Answer: The speed is equivalent to <u>159.39 kilometers per hour </u>or <u>2.65 kilometers per minute.</u>
Explanation:
Given, The speed of a race car = 99 miles/ hour
To convert the speed into kilometers per hour and kilometers per minute
Since 1 mile = 1.61 kilometers
So, Speed of car = (99 ) x (1.61 )
= 159.39 kilometers per hour.
Also, 1 hour = 60 minutes
Then, Speed of car = (159.39) ÷60
= 2.6565≈2.65 kilometer per minute.
Hence, the speed is equivalent to <u>159.39 kilometers per hour </u>or <u>2.65 kilometers per minute.</u>
Answer:
A jump occurs when a core electron is removed.
Explanation:
A jump in ionization energy occurs when a core electron is removed. A large jump in the ionization energy easily be seen from the electronic configuration of an element.
For Beryllium, the electronic configuration of is 1s2 2s2.
There are two valence electrons in the outermost shell hence the ionization energy data for beryllium will show a sudden jump or increase in going from the second to the third ionization energy owing to the removal of a core electron
The electronic configuration for Nitrogen is 1s2 2s2 2p3. Five valence electrons are found in the outermost shell so the ionization energy data for nitrogen will show a sudden jump or increase in going from the fifth to sixth ionization energy because of the removal of a core electron
The electronic configuration of oxygen is 1s2 2s2 2p4. There are six valence electrons hence ionization energy for oxygen atom will show a sudden jump or increase in going from the sixth to the seventh ionization energy because of the removal of a core electron
The electronic configuration of Lithium is 1s2 2s1
There is one valence electron in its outermost shell so its ionization energy data will show a sudden jump or increase in going from the first to the second ionization energy because of the removal of a core electron.
The given question is incomplete. The complete question is:What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances.
Isotope mass amu Relative abundance
1 77.9 14.4
2 81.9 14.3
3 85.9 71.3
Express your answer to three significant figures and include the appropriate units.
Answer: 84.2 amu
Explanation:
Mass of isotope 1 = 77.9
% abundance of isotope 1 = 14.4% = 
Mass of isotope 2 = 81.9
% abundance of isotope 2 = 14.3% = 
Mass of isotope 3 = 85.9
% abundance of isotope 2 = 71.3% = 
Formula used for average atomic mass of an element :
![A=\sum[(77.9\times 0.144)+(81.9\times 0.143)+(85.9\times 0.713)]](https://tex.z-dn.net/?f=A%3D%5Csum%5B%2877.9%5Ctimes%200.144%29%2B%2881.9%5Ctimes%200.143%29%2B%2885.9%5Ctimes%200.713%29%5D)
Therefore, the average atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances is 84.2 amu
Answer:
1. Equivalence point
2. Direct titration
3. Primary standard
4. Titrand
5. Back titration
6. Standard solution
7. Titrant
8. Indirect titration
9. End point
10. Indicator
Explanation:
1. The equivalence point is the tiration point at which the quantity or moles of the added titrant is sufficient or equal to the quantity or moles of the analyte for the neutralization of the solution of the analyte.
2. Direct titration is a method of quantitatively determining the contents of a substance
3. A primary standard is an easily weigh-able representative of the mount of moles contained in a substance
4. A titrand is the substance of unknown concentration which is to be determined
5. The titration method that uses a given amount of an excess reagent to determine the concentration of an analyte is known as back titration
6. A standard solution is a solution of accurately known concentration
7. A titrant is a solution that has a known concentration and which is titrated unto another solution to determine the concentration of the second solution
8. Indirect titration is the process of performing a titration in athe reverse order
9. The end point is the point at which the indicator indicates that the equivalent quantities of the reagents required for a complete reaction has been added
10 An indicator is a compound used to visually determine the pH of a solution.
2.91 mol Al * ( 26.982 g Al / 1 mol Al) = 78.518 grams
