Remember that any intersection of lines is a C, and that the number of hydrogens attached are the necessary to complet the 4 bonds.
1) CH3 - CH (OH) - CH (CH3) -CH3
2) CH3 - O - CH(CH3)-CH2 - CH3
I have used the parenthesis to indicate that the radical inside is in other branch, bonded by a single line -
Answer:
357 g of the transition metal are present in 630 grams of the compound of the transition metal and iodine
Explanation:
In any sample of the compound, the percentage by mass of the transition metal is 56.7%. This means that for a 100 g sample of the compound, 56.7 g is the metal while the remaining mass, 43.3 g is iodine.
Given mass of sample compound = 630 g
Calculating the mass of iodine present involves multiplying the percentage by mass composition of the metal by the mass of the given sample;
56.7 % = 56.7/100 = 0.567
Mass of transition metal = 0.567 * 630 = 357.21 g
Therefore, the mass of the transition metal present in 630 g of the compound is approximately 357 g
Answer:
Regularly test the water in residents' homes.
Explanation:
The only way to know if tap water contains lead is to do tests to determine the levels of that metal in the water. Therefore, the state is under an obligation to constantly conduct such tests in the resident´s homes and thus determine whether the water supplied is fit for human consumption.
The state after the tests must guarantee the population the treatment of the water to reduce the levels of lead. The main pipes that contain lead pipes must be changed, as well as those parts of the service connections made of lead.
Answer:
THE EMPIRICAL FORMULA OF THE SUBSTANCE IS C2H5NO
Explanation:
The steps involved in calculating the empirical formula of this substance in shown in the table below:
Element Carbon Hydrogen Nitrogen Oxygen
1. % Composition 40.66 8.53 23.72 27.09
2. Mole ratio =
%mass/ atomic mass 40.66/12 8.53/1 23.72/14 27.09/16
= 3.3883 8.53 1,6943 1.6931
3. Divide by smallest
value (0.6931) 3.3883/1.6931 8.53/1.6931 1.6943/1.6931 1.6931/1.6931
= 2.001 5.038 1.0007 1
4. Whole number ratio 2 5 1 1
The empirical formula = C2H5NO
