The answer is 42 grams of NaF are there in 1 mole.
<h3>
What is a mole ?</h3>
A mole is defined as 6.022 × 10²³ atoms, molecules, ions, or other chemical units.
and the molar mass of a substance is defined as the mass of 1 mole of that substance, expressed in grams per mole.
It is equal to the mass of 6.022 × 10²³ atoms, molecules, or formula units of that substance.
Molar Mass , i.e. mass of 1 mole of NaF is sum of molar mass of Na and F
23 + 19
42 grams
Therefore 42 grams of NaF Sodium-fluoride are there in 1 mole.
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Part 1: Potassium, and Rubidium.
Part 2: Calcium has 20 protons and 20 electrons because the atomic number for calcium is 20 and that determines how many protons there are and in an atom, the number of protons is the same number of electrons. Calcium has about 20 neutrons. I got the number of Neutrons by subtracting the mass number(40.078) and the atomic number(20), I got 20.078. Round to the nearest whole number because you cannot have half or partial neutron. So, Calcium has 20 protons, 20 electrons, and 20 neutrons,
Hope this helps and please mark as brainliest!
Answer:
Here you go!
Explanation:
Our Universe is just a small part. There are many other Universes’ that exist. We live in a Multiverse. ...
Astronomer Edwin Hubble, in 1920’s, discovered that the Universe is not static. It is expanding and contracting continuously.
There is a dark energy that is making the Universe expand and accelerate at a larger rate than it did many years ago. ...
The Universe is infinite. It has no end. Thus scientists believe that the Universe is not a closed sphere but as flat as a sheet of paper and has no ...
According to the scientists the planets, stars and galaxies include only 4% of what a Universe consists of. 96% of the things in the Universe cannot be seen.
Explanation:
The given data is as follows.
(NaCl) = 
(H-O=C-ONO) = 
(HCl) = 
Conductivity of monobasic acid is 
Concentration = 0.01 
Therefore, molar conductivity (
) of monobasic acid is calculated as follows.

= 
= 
= 
Also,
= 
= 
= 
Relation between degree of dissociation and molar conductivity is as follows.

= 
= 0.1254
Whereas relation between acid dissociation constant and degree of dissociation is as follows.
K = 
Putting the values into the above formula we get the following.
K = 
= 
= 
= 
Hence, the acid dissociation constant is
.
Also, relation between
and
is as follows.

= 
= 3.7454
Therefore, value of
is 3.7454.
Answer:
C) atmosphere → plants → animals → soil
Explanation:
The third choice provides the correct path through which carbon is cycled in nature.
Carbon passes from the atmosphere to plants then to animals and finally to the soil.
- Plants uses carbon dioxide from the atmosphere to manufacture their food.
- The food is made up of giant carbon chains which also provides nourishment for animals.
- When animals digest plant matter, they obtain energy for their process.
- The waste is passed into the soil .