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3 years ago

Iodine 131, cesium 137, and strontium 90 are all considered isotopes. What are isotopes?

Chemistry

2 answers:

Nitella [24]3 years ago

3 0

Answer:

Isotopes are variants of an element, where the number of neutrons is different but its number of protons stay the same.

Explanation:

If the number of protons had changed then it would just be a different element.

Electron number changing simply adds charge to the atom, and doesn't change it or anything.

PLS GIVE BRAINLIEST

Fudgin [204]3 years ago

3 0

Answer:

Isotopes are one of two or more forms of the same chemical element.

Explanation:

Different isotopes of an element have the same number of protons in the nucleus, giving them the same atomic number,but a different number of neutrons giving each element isotope of a different atomic weight.

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Name the reactants and products of Na3PO4 + 3KOHàNaOH + K3PO4 (Question 2 and 3)

Leona [35]

Answer:

Question 2: Na3PO4, KOH; Question 3: Na3PO4, KOH

Explanation:

Question 2

The reactants in a chemical equation are the species on the left side of the reaction arrow.

Thus the reactants are Na3PO4, KOH (sodium phosphate and potassium hydroxide).

Question 3.

The products in a chemical equation are the species on the right side of the reaction arrow.

Thus the products are NaOH, K3PO4 (sodium hydroxide and potassium phosphate).

6 0

3 years ago

The following reactions have the indicated equilibrium constants at a particular temperature: N2(g) + O2(g) ⇌ 2NO(g) Kc = 4.3 ×

Anuta_ua [19.1K]

Answer:

$Kc=~1.49x10^3^4}$

Explanation:

We have the reactions:

A: $N_2_(_g_) + O_2_(_g_) 2NO_(_g_)~~~~~~Kc = 4.3x10^-^2^5$

B: $2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~~~Kc = 6.4x10^9$

Our target reaction is:

$4NO_(_g_) N_2_(_g_) + 2NO_2_(_g_)$

We have $NO_(_g_)$ as a reactive in the target reaction and $NO_(_g_)$ is present in A reaction but in the products side. So we have to flip reaction A.

A: $2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}$

Then if we add reactions A and B we can obtain the target reaction, so:

A: $2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}$

B: $2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~Kc=6.4x10^9$

For the final Kc value, we have to keep in mind that when we have to add chemical reactions the total Kc value would be the multiplication of the Kc values in the previous reactions.

$4NO_(_g_) N_2_(_g_) + 2NO_2_(_g_)~~~Kc=\frac{6.4x10^9}{4.3x10^-^2^5}$