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Chapter 2: Problem 35

An isotope with mass number 44 has four more neutrons than protons. This is an isotope of what element?

Short Answer

Expert verified
The isotope is of the element Calcium.

Step by step solution

01

Determine the relations

Given in the problem the mass number is 44 and there are four more neutrons than protons, you need to set up a system of two equations using these relationships: The total protons and neutrons equal to 44 (represented as p + n = 44); There are 4 more neutrons than protons (represented as n = p + 4).
02

Solve the system of equations

Substitute equation two into equation one (replace n from the first equation with p + 4), yielding p + (p + 4) = 44. This simplifies to 2p + 4 = 44 which further simplifies to 2p = 40. Dividing both sides by 2, one gets p = 20.
03

Find the corresponding Element

Now that we know there are 20 protons, we look at the periodic table to find the element with an atomic number of 20 which is Calcium (Ca).

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Most popular questions from this chapter

\begin{tabular}{l} There are four naturally occurring isotopes of \\ \hline \end{tabular} chromium. Their masses and percent natural abundances are \(49.9461 \mathrm{u}, 4.35 \% ; 51.9405 \mathrm{u}, 83.79 \% ; 52.9407 \mathrm{u}\) \(9.50 \% ;\) and \(53.9389 \mathrm{u}, 2.36 \% .\) Calculate the weightedaverage atomic mass of chromium.

One oxide of rubidium has \(0.187 \mathrm{g}\) O per gram of Rb. A possible O:Rb mass ratio for a second oxide of rubidium is (a) \(16: 85.5 ;\) (b) \(8: 42.7 ;\) (c) \(1: 2.674 ;\) (d) any of these.

In one experiment, the burning of \(0.312 \mathrm{g}\) sulfur produced 0.623 g sulfur dioxide as the sole product of the reaction. In a second experiment, \(0.842 \mathrm{g}\) sulfur dioxide was obtained. What mass of sulfur must have been burned in the second experiment?

For the ion \(^{228} \mathrm{Ra}^{2+}\) with a mass of 228.030 u, determine (a) the numbers of protons, neutrons, and electrons in the ion; (b) the ratio of the mass of this ion to that of an atom of \(^{16} \mathrm{O}\) (refer to page 47 ).

Sulfur forms two compounds with oxygen. In the first compound, \(1.000 \mathrm{g}\) sulfur is combined with \(0.998 \mathrm{g}\) oxygen, and in the second, \(1.000 \mathrm{g}\) sulfur is combined with \(1.497 \mathrm{g}\) oxygen. Show that these results are consistent with Dalton's law of multiple proportions
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