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Leviafan [203]
3 years ago
10

A fertilized egg cell develops into a(n) _________

Chemistry
1 answer:
Inessa [10]3 years ago
5 0
I THINK it is d if it’s wrong i’m so sorry
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Urgent plzz help meeee thx
MakcuM [25]

Answer:

8

Explanation:

From the question given above, the following data were obtained:

t–butyl ion = (CH₃)₃C⁺

Number of valence electron =?

The valence electron(s) talks about the combining power of an element or compound as the case may be.

Considering the t–butyl ion, (CH₃)₃C⁺ we can see that it has a charge of +1 indicating that it has given out 1 electron to attain the stable octet configuration which has a valence electrons of 8. Thus, the valence electron of t–butyl ion, (CH₃)₃C⁺ is 8

5 0
2 years ago
How many grams of sodium acetate are in solution in the third beaker?
Kipish [7]

Answer:

46g of sodium acetate.

Explanation:

The data is: <em>Precipitation from a supersaturated sodium acetate solution. The solution on the left was formed by dissolving 156g of the salt in 100 mL of water at 100°C and then slowly cooling it to 20°C. Because the solubility of sodium acetate in water at 20°C is 46g per 100mL of water, the solution is supersaturated. Addition of a sodium acetate crystal causes the excess solute to crystallize from solution.</em>

The third solution is the result of the equilibrium in the solution at 20°C. As the maximum quantity that water can dissolve of sodium acetate at this temperature is 46g per 100mL and the solution has 100mL <em>there are 46g of sodium acetate in solution. </em>The other sodium acetate precipitate because of decreasing of temperature.

I hope it helps!

6 0
3 years ago
a compound has a molar mass of 129 g/mol if its empirical formula is C2H5N then what is the molecular formula
Elena-2011 [213]

Given :

A compound has a molar mass of 129 g/mol .

Empirical formula of compound is C₂H₅N .

To Find :

The molecular formula of the compound.

Solution :

Empirical mass of compound :

M_e = ( 2 \times 12 ) + ( 5 \times 1 ) + (  1  \times 14 )\\\\M_e = 43\ gram/mol

Now, n-factor is :

n = \dfrac{M}{M_e}\\\\n = \dfrac{129}{43}\\\\n = 3

Multiplying each atom in the formula by 3 , we get :

Molecular Formula, C₆H₁₅N₃

3 0
2 years ago
If the volume of a gas container at 32 degrees Celsius changes from 1.55 L to 755 mL, what will the final temperature be?
QveST [7]
So to solve this you need to know Charles’s law which is: V1/T1=V2/T2. Where T1 and V1 is the initial volume and Temperature and V2 and T2 is the temperature and volume afterwards. So first plug in the numbers you are given. V1= 1.55L T1= 32C° V2= 755mL T2=?. Since your volumes are two different units you change 755mL to be in L so that would be 0.755 L. And since your temp isn’t in Kelvin you do 273+32= 305K°. You then would rearrange your equation to solve for T2 which is V2T1/V1. Then you plug in your numbers (0.755L)(305K)/1.55L. Then you solve and would be 148.5645161 —> 1.49 x 10^2 K
4 0
3 years ago
How are electrons excited in element?
sineoko [7]

Answer:

By absorbing energy electron is jump into higher energy level. This is called excitation.

Explanation:

The electron is jumped into higher level and back into lower level by absorbing and releasing the energy.

The process is called excitation and de-excitation.

Excitation:

When the energy is provided to the atom the electrons by absorbing the energy jump to the higher energy levels. This process is called excitation. The amount of energy absorbed by the electron is exactly equal to the energy difference of orbits.  For example if electron jumped from K to L it must absorbed the energy which is equal the energy difference of these two level. The excited electron thus move back to lower energy level which is K by releasing the energy.

De-excitation:

When the excited electron fall back to the lower energy levels the energy is released in the form of radiations. this energy is exactly equal to the energy difference between the orbits. The characteristics bright colors are due to the these emitted radiations. These emitted radiations can be seen if they are fall in the visible region of spectrum.

4 0
3 years ago
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