Look on page 293 of your textbooks to fact check.The symbol for an alpha particle is 4/2 He.The super- script (4) is the mass number (the sum of the numbers of protons and neutrons).The subscript (2) is the atomic number (the number of protons).
Therefore the answer should be: Super-script 4 is the mass number, Subscript 2 is the atomic number
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Answer:
alkyl halide + 2 (ammonia) → alkyl amine + ammonium halide
Explanation:
In this type of preparation of alkyl amine this process works through substitution nucleophilic second order reaction and the nucleophile in this case is the ammonia and the leaving group will be the chloride ion
When ammonia attacks opposite to the way in which leaving group leaves, the chloride ion will be leaved and the nitrogen attains positive charge
So one of the hydrogen attached to the nitrogen will be removed and as there is another mole of ammonia and as ammonia is a base, it takes that hydrogen and it will form ammonium ion and to this ion chloride ion will get attracted as both have opposite charges and at last alkyl amine and ammonium halide will be formed
It would be A bc carbon is NOT usually a product. You can find out more about the by searching combustion reactions, then u should be able to answer the questions on your own.
Answer:
Water waste is produced by a fuel cell.
Explanation:
Answer:
0.313 M NaOH
Explanation:
Remember that Molarity is the concentration of a solution and is represented by:
Let's take your problem into account. The given is the following:
30.0 g of NaOH
2.40 L of solution
Molarity = ?
You are given the solute in grams, so you need to figure out how many moles there are in the given mass of solute. To do this, you need to first solve the molar mass of the solute.
NaOH
Element # of atoms Atomic Mass Total
Na 1 x 22.99 = 22.99
O 1 x 16 = 16.00
H 1 x 1 = <u> 1.00 </u>
39.99 g
So 1 mole of NaOH there is 39.99g of NaOH
We can now use this to convert the 30.00g of of NaOH to moles.
This means that there are 0.75 moles of NaOH in 30.0g of NaOH. We can use this in our molarity equation: