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Anuta_ua [19.1K]
3 years ago
9

Help please answer show proof

Chemistry
1 answer:
slavikrds [6]3 years ago
5 0

Answer:

d or c

Explanation:

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PPRR, PpRR, PPRr, and PpRr
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2 years ago
One cup of kidney beans contains 15 g of protein, 1 g of fat, and 42 g of carbohydrate. how many kilocalories, to two significan
ollegr [7]
 we need to get the total calories, we can get it by getting the average of calories of the protein, fat, and carbohydrates:
calories of protein = 15g x 4 KCal/g
                               = 60 KCal
calories of fat = 1 g x 9 KCal/g
                        = 9 kCal
calories of carbohydrates = 42 g x 4 KCal/g 
                                            = 168 Kcal
∴ the total calories = 60 + 9 + 168 = 237Kcal = 0.237 Cal
to get the resualt to two significant figures 
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4 0
3 years ago
108/49 In ? 108/48 Cd + ? In the equation above, what particle or type of radiation needs to be included on the right hand side
Oksanka [162]

Answer: b) positron

Explanation:

The antiparticle or the antimatter counterpart of the electron is called positron.

The positron has an electric charge of +1 e and same mass as an electron. When a positron collides with an electron, annihilation takes place.

Below is an attachnent that further explains this

7 0
3 years ago
WILL GIVE THE BRAINLIEST
stich3 [128]

Answer:

Oxygen: Carbon Dioxide

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1:2

3 0
3 years ago
A 42.0g sample of a compound containing only C and H was analyzed. The results showed that the sample contained 36.0g of C and 6
SashulF [63]

Answer:

The correct answer is option D.

Explanation:

Mass of compound = 42.0 g

Mass of carbon in compound = 36.0 g

Mass of hydrogen in compound = 6.0 g

Moles of carbon = \frac{36.0 g}{12 g/mol}=3.0 mol

Moles of hydrogen = \frac{6.0 g}{1 g/mol}=6.0 mol

Empirical formula of the compound, divide least number of moles from each element.

Carbon = \frac{3.0}{3.0}=1

Hydrogen = \frac{6.0}{3.0}=2

Empirical formula of compound = CH_2

The empirical formula of the compound can be calculated from the given data.

4 0
2 years ago
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