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3 years ago

How is iodine-131 the same as iodine-126

1 answer:

8 0

Iodine 131 and iodine 126 are the same in the sense that, they both have the same number of electrons and protons in their atoms, it is only the number of their neutrons that is different. Iodine 131 has 78 neutrons while iodine 126 has 73 neutrons.

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Consider the balanced equation for the following reaction:

Zanzabum

Answer: The amount of carbon dioxide formed in the reaction is 5.663 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of oxygen gas = 8 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{8g}{32g/mol}=0.25mol$

For the given chemical equation:

$7O_2(g)+2C_2H_6(g)\rightarrow 4CO_2(g)+6H_2O(l)$

By Stoichiometry of the reaction:

7 moles of oxygen gas produces 4 moles of carbon dioxide

So, 0.25 moles of oxygen gas will produce = $\frac{4}{7}\times 0.25=0.143mol$ of carbon dioxide

Now, calculating the mass of carbon dioxide from equation 1, we get:

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = 0.143 moles

Putting values in equation 1, we get:

$0.143mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(0.143mol\times 44g/mol)=6.292g$

To calculate the experimental yield of carbon dioxide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Percentage yield of carbon dioxide = 90 %

Theoretical yield of carbon dioxide = 6.292 g

Putting values in above equation, we get:

$90=\frac{\text{Experimental yield of carbon dioxide}}{6.292g}\times 100\\\\\text{Experimental yield of carbon dioxide}=\frac{90\times 6.292}{100}=5.663g$

Hence, the amount of carbon dioxide formed in the reaction is 5.663 grams

7 0

3 years ago

Consider the following equilibrium system: 3O2(g) 2O3(g); Keq = 1 Which equation compares the concentration of oxygen and ozone

Rashid [163]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

3O2(g) <--> 2O3(g);

Keq = 1 = [O3]^2/[O2]^3

So [O2]^3 = [O3]^2

Thus A) is correct

7 0

3 years ago

Read 2 more answers

Hi can someone please help me find the average for these temperatures? I need them asap

goblinko [34]

Answer:

55 degrees F - cold water

120 degrees F

the last on I don't know

5 0

2 years ago

If 5.32 mols N2 and 15.8 mols H2 react together, what mass NH3 can be

hodyreva [135]

Answer:

2.87 gram

N2 is the limiting agent

Explanation:

We will find out if there is sufficient N2 and h2 to produce NH3

a) For 2.36 grams of N2

Molar mass of N2 = 28.02

Number of moles of N2 in 2.36 grams = 2.36/28.02

Mass of NH3 = 17.034 g

Now NH3 produced form 2.36 grams of N2 =

2.36/28.02 * 2 * 17.034 = 2.87 g NH3

b) For 1.52 g of H2

NH3 produced = 1.52/2.016 * (2/3) * 17.034 = 8.56

N2 Is not enough to produce 2.87 g of NH3 and also H2 is not enough to make 8.56 g of NH3.

N2 is the limiting agent as it has smaller product mass

3 0

2 years ago

When a solution of 0.1 M Mg(NO3)2 was mixed with a limited amount of aqueous ammonia, a light white, wispy solid was observed, i

Ipatiy [6.2K]

Answer: The net ionic equation is written below.

Explanation:

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of magnesium nitrate and aqueous ammonia (ammonium hydroxide) is given as:

$Mg(NO_3)_2(aq.)+2NH_4OH(aq.)\rightarrow Mg(OH)_2(s)+2NH_4NO_3(aq.)$

A white precipitate of magnesium hydroxide is formed in the above reaction.

Ionic form of the above equation follows:

$Mg^{2+}(aq.)+2NO_3^-(aq.)+2NH_4^+(aq.)+2OH^-(aq.)\rightarrow Mg(OH)_2(s)+2NH_4^+(aq.)+2NO_3^-(aq.)$

As, ammonium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$Mg^{2+}(aq.)+2OH^-(aq.)\rightarrow Mg(OH)_2(s)$

Hence, the net ionic equation is written above.

8 0

3 years ago