All categories

3 years ago

How many electron shells/orbitals will be around the nucleus of an atom of aluminum?

2 answers:

7 0

C.
There is 13 electrons in a aluminum atom. Looking at the picture, you can see there are two electrons in shell one, eight in shell two, and three in shell three.

Romashka-Z-Leto [24]3 years ago

6 0

C would be the correct answer I believe

You might be interested in

I think it’s D but I’m not sure.. Please help!

alisha [4.7K]

Answer: I think it’s D?

Explanation:

7 0

3 years ago

Read 2 more answers

what is the pH of a solution prepared from solid, neutral 2-nitrophenol providing a fromal concentration of 0.0353M, given that

REY [17]

Answer:

pH = 4.34

Explanation:

pH= -1/2(logKa) -1/2(log C)

= -1/2( log 5.98*10^-8) -1/2(log 0.0353)

=-1/2(-7.22)-1/2(-1.45)

=3.61+0.725= 4.34

7 0

3 years ago

How is ozone produced?

zubka84 [21]

The answer is A. the solar ultraviolet ray breaks the molecule apart

3 0

3 years ago

Read 2 more answers

400 liters of a certain gas is collected at STP. What will the volume be at 273 C and 190 torr pressure?

Firlakuza [10]

Answer:

2.00 L of a gas is collected at 25.0°C and 745.0 mmHg. What is the volume at STP? STP is a common abbreviation for "standard temperature and pressure." You have to recognize that five values are given in the problem and the sixth is an x. Also ... 273 1. A gas has a volume of 800.0 mL at minus 23.00 °C and 300.0 torr.

Explanation:

3 0

3 years ago

If 20.0 g of NaOH is added to 0.750 L of 1.00 M Cd(NO₃)₂, how many grams of Cd(OH)₂ will be formed in the following precipitatio

bulgar [2K]

Answer:

$m_{Cd(OH)_2}=36.6 gCd(OH)_2$

Explanation:

Hello.

In this case, for the given chemical reaction, in order to compute the grams of cadmium hydroxide that would be yielded, we must first identify the limiting reactant by computing the yielded moles of that same product, by 20.0 grams of NaOH (molar mass = 40 g/mol) and by 0.750 L of the 1.00-M solution of cadmium nitrate as shown below considering the 1:2:1 mole ratios respectively:

$n_{Cd(OH)_2}^{by\ NaOH}=20.0gNaOH*\frac{1molNaOH}{40gNaOH} *\frac{1molCd(OH)_2}{2molNaOH} =0.25molCd(OH)_2\\\\n_{Cd(OH)_2}^{by\ Cd(NO_3)_2}=0.750L*1.00\frac{molCd(NO_3)_2}{L}*\frac{1molCd(OH)_2}{1molCd(NO_3)_2} =0.75molCd(OH)_2$

Thus, since 20.0 grams of NaOH yielded less of moles of cadmium hydroxide, NaOH is the limiting reactant, therefore the mass of cadmium hydroxide (molar mass = 146.4 g/mol) is:

$m_{Cd(OH)_2}=0.25molCd(OH)_2*\frac{146.4gCd(OH)_2}{1molCd(OH)_2} \\\\m_{Cd(OH)_2}=36.6 gCd(OH)_2$

Best regards.

4 0

3 years ago