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anyanavicka [17]
2 years ago
13

1 cholo 2 floro 4 nitro + Na^+​

Chemistry
2 answers:
N76 [4]2 years ago
4 0

Answer:

Incomplete Q.

GOOD LUCK FOR THE FUTURE! :)

REY [17]2 years ago
3 0

Answer:

What is the question ? Maybe I can help if you make a snip or take a picture  of the question

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in heating a kettle of water on an electric stove, 3.34×10^3 J of thermal energy was provided by the element of the stove. yet,
insens350 [35]

Answer:

The percentage efficiency of the electrical element is approximately 82.186%

Explanation:

The given parameters are;

The thermal energy provided by the stove element, H_{supplied} = 3.34 × 10³ J

The amount thermal energy gained by the kettle, H_{absorbed}  = 5.95 × 10² J

The percentage efficiency of the electrical element in heating the kettle of water, η%, is given as follows;

\eta \% = \dfrac{H_{supplied} - H_{absorbed} }{H_{supplied}}  \times 100

Therefore, we get;

\eta \% = \dfrac{3.34 \times 10^3 - 5.95 \times 10^2}{3.34 \times 10^3}  \times 100 = \dfrac{549}{668} \times 100 \approx 82.186 \%

The percentage efficiency of the electrical element, η% ≈ 82.186%.

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