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valkas [14]
3 years ago
10

Sarah is learning about human-harnessed electricity as well as naturally-occurring electricity. Sarah comes in contact with many

forms of electricity each day. Which is an example of Sarah coming in contact with naturally-occurring electricity?
A) Sarah walks quickly across the carpet in socks and feels a small shock on her foot.
B) Sarah felt a small sting on her finger when she changed the light bulb on her lamp.
C) Sarah was lightly shocked when she inserted the plug of her lamp in the wall socket.
D) Sarah accidentally left her curling iron plugged in and burned her hand on the metal.
Chemistry
1 answer:
VladimirAG [237]3 years ago
8 0
Hello there! 

Your answer is: <span>A) Sarah walks quickly across the carpet in socks and feels a small shock on her foot.

This is because this could happen with out man-made electrical appliances. Light bulbs, wall sockets, and curling irons are all man-made, therefor not naturally-occurring. </span>


I hope I helped!

Let me know if you need anything else!

~ Zoe
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Answer:

Acids:  HNO₃ and HI

Base: KOH

Explanation:

For the Arrhenius concept, acid is a substance that, in water, produces the hydrogen ion (H⁺), and base in the substance that, in water, produces the hydroxyl ion (OH⁻).

For the first reaction then KOH must produce OH⁻ so its a base, and HNO₃ must produce H⁺ so it's an acid.

For the second one, HI must produce H⁺ so it's an acid. For the concept of Arrhenius, (CH₃)₃N can't be classified as a base or as an acid.

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During a dilution, what happens to the concentration of a solution?
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B

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Calculate how to prepare 250.00 ml of approximately 1.0 m hcl solution from the 2.5 m hcl solution. show your work.
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The sum of the first 10 terms of an arithmetic progression is 120 and the sum of first twenty is 840. find sum of first 30 terms
STatiana [176]

Answer:

The sum of first 30 terms of the arithmetic progression is <u>2160.</u>

Explanation:

For an arithmetic progression, the sum of first n terms with first term as a and common difference d is given as:

S_n=\frac{n}{2}(2a+(n-1)d)

Now, it is given that:

For\ n=10,S_n=120\\For\ n=20,S_n=840

Now, plug in these values and frame two equations in a\ and\ d

S_{10}=\frac{10}{2}(2a+(10-1)d)\\120=5(2a+9d)\\2a+9d=\frac{120}{5}\\2a+9d=24------------1

S_{20}=\frac{20}{2}(2a+(20-1)d)\\840=10(2a+19d)\\2a+19d=\frac{840}{10}\\2a+19d=84-----------2

Now, we solve equations (1) and (2) for a\ and\ d. Subtract equation (1) from equation (2). This gives,

2a+19d-2a-9d=84-24\\19d-9d=60\\10d=60\\d=\frac{60}{10}=6

Now, plug in the value of d=6 in equation (1) and solve for a.

2a+9(6)=24\\2a+54=24\\2a=24-54\\2a=-30\\a=\frac{-30}{2}=-15

Plug in the values of a=-15,\ n=30\ and\ d=6 in the sum formula to find the sum of first 30 terms.

Now, the sum of first 30 terms is given as:

S_{30}=\frac{30}{2}(2(-15)+(30-1)(6))\\S_{30}=15(-30+29(6))\\S_{30}=15(-30+174)\\S_{30}=15(144)=2160

Therefore, the sum of first 30 terms of the arithmetic progression is 2160.

4 0
3 years ago
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