Practice

Chemistry

1 answer:

krek1111 [17]2 years ago

4 0

The number of significant figures in each of the following numbers are as follows:

640 cm³ - number of significant figures: 2
200.0 mL - number of significant figures: 4
0.5200 g - number of significant figures: 4
1.005 kg - number of significant Figures: 4
10,000 L - number of significant figures: 1
20.900 cm - number of significant figures: 5

<h3>What is significant figures?</h3>

Significant figures are those figures that are meaningful with respect to the precision of a measurement.

In other words, a digit that is nonzero, followed by a nonzero digit,or (for trailing zeroes) justified by the precision of the derivation or measurement is a significant figure.

Based on the question, the number of significant figures in each of the following numbers are as follows:

640 cm³ - number of significant figures: 2
200.0 mL - number of significant figures: 4
0.5200 g - number of significant figures: 4
1.005 kg - number of significant figures: 4
10,000 L - number of significant figures: 1
20.900 cm - number of significant figures: 5

Learn more about significant figures at: brainly.com/question/14359464

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4 years ago

A plutonium atom undergoes nuclear fission. Identify the missing element in the nuclear equation.

horrorfan [7]

Answer:

$_{92}^{226}U$

Explanation:

Let's firstly identify the atomic number (the number of protons) of Pu. This is done by referring to the periodic table and finding Pu. The atomic number of Pu is:

$Z=94$

In order to identify the type of a nuclear decay, we need to find the N/Z ratio. This is the ratio between the number of neutrons and the atomic number of an isotope. The number of neutrons is found by subtracting the number of protons from the mass number:

$N=M-Z$

That said, the N/Z ratio equation becomes:

$N/Z=\frac{M-Z}{Z}=\frac{M}{Z}-1=\frac{230}{94}-1=1.45$

This is a relatively high number thinking about the belt of stability of isotopes. Ideally, stable isotopes with a low Z value have an N/Z ratio of 1. Heavier isotopes with Z > 50 would have a slightly higher N/Z ratio and would be stable around N/Z = 1.25. This means we wish to decrease the N/Z ratio as much as possible.

Among all the decays, alpha-decay is preferred to decrease the N/Z ratio significantly (1.45 is much higher than 1.25). That said, we'll release an alpha particle with some nucleotide X of mass M and atomic number Z:

$_{94}^{230}Pu\rightarrow _Z^MX+_2^4\alpha$