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NISA [10]
3 years ago
7

The atomic mass of an imaginary element X is 30.7377 amu. Element X has three isotopes. One isotope has an isotopic mass of 30.5

734 amu and an abundance of 85%. Another isotope has an isotopic mass of 31.2892 and an abundance of 10%. The last isotope has an abundance of 5%. What is the isotopic mass of the last isotope? Round your answer to the nearest hundredth.
Chemistry
1 answer:
ryzh [129]3 years ago
8 0

Answer:

wuhwwisbsjsbsisjsbsjsjiw

Explanation:

bhajwjbw wjsjw bwuebe e

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A metallic material that is made by uniformly mixing copper with zinc to form bronze.
Airida [17]

YOU SUCK YOUR MOMS CHIVKEN NUGGA              YOU WHODDDDDDD :))))))))) \\\

<h2>it is not dirty so do try it >:((((((</h2>
8 0
3 years ago
Consider the following genotype Yy Ss Hh
Ilya [14]

From: Yy Ss Hh 8 different gametes can be formed

  • 5 Eye Color Genes = 243 genotypes
  • 10 Eye Color Genes= 59049 genotypes
  • 20 Eye Color Genes= 3,486,784,401 genotypes

This is further explained below.

<h3>What is a gamete?</h3>

Generally, Gametes are the cells of an organism that are responsible for reproduction. In certain contexts, they are also referred to as egg cells and sperm cells.

The popular word for female gametes is ova, whereas the common name for male gametes is sperm. Ovum and egg cells are other frequent names for female gametes.

Gametes are instances of haploid cells since they only contain a single copy of each chromosome. Haploid cells are described as having only one copy of each chromosome.

In conclusion, For 5 Eye Color Genes

3^n is implored hence

3^5=243 genotypes

Repeating said pattern e have

  • 10 Eye Color Genes= 59049 genotypes
  • 20 Eye Color Genes= 3,486,784,401 genotypes

Read more about gamete

brainly.com/question/2569962

#SPJ1

CQ

4. Consider the following genotype: Yy Ss Hh. We have now added the gene for height: Tall (H) or Short (h).

a. How many different gamete combinations can be produced?

b. Many traits (phenotypes), like eye color, are controlled by multiple genes. If eye color were controlled by the number of genes indicated below, how many possible genotype combinations would there be in the following scenarios?

a. 5 Eye Color Genes:

b. 10 Eye Color Genes:

c. 20 Eye Color Genes:

6 0
1 year ago
If there are 40 mol of NBr3 and 48 mol of NaOH, what is the excess reactant
Vanyuwa [196]
<h3>Answer:</h3>

              Excess Reagent =  NBr₃

<h3>Solution:</h3>

The Balance Chemical Equation for the reaction of NBr₃ and NaOH is as follow,

                       2 NBr₃ + 3 NaOH   →   N₂ + 3 NaBr + 3 HBrO

Calculating the Limiting Reagent,

According to Balance equation,

               2 moles NBr₃ reacts with  =  3 moles of NaOH

So,

           40 moles of NBr₃ will react with  =  X moles of NaOH

Solving for X,

                       X  =  (40 mol × 3 mol) ÷ 2 mol

                       X  =  60 mol of NaOH

It means 40 moles of NBr₃ requires 60 moles of NaOH, while we are provided with 48 moles of NaOH which is Limited. Therefore, NaOH is the limiting reagent and will control the yield of products. And NBr₃ is in excess as some of it is left due to complete consumption of NaOH.

6 0
3 years ago
Consider the balanced equation below
Svet_ta [14]
PCl₃ + Cl₂ = PCl₅

1:1
3 0
3 years ago
Read 2 more answers
hurry please! avogadro's law relates the volume of a gas to the number of moles of gas when temperature and pressure are constan
DIA [1.3K]

Answer:

Option B. 4 moles of the gaseous product

Explanation:

Data obtained from the question include:

Initial volume (V1) = V

Initial number of mole (n1) = 2 moles

Final volume (V2) = 2V

Final number of mole (n2) =..?

Applying the Avogadro's law equation, we can obtain the number of mole of the gaseous product as follow:

V1/n1 = V2/n2

V/2 = 2V/n2

Cross multiply

V x n2 = 2 x 2V

Divide both side by V

n2 = (2 x 2V)/V

n2 = 2 x 2

n2 = 4 moles

Therefore, 4 moles of the gaseous product were produced.

5 0
3 years ago
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