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MrMuchimi
3 years ago
7

A sample of ethanol (C2H6O) has a mass of

Chemistry
2 answers:
kupik [55]3 years ago
7 0

Answer:

6.52 kj

Explanation:

Just got it correct on edge 2020 :)

Tems11 [23]3 years ago
3 0

Answer:

6.52 kJ

Explanation:

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L20_UENAxleFsh-VHfg9rLfZsitva3L3UKGjmF1B299/vie Settle a dispute. X Anya claims that this challenge can be solved with a single
Fynjy0 [20]

Answer:

Neither arre correct

Explanation:

Neither Anya nor Braden are correct. This is because if you use 90 degrees, 180 degrees, or even 270 degrees you will not get the exact image, which means that the image will not be found by just a rotation because there will be a curve in the image. You can solve it if you can do 90 degree rotation and translation.

4 0
3 years ago
Middle School science 3 1. Illustrate the position of the Earth, moon, and sun during a lunar eclipse. You may do so by typing a
lilavasa [31]
I hope this helps you

7 0
3 years ago
Use the given half reactions to "construct" an electrolytic cell. Zn^2+ + 2 e^--------->Zn E°cell = -0.76 V Cu^2+ + 2 e^-----
Neko [114]

Answer:

See explanation and image attached

Explanation:

The standard cell potential at 298 K is given by;

E°cathode - E°anode

Hence;

E°cell = 0.34 V - (-0.76 V)

E°cell = 0.34 V + 0.76 V

E°cell = 1.1 V

To reduce Zn^2+ to Zn then Zn must be the cathode, hence;

E°cell = (-0.76 V) - 0.34 V

E°cell = -1.1 V

5 0
3 years ago
An unknown radioactive substance has a half-life of 3.20hours . If 46.2g of the substance is currently present, what mass A0 was
iogann1982 [59]

Answer:

a) a0 was 46.2 grams

b) It will take 259 years

c) The fossil is 1845 years old

Explanation:

<em>An unknown radioactive substance has a half-life of 3.20hours . If 46.2g of the substance is currently present, what mass A0 was present 8.00 hours ago?</em>

A = A0 * (1/2)^(t/h)

⇒ with A = the final amount = 46.2 grams

⇒ A0 = the original amount

⇒ t = time = 8 hours

⇒ h = half-life time = 3.2 hours

46.2 = Ao*(1/2)^(8/3.2)

Ao = 261.35 grams

<em>Americium-241 is used in some smoke detectors. It is an alpha emitter with a half-life of 432 years. How long will it take in years for 34.0% of an Am-241 sample to decay?</em>  

t = (ln(0.66))-0.693) * 432 = 259 years

It will take 259 years

<em>A fossil was analyzed and determined to have a carbon-14 level that is 80% that of living organisms. The half-life of C-14 is 5730 years. How old is the fossil?</em>

<em />

t = (ln(0.80))-0.693) * 5730 = 1845

The fossil is 1845 years old

7 0
3 years ago
Choose all the answers that apply.
vovangra [49]
Diabetes,  ALS, and Alzheimeir's.

If this was the appropriate answer make sure to mark as the brainliest!
-procklown
7 0
3 years ago
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