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Dafna11 [192]
4 years ago
15

18. Explain how an organism with a genotype Gg and an organism with the genotype GG

Chemistry
2 answers:
o-na [289]4 years ago
8 0

G is the dominant trait, so even though the recesseive trait (g) is present in one organism, they both contain the dominant trait, which hides the recessive trait when it is present. They both display the dominent phenotype

My name is Ann [436]4 years ago
3 0

Answer:

because they both have the dominant gene which is G

Explanation:

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The diagram below is an artist’s impression of a single atom of element Be. The neutrons are shown with stripes, the protons are
Ahat [919]

Answer is: beryllium-10.

The diagram shows that atom has 4 protons, 5 neutrons and 2 valence electrons.

Atomic number is the number of protons, which is characteristic of a chemical element, beryllium (Be) is an element with atomic number 4.

Two valence electrons means that atom is from 2. group of periodic table, only beryllium is from that group; sodium (1. group), boron (13. group) and carbon (14. group).

Beryllium-10 has 6 neutrons, so it is isotope (different number of neutrons or mass number).

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a chemist determined by measurements that 0.015 moles of tungsten participated in a chemical reaction. Calculate the mass of tun
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Explanation:

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What is 4.9 + 3.822 is sig fig?
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A) Find the gas speed of ammonia at 25.0 degrees Celsius. ______________
Cloud [144]

a. 661.23 m/s

b.  the rate of effusion of Ammonia = 4.5 faster than Silicon tetra bromide

<h3> Further explanation </h3>

Given

T = 25 + 273 = 298 K

Required

a. the gas speed

b. The rate of effusion comparison

Solution

a.

Average velocities of gases can be expressed as root-mean-square averages. (V rms)  

\large {\boxed {\bold {v_ {rms} = \sqrt {\dfrac {3RT} {Mm}}}}

R = gas constant, T = temperature, Mm = molar mass of the gas particles  

From the question  

R = 8,314 J / mol K  

T = temperature  

Mm = molar mass, kg / mol  

Molar mass of Ammonia = 17 g/mol = 0.017 kg/mol

\tt v=\sqrt{\dfrac{3\times 8.314\times 298}{0.017} }=661.23~m/s

b. the effusion rates of two gases = the square root of the inverse of their molar masses:  

\rm \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }

M₁ = molar mass Ammonia NH₃= 17

M₂ =  molar mass Silicon tetra bromide SiBr₄= 348

\rm \dfrac{r_1}{r_2}=\sqrt{\dfrac{348}{17} }=4.5

the rate of effusion of Ammonia = 4.5 faster than Silicon tetra bromide

3 0
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