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Lelechka [254]
4 years ago
15

The number of hydrogen atoms in 6NH3 is

Biology
2 answers:
Vika [28.1K]4 years ago
4 0

Answer: 18

Explanation:

This molecule, called ammonia, is a group of atoms held together by chemical bonds. In this example, the atoms (smallest pieces of matter) are nitrogen (N) and hydrgoen (H)

The number 6 indicates<u> the number of moles of the NH3 molecule</u>. A mole  is the unit of measurement used to measure the number of atoms or molecules. One mole is equal to 6.02214078×10^23 of the same things.

<u>And since there are 3 Hydrogen atoms found in each mole of ammonia,  and there are 6 moles of amminia, then 6 x 3 = 18 atoms of hydrogen.</u>

<u />

mamaluj [8]4 years ago
3 0

No 6 is common for N and H. 3 is only for H(Hydrogen).

For finding the number of hydrogen atoms we should multiply 6 with 3

 So, 6*3 = 18. So there will be 18 hydrogen atoms in 6NH₃

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3 years ago
What is the probability that each of the following pairs of parents will produce the indicated offspring? (Assume independent as
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Answer:

(a) AABBCC × aabbcc->AaBbCc     → 1

(b) AABbCc × AaBbCc->AAbbCC   → 1/32

(c) AaBbCc × AaBbCc->AaBbCc     → 1/8

(d) aaBbCC × AABbcc->AaBbCc    → 1/2

Explanation:

In such cases we calculate the probability of each allelic pair separately.

(a) AABBCC × aabbcc -> AaBbCc  

Aa  Aa   Aa  Aa           Bb  Bb  Bb  Bb         Cc  Cc  Cc  Cc

↓                                   ↓                                ↓  

4/4 = 1                          4/4 = 1                       4/4 = 1    

In case of gene A, out of the 4 probable allelic combinations, 4 will be Aa so the probability is 4/4 = 1.

In case of gene B, out of the 4 probable allelic combinations, 4 will be Bb so the probability is 4/4 = 1.

In case of gene C, out of the 4 probable allelic combinations, 4 will be Cc so the probability is 4/4 = 1.

So, the total probability of getting AaBbCc   = 1 x 1 x 1 = 1.                

(b) AABbCc × AaBbCc -> AA bb CC

AA  AA   Aa  Aa           BB  Bb  Bb  bb         CC  Cc  Cc  cc

↓                                                        ↓           ↓  

2/4                                                    1/4          1/4

In case of gene A, out of the 4 probable allelic combinations, 2 will be AA so the probability is 2/4.

In case of gene B, out of the 4 probable allelic combinations, 1 will be bb so the probability is 1/4.

In case of gene C, out of the 4 probable allelic combinations, 1 will be CC so the probability is 1/4.

So, the total probability of getting AA bb CC = 2/4 x 1/4 x 1/4 = 1/32.

(c) AaBbCc × AaBbCc -> AaBbCc

AA  Aa   Aa  aa           BB  Bb  Bb  bb         CC  Cc  Cc  cc

        ↓                                  ↓                               ↓  

       2/4                               2/4                           2/4

In case of gene A, out of the 4 probable allelic combinations, 2 will be Aa so the probability is 2/4.

In case of gene B, out of the 4 probable allelic combinations, 2 will be Bb so the probability is 2/4.

In case of gene C, out of the 4 probable allelic combinations, 2 will be Cc so the probability is 2/4.

So, the total probability of getting AaBbCc = 2/4 x 2/4 x 2/4 = 1/8.

(d) aaBbCC × AABbcc->AaBbCc

Aa  Aa   Aa  Aa               BB  Bb  Bb  bb          Cc  Cc  Cc  Cc

      ↓                                        ↓                                ↓  

4/4 = 1                                       2/4                          4/4 = 1    

In case of gene A, out of the 4 probable allelic combinations, 4 will be Aa so the probability is 4/4 = 1.

In case of gene B, out of the 4 probable allelic combinations, 2 will be Bb so the probability is 2/4.

In case of gene C, out of the 4 probable allelic combinations, 4 will be Cc so the probability is 4/4 = 1.

So, the total probability of getting AaBbCc = 1 x 2/4 x 1 = 1/2.

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