A law is statement about an observed concept. A theory involves the explanation of scientific concepts or principles. A hypothesis is the predicted explanation about some concepts that has to be tested in order to prove it to be right. An observation is the observing the results of a scientific experiment carried out to test an hypothesis.
Here the given statement 'A chemist mixes sodium with water and witnesses a violent reaction between the metal and water,' can be classified as an observation as it explains what the chemists observes as a result of his chemical experiment or test..
<span>Answer: B. Ionic solids have higher melting points than molecular solids.
</span>
This is because the rest are false, as solids are able to melt, and do have melting points. Also, not all solids have the same melting points.
Answer:
1 mole represents 6.023×1023 particles.
1 mole of iodine atom= 6.023×1023
Given 127.0g of iodine.
no. of iodine atom = 1 mole of iodine
1mole of magnesium = 24g of Mg = 6.023×1023no.of Mg
Given 48g of Mg = 2×6.023×1023
no. of Mg = 2 moles of Mg
1 mole of chlorine atom= 6.023× 1023
no. of chlorine atom = 35.5g of chlorine atom
Given 71g of chlorine atom=2× 6.023× 1023
no. of chlorine atom = 6.023×1023
2 moles of chlorine atom.
Given that 4g of hydrogen atom.
will be equal to 4 × 6.023 × 1023
no. of atoms of hydrogen= 4 moles of hydrogen atom.
<u>Answer:</u> The percentage abundance of
and
isotopes are 77.5% and 22.5% respectively.
<u>Explanation:</u>
Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.
Formula used to calculate average atomic mass follows:
.....(1)
Let the fractional abundance of
isotope be 'x'. So, fractional abundance of
isotope will be '1 - x'
- <u>For
isotope:</u>
Mass of
isotope = 35 amu
Fractional abundance of
isotope = x
- <u>For
isotope:</u>
Mass of
isotope = 37 amu
Fractional abundance of
isotope = 1 - x
Average atomic mass of chlorine = 35.45 amu
Putting values in equation 1, we get:
![35.45=[(35\times x)+(37\times (1-x))]\\\\x=0.775](https://tex.z-dn.net/?f=35.45%3D%5B%2835%5Ctimes%20x%29%2B%2837%5Ctimes%20%281-x%29%29%5D%5C%5C%5C%5Cx%3D0.775)
Percentage abundance of
isotope = 
Percentage abundance of
isotope = 
Hence, the percentage abundance of
and
isotopes are 77.5% and 22.5% respectively.