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Tom [10]
3 years ago
15

Which of the following is a benefit of tidal energy

Chemistry
2 answers:
wariber [46]3 years ago
8 0

Answer:

A i took the test

Explanation:

Hoochie [10]3 years ago
6 0

I believe it’s A but I could be wrong

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A single serving bag of snack chips contains 65.0 Cal. Assuming that all of the energy from eating these chips goes toward keepi
AveGali [126]

Answer:

= 62.1 hours

Explanation:

Energy provide by the serving is 65 cal

= 65 cal  × 4.184 Kj = 271.96 kJ

271.96 KJ = 271960 J

Energy required for 1minute of energy

= 73 x 1

= 73 J/min

So, 271960 joules will be required for 271960 heart beat

Minutes = 271960 / 73

= 3593.94 minutes  

Time in hours = 3725.429 / 60

= 62.1 hours

5 0
3 years ago
Humans have used rocks in the creation of tools and building materials true or flase
Anuta_ua [19.1K]
True, if you would like an example look at Indian arrow heads or early architecture all use rocks.
4 0
3 years ago
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Find the total surface area of the rectangular prism in the figure. A) 174 cm^ 2 B) 522c * m ^ 2; 348c * m ^ 2 D) 432c * m ^ 2
WARRIOR [948]
Surface area: 348 m^2
8 0
2 years ago
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balance the following equation by oxidation number method KMnO4 + Na2So3 gives Mno2 + Na2 So4 + Koh​
belka [17]

Answer:

good i think

Explanation:

yes correct

7 0
3 years ago
Using only the 1h nmr spectrum of the crude nitration product, determine if the product is mostly ortho-, meta-, or para- substi
tatiyna
<span>The nitartion of methyl benzoate is expected to proceed as given in the equation below:

</span>

In methyl benzoate there are 3 types of 1 H proton. The two ortho to the C=O group is a doublet at 8 ppm the 2 metal to the C=O is a multiple at 7.5 ppm and one para to the C=O is a multiplet at 7.5 ppm.

On nitration the ortho will probably show two signal one being a single with 3 proton integration and one a doublet with 1 H integration

The meta will show a highly down field singlet (coresponding to 1 proton), two unequal doublets (corresponding to 1 H each) and one multiplets (corresponding to 1H). This is the major product as seen from the 1H NMR.

The para isomer will come as two doublets which will be very close to each other there is a small signal for this set between 8.2 and 8.3 ppm.

7 0
3 years ago
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