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2 years ago

How to learn stoichiometry

1 answer:

3 0

If you go on youtube and do a quick search about stoichiometry it really helps, or you can google "stoichiometry practice" and get wonderful worksheets that walk you through everything

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Bismuth-210 has a half life of 5.0 days if you had a sample of 60.00g of Bismuth-210 how much would be left after 15 days? (Show

Zielflug [23.3K]

Answer:

7.5 gm left

Explanation:

Bismuth-210 has a half life of 5 days

15 days is 15/5 = 3 half lives

since half the amount is left in 5 days or 1 half life

(1/2) x (1/2) x (1/2) the staring amount would be left in

3 half lives. so 1/8 is left

(1/8) x 60.0 = 7.5 gm left

5 0

2 years ago

What is it??!!!!!??!??!?

Softa [21]

Answer:

other the effect of temperature on plants

4 0

2 years ago

Identify the Bronsted-Lowry acid, the Bronsted-Lowry base, the conjugate acid and the conjugate base for each of the following r

Vlad1618 [11]

Answer:

Acids → H₂CO₃ from equilibrium 1 and water, from equilibrium 2.

Bases → Water from equilibrium 1 and ammonia from equilibrium 2.

In 1st equilibrium, H₃O⁺ is the conjugate acid and HCO₃⁻ the conjugate base.

In 2nd equilibrium, NH₄⁺ is the conjugate acid, and OH⁻, the conjugate base.

Explanation:

By the Bronsted-Lowry you know that acids are the one that release protons and base are the ones that catch them.

For the first equilibrium:

H₂CO₃(aq) + H₂O(l) ⇄ H₃O⁺(aq) + HCO₃⁻(aq)

Carbonic acid is the acid → It donates the proton to water, so the water becomes the base. As H₂CO₃ is the acid, the bicarbonate is the conjugate base (it can accept the proton from water to become carbonic acid, again) and the hydronium is the conjugate acid (it would release the proton to become water).

For the second equilibrium:

NH₃(aq) + H₂O(l) ⇄ NH₄⁺ (aq) + OH⁻(aq)

This is the opposite situation → Water relase the proton to ammonia, that's why water is the acid and NH₃, the base (it accepted to become ammonium). The NH₄⁺ is the conjugate acid (it can release the H⁺ to become ammonia) and the OH⁻ is the conjugate base (It can accept the proton to become water, again).

5 0

3 years ago

Write a balanced equation for the complete oxidation reaction that occurs when methane (CH4) burns in air.

Lelechka [254]

Answer: The balanced equation for the complete oxidation reaction that occurs when methane (CH4) burns in air is $CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O$ .

Explanation:

When a substance tends to gain oxygen atom in a chemical reaction and loses hydrogen atom then it is called oxidation reaction.

For example, chemical equation for oxidation of methane is as follows.

$CH_{4} + O_{2} \rightarrow CO_{2} + H_{2}O$

Number of atoms present on reactant side are as follows.

C = 1
H = 4
O = 2

Number of atoms present on product side are as follows.

C = 1
H = 2
O = 3

To balance this equation, multiply $O_{2}$ by 2 on reactant side. Also, multiply $H_{2}O$ by 2 on product side. Hence, the equation can be rewritten as follows.

$CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O$

Now, the number of atoms present on reactant side are as follows.

C = 1
H = 4
O = 4

Number of atoms present on product side are as follows.

C = 1
H = 4
O = 4

Since, the atoms present on both reactant and product side are equal. Therefore, this equation is now balanced.

Thus, we can conclude that balanced equation for the complete oxidation reaction that occurs when methane (CH4) burns in air is $CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O$ .

4 0

2 years ago

How many hybrid orbitals are found in CCl4? o one o two o three o four

BlackZzzverrR [31]

Answer:

I Believe it is 4 orbitals s,p,p,p or aka sp^3

Explanation:

4 0

2 years ago

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