.2 L :

= 53.2/22.4

= 2.375 moles

1 mol = 6.02 x 10²³ particles (molecules, atoms, ions)

The number of paricles for 2.375 moles gas :

= 2.375 x 6.02 x 10²³

= 1.43 x 10²⁴ particles

6 0

3 years ago

A mixture of water and acetone at 756 mm boils at 70.0°C. The vapor pressure of acetone is

Tatiana [17]

Given :

A mixture of water and acetone at 756 mm boils at 70.0°C.

The vapor pressure of acetone is 1.54 atm at 70.0°C, while the vapor pressure of water is 0.312 atm at the same temperature.

To Find :

The percentage composition of the mixture.

Solution :

By Raoult's law :

$P=x_{acetone}P^o_{acetone}+x_{water}P^o_{water}\\\\x_{acetone}1.58+x_{water}0.312=\dfrac{756}{760}=0.995\ atm\\\\1.58x_a+0.312x_w=0.995$ ......( 1 )

Also , $x_a+x_b=1$ ......( 2 )

Solving equation 1 and 2 , we get :

$x_a=0.54\ and \ x_w=0.46$ .

Mass of acetone ,

$m_a=x_a\times MM_a\\\\m_a=0.54\times 58\\\\m_a=31.32\ g$

Mass of water ,

$m_w=x_w\times MM_w\\\\m_w=0.46\times 18\\\\m_a=8.28\ g$

$\%water =\dfrac{8.28}{8.28+31.32}\times 100\\\\\%water =20.9\%\\\\\%acetone =79.1\%$

Hence , this is the required solution.

6 0

3 years ago