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serious [3.7K]
3 years ago
10

What is a mutation

Chemistry
2 answers:
Lubov Fominskaja [6]3 years ago
6 0

the second one, any change in a DNA gene or chromosome

Rzqust [24]3 years ago
5 0

Answer:

2nd one

Explanation:

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If D+2 would react with E!, what do you predict to be the formula?
kherson [118]

Answer:

I think the answers D. Hope this helps you.

Explanation:

6 0
3 years ago
What are four chemical and physical properties of phosphorus?
DedPeter [7]

Answer:

Properties: The melting point of phosphorus (white) is 44.1°C, boiling point (white) is 280°C, specific gravity (white) is 1.82, (red) 2.20, (black) 2.25-2.69, with a valence of 3 or 5. There are four allotropic forms of phosphorus: two forms of white (or yellow), red, and black (or violet).Oct 7, 2019

Atomic Number: 15

Atomic Weight: 30.973762

Explanation:

: )

8 0
3 years ago
Read 2 more answers
using the same balanced equation, what is the number of moles of carbon dioxide produced from 3.750 moles of propane?
kondor19780726 [428]

Answer:

11.25moles of CO2

Explanation:

First, let us generate a balanced equation for the reaction of propane to produce CO2. This reaction called Combustion. It is a reaction in which propane burns in air (O2) to produce CO2 and H20. The equation is given below:

C3H8 + 5O2 —> 3CO2 + 4H2O

From the equation,

1mole of C3H8 produced 3moles of CO2.

Therefore, 3.750 moles of C3H8 will produce = 3.750 x 3 = 11.25moles of CO2

5 0
3 years ago
Molybdenum (Mo) has a body centered cubic unit cell. The density of Mo is 10.28 g/cm3. Determine (a) the edge length of the unit
ser-zykov [4K]

<u>Answer:</u>

<u>For a:</u> The edge length of the unit cell is 314 pm

<u>For b:</u> The radius of the molybdenum atom is 135.9 pm

<u>Explanation:</u>

  • <u>For a:</u>

To calculate the edge length for given density of metal, we use the equation:

\rho=\frac{Z\times M}{N_{A}\times a^{3}}

where,

\rho = density = 10.28g/cm^3

Z = number of atom in unit cell = 2  (BCC)

M = atomic mass of metal (molybdenum) = 95.94 g/mol

N_{A} = Avogadro's number = 6.022\times 10^{23}

a = edge length of unit cell =?

Putting values in above equation, we get:

10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm

Conversion factor used:  1cm=10^{10}pm  

Hence, the edge length of the unit cell is 314 pm

  • <u>For b:</u>

To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

R=\frac{\sqrt{3}a}{4}

where,

R = radius of the lattice = ?

a = edge length = 314 pm

Putting values in above equation, we get:

R=\frac{\sqrt{3}\times 314}{4}=135.9pm

Hence, the radius of the molybdenum atom is 135.9 pm

4 0
3 years ago
An astronaut weighs 104 newtons on the moon,where the strength of gravity is 1.6 newtons per kilogram
belka [17]
I'd say he ways about 35 kilograms, but I'm probably wrong, xD

6 0
3 years ago
Read 2 more answers
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