Answer:
Endothermic hope that helps
Explanation:
Answer:
Solution - (a) Brine . Suspension - (c) sand and water, (g) chalk and water Colloid - (e) air, (f) smoke , (d) soda , (b) milk
Explanation:
A solution is an homogeneous mixture of two or more compounds.
A suspension is a heterogeneous mixture of two or more compounds while a colloid is a homogeneous mixture of two or more compounds with suspended particles which do not settle.
So, under these definitions, the classifications are as follows-
Solution - (a) Brine .
Suspension - (c) sand and water, (g) chalk and water
Colloid - (e) air, (f) smoke , (d) soda , (b) milk
Answer:
In the previous section, we discussed the relationship between the bulk mass of a substance and the number of atoms or molecules it contains (moles). Given the chemical formula of the substance, we were able to determine the amount of the substance (moles) from its mass, and vice versa. But what if the chemical formula of a substance is unknown? In this section, we will explore how to apply these very same principles in order to derive the chemical formulas of unknown substances from experimental mass measurements.
Explanation:
tally. The results of these measurements permit the calculation of the compound’s percent composition, defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The percent composition of this compound could be represented as follows:
\displaystyle \%\text{H}=\frac{\text{mass H}}{\text{mass compound}}\times 100\%%H=
mass compound
mass H
×100%
\displaystyle \%\text{C}=\frac{\text{mass C}}{\text{mass compound}}\times 100\%%C=
mass compound
mass C
×100%
If analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C, the percent composition would be calculated to be 25% H and 75% C:
\displaystyle \%\text{H}=\frac{2.5\text{g H}}{10.0\text{g compound}}\times 100\%=25\%%H=
10.0g compound
2.5g H
×100%=25%
\displaystyle \%\text{C}=\frac{7.5\text{g C}}{10.0\text{g compound}}\times 100\%=75\%%C=
10.0g compound
7.5g C
×100%=75%
Molarity = moles / liter of solution
Given, Moles of Fe³⁺ = 0.200
Volume of solution = 725 ml = 0.725 L
Conversion factor: 1000 ml = 1L
Molarity = 0.200 / 0.725 L = 0.275 M
The dissociation of Fel₃ in water is ad follows:
Fel₃ → Fe³⁺ + 3l⁻
1 mole of Fel₃ gives 1 mole of Fe³⁺ ions and 3 moles of l⁻
Since the solution is 0.275 M of Fel₃, so there are 0.275 M of Fe³⁺ ions,
and (3 x 0.275 M) = 0.825 M of l⁻ ions.
