Answer:
See explanation below
Explanation:
You forgot to put the picture to do so. In this case, I manage to find one, and I hope is the one you are looking for. If not, then post it again and I'll gladly help you out again.
According to the picture with the answer, we have a cyclohexane with 4 methyl groups there. Two of them are facing towards the molecule with a darker bond. This means that the alkyl bromide, should have a bromine in one of the bonds, and in order to produce an E2 reaction, this bromine should be facing in the opposite direction of the methyl groups which are facing towards. This is because an E2 reaction occurs with the less steric hindrance in the molecule. If the bromine is in the same direction as the methyl group, it will cause a lot more of work to do a reaction, and therefore, an E2 reaction. I will promote instead a E1 or a sustitution product.
Therefore the alkyl bromide should be like the one in the picture 2.
The chemical reaction would be written as follows:
2Al + 3Cl2 = 2AlCl3
We are given the amount of aluminum to be used in the reaction. This will be the starting point of the calculations. We do as follows:
19.0 g Al ( 1 mol / 29.98 g ) ( 2 mol AlCl3 / 2 mol Al ) = 0.63 mol AlCl3
Answer:
The rate at which the solute dissolves will increase.
Explanation:
If a solution is stirred, the rate at which a solute dissolves would increase substantially provided the solution is not yet saturated.
Stiring would cause more of the solution to come in contact with every part of the solute. It will increase the surface area of contact for the solution to act which will shoot up the rate of reaction. Stiring helps to bring solutes in solutions into a more close contact with the molecules or compounds of the medium.
Answer:
There is 54.29 % sample left after 12.6 days
Explanation:
Step 1: Data given
Half life time = 14.3 days
Time left = 12.6 days
Suppose the original amount is 100.00 grams
Step 2: Calculate the percentage left
X = 100 / 2^n
⇒ with X = The amount of sample after 12.6 days
⇒ with n = (time passed / half-life time) = (12.6/14.3)
X = 100 / 2^(12.6/14.3)
X = 54.29
There is 54.29 % sample left after 12.6 days
<u>Answer:</u> The solubility product of silver (I) phosphate is 
<u>Explanation:</u>
We are given:
Solubility of silver (I) phosphate = 1.02 g/L
To convert it into molar solubility, we divide the given solubility by the molar mass of silver (I) phosphate:
Molar mass of silver (I) phosphate = 418.6 g/mol
Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.
The chemical equation for the ionization of silver (I) phosphate follows:
3s s
The expression of 
for above equation follows:
We are given:
Putting values in above expression, we get:
Hence, the solubility product of silver (I) phosphate is
