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siniylev [52]
3 years ago
14

Given the noble gas configuration of an element: [Ar] 4s2, 3d5, what is the element?

Chemistry
1 answer:
monitta3 years ago
6 0

Question

<em>Given the noble gas configuration of an element: [Ar] 4s2, 3d5, what is the element? </em>

Answer:

<em>B.) Argon</em>

Hope this helps!

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NEED HELP FAST!!
koban [17]

Answer:

I'm pretty sure that's right.

Explanation:

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8 0
3 years ago
What do you call the procedure that helps you determine the volume of an irregularly shaped object, whole using a graduated cyli
ki77a [65]
The method is called the displacement method.

You place some water in the graduated cylinder and measure its volume.
Then you add your object and measure the new volume.
The difference between the two volumes is the volume of your object.

6 0
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Help asap <br><br> i have to get a good grade on this
baherus [9]
I think The answer is 34.5l
4 0
3 years ago
A solution is prepared by dissolving 27.7 g of cacl2 in 375 g of water. the density of the resulting solution is 1.05 g/ml. the
bagirrra123 [75]
The answer is 6.88.
Solution:
We can calculate for the percent composition of CaCl2 by mass by dividing the mass of the CaCl2 solute by the mass of the solution and then multiply by 100. The total mass of the resulting solution is the sum of the mass of CaCl2 solute and the mass of water solvent. Therefore, the percent composition of CaCl2 by mass is 
     % by mass = (mass of the solute / mass of the solution)*100 
                        = mass of solute / (mass of the solute + mass of the solvent)*100
                        = (27.7 g CaCl2 / 27.7g + 375g) * 100 
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5 0
3 years ago
Read 2 more answers
Suppose Gabor, a scuba diver, is at a depth of 15m15m. Assume that: The air pressure in his air tract is the same as the net wat
alexandr1967 [171]

Answer:

The ration of molar concentration is "2.5".

Explanation:

The given values are:

Average density of salt water,

= 1.03 \ g/cm^3

Net pressure,

= 2.00 \ atm

Increase in pressure,

= 1.00 \ atm

Now,

The under water pressure will be:

=  \frac{15 \ m}{10}\times 1 \ atm +1 \ atm

=  1.5\times 1+1

=  1.5+1

=  2.5 \ atm

hence,

The ratio will be:

=  \frac{(\frac{n}{V})_{15m} }{(\frac{n}{V})_{surface} }

or,

=  \frac{P}{P_s}

=  \frac{2.5}{1}

=  2.5

7 0
3 years ago
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