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baherus [9]
3 years ago
14

I need the answer in 15 mins pls urjent

Chemistry
1 answer:
Sedbober [7]3 years ago
7 0

Answer:

I don't know I'm sorry I will tell you another answer asks me to

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How many orbitals are contained in the third main energy level and in the fifth energy level
antoniya [11.8K]

There are 9 orbitals in the third energy level and 25 orbitals in the fifth energy level.

I hope this helps you.

5 0
3 years ago
I will make you brainiest! In this analogy, which part of the lunar cycle is waning - from the new moon to the full moon or the
Veronika [31]

Answer:

Explanation:

After the glorious appearance of Full Moon, the lunar shape starts to wane, meaning it gets smaller. It's visible later at night and into the early morning, and we see a steadily shrinking shape of the lunar surface that's being lit up.

7 0
2 years ago
What mass of Cu(s) is electroplated by running 24.5A of current through a Cu2+(aq)solution for 4.00 h?Express your answer to thr
Masja [62]

Answer: 116 g of copper

Explanation:

Q=I\times t

where Q= quantity of electricity in coloumbs

I = current in amperes = 24.5A

t= time in seconds =  4.00 hr = 4.00\times 3600s=14400s  (1hr=3600s)

Q=24.5A\times 14400s=352800C

Cu^{2+}+2e^-\rightarrow Cu

2\times 96500C=193000C  of electricity deposits 63.5 g of copper.

352800 C of electricity deposits = \frac{63.5}{193000}\times 352800=116g of copper.

Thus 116 g of Cu(s) is electroplated by running 24.5A of current

Thus  remaining in solution = (0.1-0.003)=0.097moles

8 0
2 years ago
Which example best represents an interaction of atmosphere and hydrosphere. A clouds form, b sunlight is reflected off the upper
olga55 [171]
B.sunligh 
is reflected off the upper atmosphere

3 0
3 years ago
Calculate the percentage of each element in acetic acid, hc2h3o2, and glucose, c6h12o6.
My name is Ann [436]
<em>Acetic acid, HC2H3O2</em>

First, calculate for the molar mass of acetic acid as shown below.
    M = 1 + 2(12) + 3(1) + 2(16) = 60 g

Then, calculating for the percentages of each element.
<em> Hydrogen:</em>
    P1 = ((4)(1)/60)(100%) = <em>6.67%</em>

<em> Carbon:</em>
   P2 = ((2)(12)/60)(100%) = <em>40%</em>

<em>Oxygen</em>
  P3 =((2)(16) / 60)(100%) = <em>53.33%</em>

<em>Glucose, C6H12O6</em>

The molar mass of glucose is as calculated below,
   6(12) + 12(1) + 6(16) = 180

The percentages of the elements are as follow,
 <em> Hydrogen:</em>
   P1 = (12/180)(100%) = <em>6.67%</em>

<em>Carbon:</em>
  P2 = ((6)(12) / 180)(100%) = <em>40%</em>

<em>Oxygen:</em>
  P3 = ((6)(16) / 180)(100%) = <em>53.33%</em>

b. Since the empirical formula of the given substances are just the same and can be written as CH2O then, the percentages of each element composing them will just be equal. 
6 0
3 years ago
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