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3 years ago

What is the difference between nuclear fusion and nuclear fission?

Chemistry

1 answer:

True [87]3 years ago

4 0

Answer:

[Fusion: Nuclei Fuse together and release energy] [Fission: Pressure and energy is released. This release of energy, is what causes the unstable atom to split to an isotope. I hope this helps!]

[isotopes can split too. At least some.]

Explanation:

Imagine you are trying to push something heavy, fusion is how you can join together and help each other to push this heavy material.

Fun fact! Heavier materials have more inertia against other forces acting on them.

You push harder, harder and harder! Until its pushed.

This will release a lot of energy which escalates to an extended length.

It might not be the finest example, it helped me to understand the basics.

Fission. Imagine having so much energy, in this case, try thinking of so many people in a tight space. What do you do? You will eventually start to escape and leave the room. This is the act of splitting the atom.

This won't be the same than how it was before.

Again, Energy is released (energy!!).

For anyone struggling with this, although it might not be the best idea, I would recommend researching the more advaced options for this topic.

This will give you a wide range of information which can develop your current knowledge about, "Nuclear Fission and Fusion."

It will be difficult, it definitely was for me when I was a child (starting to study science).

Just keep trying, you'll get it.

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Convert 5.4g of NaCl to moles

sammy [17]

Answer:

.0924 moles of NaCl

Explanation:

So you know you have 5.4 g of NaCl and you need to know how many moles there are in this amount of NaCl

You'll need to find the atomic mass of the compound NaCl to help you solve for moles

Sodium (Na) on the periodic table has a mass of 22.99

Chlorine (Cl) on the periodic table has a mass of 35.45

Add these two together----> 22.99 + 35.45 = 58.44

Now you can calculate for moles

Written-out method:

5.4 grams of NaCl | 1 mole of NaCl

| 58.44 grams NaCl = .0924 moles of NaCl

Plug into calculator method:

(5.4 g of NaCl/ 58.44g NaCl= .0925 moles)

7 0

3 years ago

Suppose of copper(II) acetate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of acetate

uranmaximum [27]

Answer:

0.0714 M for the given variables

Explanation:

The question is missing some data, but one of the original questions regarding this problem provides the following data:

Mass of copper(II) acetate: $m_{(AcO)_2Cu} = 0.972 g$

Volume of the sodium chromate solution: $V_{Na_2CrO_4} = 150.0 mL$

Molarity of the sodium chromate solution: $c_{Na_2CrO_4} = 0.0400 M$

Now, when copper(II) acetate reacts with sodium chromate, an insoluble copper(II) chromate is formed:

$(CH_3COO)_2Cu (aq) + Na_2CrO_4 (aq)\rightarrow 2 CH_3COONa (aq) + CuCrO_4 (s)$

Find moles of each reactant. or copper(II) acetate, divide its mass by the molar mass:

$n_{(AcO)_2Cu} = \frac{0.972 g}{181.63 g/mol} = 0.0053515 mol$

Moles of the sodium chromate solution would be found by multiplying its volume by molarity:

$n_{Na_2CrO_4} = 0.0400 M\cdot 0.1500 L = 0.00600 mol$

Find the limiting reactant. Notice that stoichiometry of this reaction is 1 : 1, so we can compare moles directly. Moles of copper(II) acetate are lower than moles of sodium chromate, so copper(II) acetate is our limiting reactant.

Write the net ionic equation for this reaction:

$Cu^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow CuCrO_4 (s)$