What is the solution to the problem rounded to the correct number of significant figures? 8.01 × 4.1 = ? A. 32.8 B. 32.84 C. 32.
2 answers:
<u>Answer:</u> The correct option is Option D.
<u>Explanation:</u>
Significant figures are defined as the figures that represent the digits of a number which carry an important contribution to the numerical value, starting with the first non-zero digit. For Example: 105.268 has 6 significant figures because every digit carry its own contribution towards the numerical value.
Whenever there is multiplication, the answer will contain the same number of significant figures as there are in the least precise numerical value.
Significant figures in 8.01 are 3 and that in 4.1 are 2.
So, the answer of these two numerical value will contain 2 significant figures.
Hence, the correct option is Option D.
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0.85 moles formula units of lead nitrate will produce 0.57 moles formula units of chromium (III) nitrate.
<h3>Explanation</h3>Typically, the oxidation state of Pb in lead nitrate tend to be +2. In other words, Pb in lead nitrate tends to exist as ions. The formula for a nitrate ion is
. The charge on each of the nitrate ion is -1. The charge on the two ions should balance. As a result, each
ion in lead nitrate would pair up with two
ions. The formula for lead nitrate will be
. Each formula unit of lead nitrate will contain one
ion and two
ions.
The "III" in the name "chromium (III) nitrate" is a Roman Numeral. It indicates that the oxidation state of Cr in chromium (III) nitrate is +3. The Cr in that compound will exist as . Similarly, each
will pair up with three
ions. The formula for chromium (III) nitrate will be
. Each formula unit of chromium (III) nitrate will contain one
ion and three
ions.
0.85 moles formula units of lead nitrate will contain 0.85 × 2 = 1.7 moles of ions. Those nitrate ions will end up in 1.7 / 3 = 0.57 moles formula units of chromium (III) nitrate. As a result, the reaction will produce 0.57 moles formula units of chromium (III) nitrate.