Answer:
the answer is most likely b, sorry if im wrong.
Explanation:
.....
To obey the Law of Conservation of Mass, the sum of all individual elements of a compound is equal to the mass of the compound. So, if HCN has a mass of 7.83 grams, then
7.83 g = mass of H + mass of C + mass of N
We know the masses of H and N to be 0.290 g and 4.06 g, respectively. Hence, we can find for the mass of C:
7.83 = 0.29 + mass of C + 4.06
mass of C = 3.48 g
As an extension to the Law of Conservation of Mass, there is also a Law of Definite Proportions. According to Dalton's atomic theory, a compound is formed from a fixed ratio of its individual elements. From our previous calculations, we know that the mass ratio of H to C to N is 0.29 g: 3.48 g:4.06 grams. The ratio could also be expressed in percentages. Let's find the mass percentage of Carbon in HCN to be used later:
mass % of Carbon = (3.48 g/7.83 g)*100
mass % of Carbon = 44.44%
So, if you collect a different mass of HCN, say 3.37 g, the corresponding mass of Carbon is equal to:
Mass of Carbon = (3.37)(44.44%)
Mass of Carbon = 1.498 g
Answer:
To see if the water is safe to drink.
Explanation:
It is important for local water to be tested to see how the water quality is and to check if there are any pollutants (such as fertilizers or oil). Most of the world's water is not safe to consume, it would be useful to know what your local water is like.
The base conjugate acid pair when strong acid is added to methylamine CH3NH2 is
base CH3NH2; conjugate acid: CH3NH3 +
Explanation
When a base CH3CH2 accept proton it form its conjugate acid which known as methyl ammonium ion. Therefore conjugate acid of CH3CH2 is CH3NH3+ ion
For example when CH3CH2 react with HCl
CH3CH2 +HCl → CH3CH3+ + Cl-
Answer:
A solid residue of limestone and some gypsum.
Explanation:
In step 2, some of the limestone and gypsum in the chalk dissolve in the water. When the mixture is filtered, the dissolved substances remain in the water. When the water is boiled off or evaporated, the solid substances remain behind.
