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borishaifa [10]
3 years ago
9

3. Rank each of the following in order of DECREASING atomic radius a. Cl, Br, Ga

Chemistry
1 answer:
netineya [11]3 years ago
7 0

Answer:

Br,Ga,Cl?

Explanation:

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How many atoms are present in 4.0 Mol of sodium
LekaFEV [45]
1 mole = 6.02 x 10^23 atoms (of any element)

So we are going to take our known value of 4 moles, and multiply by 6.02 x 10^23 (Avogadro's number) and we will get the number of atoms that are in 4 moles. 

(4.0 moles of Na) x (6.02 x 10^23) / (1 mole) = 2.4 x 10^24 atoms of Na
There are 2.4 x 10^24 atoms of Na in 4.0 moles. 


8 0
3 years ago
Helppppp me please. chem question
yKpoI14uk [10]
Oxygen is the correct answer hope this helps
5 0
4 years ago
Please help!<br><br> Calculate the percent by mass of water in FeCl2 · 4H20.
Sedaia [141]

Answer:

36.23 %

Explanation:

Let's <em>assume we have 1 mol of FeCl₂ · 4H₂O</em>. In that case we would have:

  • 1 mol of FeCl₂, weighing 126.75 g (that's the molar mass of FeCl₂), and
  • 4 moles of H₂O, weighing (4 * 18 g/mol) 72 g.

Now we can <u>calculate the percent by mass of water</u>:

  • % mass = mass of water / total mass * 100%
  • % mass = \frac{72}{72+126.75} * 100% = 36.23 %
5 0
3 years ago
Do plants uptake phosphorus as P?
salantis [7]
In general, roots absorb phosphorus in the form of orthophosphate, but can also absorb certain forms of organic phosphorus. Phosphorus moves to the root surface through diffusion.
3 0
3 years ago
The atomic masses of 151eu and 153eu are 150.919860 and 152.921243 amu, respectively. The average atomic mass of europium is 151
vitfil [10]

Answer:-  The natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .

Solution:- Average atomic mass of an element is calculated from the atomic masses of it's isotopes and their abundances using the formula:

Average atomic mass = mass of first isotope(abundance) + mass of second isotope(abundance)

We have been given with atomic masses for ^1^5^1_E_u and ^1^5^3_E_u as 150.919860 and 152.921243 amu, respectively.  Average atomic mass of Eu is 151.964 amu.

Sum of natural abundances of isotopes of an element is always 1. If we assume the abundance of ^1^5^1_E_u as n then the abundance of ^1^5^3_E_u would be 1-n .

Let's plug in the values in the formula:

151.964=150.919860(n)+152.921243(1-n)

151.964=150.919860n+152.921243-152.921243n

on keeping similar terms on same side:

151.964-152.921243=150.919860n-152.921243n

-0.957243=-2.001383n

negative sign is on both sides so it is canceled:

0.957243=2.001383n

n=\frac{0.957243}{2.001383}

n=0.478

The abundance of ^1^5^1_E_u is 0.478 which is 47.8%.  

The abundance of ^1^5^3_E_u is = 1-0.478

= 0.522 which is 52.2%

Hence, the natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .


3 0
3 years ago
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