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3 years ago

5

Is oxygen radioactive yes or no?

2 answers:

Assoli18 [71]3 years ago

7 0

Answer:

ima go with no.....

Explanation:

Art [367]3 years ago

3 0

Answer:

Stable isotopes are variants of atoms that differ in mass but do not decay over time, that is, they are not radioactive. The element oxygen (O) is found in three naturally occurring stable isotopes. So oxygen is not radioactive.

Explanation:

You might be interested in

How many O atoms are in 4C₆H₁₂O₆

ipn [44]

Number of O atoms : 24

<h3>Further explanation</h3>

Given

C₆H₁₂O₆ compound

Required

Number of atoms

Solution

A molecular formula shows the number of atomic elements in compound.

The empirical formula is the smallest comparison of the atoms

Glucose-C₆H₁₂O₆ is composed of 3 elements, namely C, H, and O.

The number of atoms in a compound can usually be seen from the subscript number after the atom and the reaction coefficient shows the number of molecules

So number of O atoms :

= 4 x 6 = 24 atoms

6 0

3 years ago

The stomach and intestines are organs of the body system which _______.

yulyashka [42]

Answer:

A. Digests food

Explanation:

Look about diggestive process in Google

Best regards

3 0

3 years ago

Read 2 more answers

Choose all the answers that apply.

iVinArrow [24]

Answer:it all above

Explanation:

it all above because all the answer are truth so it all above

8 0

3 years ago

Read 2 more answers

Determine the value of the equilibrium constant, KgoalKgoalK_goal, for the reaction CO2(g)⇌C(s)+O2(g)CO2(g)⇌C(s)+O2(g), Kgoal=?

NISA [10]

Answer:

The value of the equilibrium constant for reaction asked is $1.24\times 10^{-7}$ .

Explanation:

$CO_2(g)\rightleftharpoons C(s)+O_2(g)$

$K_{goal}=?$

$K_{goal}=\frac{[C][O_2]}{[CO_2]}$

$2CO_2(g)+2H_2O(l)\rightleftharpoons CH_3COOH(l)+2O_2(g)$ ..[1]

$K_1=\frac{[CH_3COOH][O_2]^2}{[CO_2]^2[H_2O]^2}$

$2H_2(g)+O2(g)\rightleftharpoons 2H_2O(l)$ ..[2]

$K_2=\frac{[H_2O]^2}{[H_2]^2[O_2]}$

$CH_3COOH(l)\rightleftharpoons 2C(s)+2H_2(g)+O_2(g)$ ..[3]

$K_3=\frac{[C]^2[H_2]^2[O_2]}{[CH_3COOH]}$

[1] + [2] + [3]

$2CO_2(g)\rightleftharpoons 2C(s)+2O_2(g)$

( on adding the equilibrium constant will get multiplied with each other)

$K=K_1\times K_2\times K_3$

$K=5.40\times 10^{-16}\times 1.06\times 10^{10}\times 2.68\times 10^{-9}$

$K=1.53\times 10^{-14}$

$K=\frac{[C]^2[O_2]^2}{[CO_2]^2}$

On comparing the K and $K_{goal}$ :

$K^2=K_{goal}$

$K_{goal}=\sqrt{K}=\sqrt{1.53\times 10^{-14}}=1.24\times 10^{-7}$

The value of the equilibrium constant for reaction asked is $1.24\times 10^{-7}$ .

4 0

3 years ago

Because the moon travels around Earth,it is a an blank

Lunna [17]

The answer could be a satellite

6 0

3 years ago