If two different elements combine separately with a fixed mass of a third element, the ratio of the masses in which they do so are either the same as or a simple multiple of the ratio of the masses in which they combine with each other.
Answer:
Because no two elements have the same set of energy levels, different elements emit different colors of light. Energy is released when electrons move from higher energy levels to lower ones (visible light).
Explanation:
Answer:
10.00 mL
Explanation:
Please, do not forget to write the dot and 2 of Zeros after the number 10 in the answer!
Please see the step-by-step solution in the picture attached below.
Hope this answer can help you. Have a nice day!
Answer:
4.8 %
Explanation:
We are asked the concentration in % by mass, given the molarity of the solution and its density.
0.8 molar solution means that we have 0.80 moles of acetic acid in 1 liter of solution. If we convert the moles of acetic acid to grams, and the 1 liter solution to grams, since we are given the density of solution, we will have the values necessary to calculate the % by mass:
MW acetic acid = 60.0 g/mol
mass acetic acid (the solute) = 0.80 mol x 60 g / mol = 48.00 g
mass of solution = 1000 cm³ x 1.010 g/ cm³ (1l= 1000 cm³)
= 1010 g
% (by mass) = 48.00 g/ 1010 g x 100 = 4.8 %
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.