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STALIN [3.7K]
3 years ago
7

Eye and hair color are examples of what?

Chemistry
1 answer:
NemiM [27]3 years ago
3 0
 the anwser is genotypes which is a personal characsteristic such as skin color eye or hair color or height
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Which model would represent asexual reproduction? explain you reasoning
matrenka [14]

Answer:

Answer)   The first box represent the budding type of asexual reproduction . In budding due to the cell divison at the particular place in the organism , new  individual will produced.

Second picture represent the budding in hydra. Budding is also a asexual reproduction.

Third type of asexual reproduction is found in plants  it is vegetative reproduction.

Fourth is Binary fission . This is also a type of asexual reproduction, in which nucleus dicied into two new daughter cell via mitosis.Explanation:

8 0
3 years ago
A gas station charges 1.299 per gallon of gas. What would be the price for a liter of gas?
Fudgin [204]

Answer:

the price for a liter of gas will be of  $ 0.3432

Explanation:

⇒ $ gas / L = $ 1.299 / Gal * ( Gal / 3.785 L )

⇒ $ gas / L = $ 0.3432 / L

5 0
3 years ago
Electrons are mapped beginning with the ____ electrons.
kifflom [539]
D the lowest amount of energy
7 0
3 years ago
calculate the molar mass of nickel (iii)oxide to the nearest tenth place. Give both the numerial value and the units
maksim [4K]
Nickel (Ni) has the charge as +3 while oxygen (O) has -2. Hence, the chemical formula for the nickel (iii) oxide is Ni₂O₃. 

molar mass of Ni = 58.69 g/mol
molar mass of O = <span>15.99 g/mol

number of Ni atoms in </span>Ni₂O₃ = 2
Molar mass of Ni in Ni₂O₃ = 2 x 58.69 g/mol = 117.38 g/mol

number of O atoms in Ni₂O₃ = 3
Molar mass of O in Ni₂O₃  = 3 x 15.99 g/mol = 47.97 g/mol


Hence, molar mass of compound = 117.38 g/mol + 47.97 g/mol
                                                          = 165.35 g/mol
                                                          =
<span> 165.4 g/mol</span>
8 0
4 years ago
A soluble iodide was dissolved in water. Then an excess of silver nitrate, AgNO3, was added to precipitate all of the io- dide i
Sergeeva-Olga [200]

Answer:

m_{I^-}=1.18gI^-

Explanation:

Hello,

In this case, the reaction is given as:

I^-+AgNO_3\rightarrow AgI+NO_3^-

Thus, starting by the yielded grams of silver iodide, we obtain:

n_{I^-}=2.185gAgI*\frac{1molAgI}{234.77gAgI}*\frac{1molI}{1molAgI}=9.31x10^{-3}molI^-

Which correspond to the iodide grams in the silver iodide. In such a way, by means of the law of the conservation of mass, it is known that the grams of each atom MUST remain constant before and after the chemical reaction whereas the moles do not, therefore, the mass of iodine from the silver iodide will equal the mass of iodine present in the soluble iodide, thereby:

m_{I^-}=9.31x10^{-3}molI^-\frac{127gI^-}{1molI^-} =1.18gI^-

And the rest, correspond to the iodide's metallic cation which is unknown. Such value has sense since it is lower than the initial mass of the soluble iodide which is 1.454g, so 0.272 grams correspond to the unknown cation.

Best regards.

7 0
4 years ago
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