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2 years ago

What is the ionsphere in the layers of the atmosphere?

1 answer:

5 0

Hi!

The Ionosphere, also known as the Ozone Layer, is a layer in the atmosphere that is home to Ozone gas. This layer protects earth from ultraviolet (UV) radiation.

Hope this helps :)

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Which of the following contribute(s) to most of the mass of an atom?

xenn [34]

Answer:

B) protons and newtrons

6 0

2 years ago

Two atoms of calcium and one molecule of oxygen gas react and yield two molecules of calcium oxide.

Vitek1552 [10]

2Ca + O2 ——————-> 2CaO2

3 0

2 years ago

What is the net force acting on a teddy bear being pulled to the right 67N and to ghe left 112N

noname [10]

|F| = 112 - 67 = 45 N

F = 45 N ( to the left)

3 0

2 years ago

Read 2 more answers

Calculate the work, w, gained or lost by the system when a gas expands from 15 L to 45 L against a constant external pressure of

marta [7]

Answer:

Work done by the system = 4545 J

Explanation:

The expression for the calculation of work done is shown below as:

$w=-P\times \Delta V$

Where, P is the pressure

$\Delta V$ is the change in volume

From the question,

$\Delta V$ = 45 - 15 L = 30 L

P = 1.5 atm

$w=-1.5\times 30\ atmL$

Also, 1 atmL = 101 J

So,

$w=-1.0\times0.75\times 101\ J=-4545\ J$ (negative sign implies work is done by the system)

<u>Work done by the system = 4545 J</u>

6 0

2 years ago

A wooden artifact from a Chinese temple has a 14C activity of 42.8 counts per minute as compared with an activity of 58.2 counts

Arlecino [84]

Answer : The age of the artifact is, $2.54\times 10^3\text{ years}$

Explanation :

Half-life = 5715 years

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

$k=\frac{0.693}{5715\text{ years}}$

$k=1.21\times 10^{-4}\text{ years}^{-1}$

Now we have to calculate the time taken to decay.

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 by sample = ?

a = initial activity of the reactant = 58.2 counts per minute

a - x = activity left after decay process = 42.8 counts per minute

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

$t=\frac{2.303}{1.21\times 10^{-4}}\log\frac{58.2}{42.8}$

$t=2540.5\text{ years}=2.54\times 10^3\text{ years}$

Therefore, the age of the artifact is, $2.54\times 10^3\text{ years}$

4 0

2 years ago