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Feliz [49]
3 years ago
10

What is 2-2? When the cow is green

Chemistry
2 answers:
bija089 [108]3 years ago
6 0
0 with a green cow lol
skad [1K]3 years ago
5 0
0 and the cow remains green (:
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Which group, metals or nonmetals, has the greatest chance to gain electrons and to become negative ions?
tatyana61 [14]
It would be non-metals
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Can you name some chemical elements you use or see on a regular basis?
kaheart [24]

Answer:

Hydrogen. Symbol: H. Atomic Weight: 1.008. ...

Magnesium. Symbol: Mg. Atomic Weight: 24.305. ...

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4 0
3 years ago
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How many total electrons can there be in the n=4 orbital?
Arturiano [62]

Answer:

the total electron that can be in the n=4 orbital is 32

Explanation:

it is because the formula is 2n^2,if n=4then it is 2(4)^2=32

7 0
4 years ago
Assume that a P orbital is dumbbell shaped the two major lobes in a certain atom contains one electron. In which lobe is electro
Tanya [424]

Answer:

See explanation

Explanation:

The p orbital is threefold degenerate. This implies that the p-sublevel is composed of three orbitals; px, py and pz.

According to Hund's rule, electrons occur singly when filling degenerate orbitals before pairing takes place. Since the three orbitals are degenerate, any of px, py or pz may be first filled.

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5 0
3 years ago
Assuming that you start with 21.4 g of ammonia gas and 18.0 g of sodium metal and assuming that the reaction goes to completion,
Norma-Jean [14]

Answer:

m_{NaNH_2}=30.42gNaNH_2

m_{H_2}=0.783gH_2

Explanation:

Hello,

In this case, the reaction between sodium and ammonia is:

2Na+2NH_3\rightarrow 2NaNH_2+H_2

Thus, as we know the initial masses of both sodium and ammonia, we should first identify the limiting reactant, for which we firstly compute the available moles of sodium:

n_{Na}=18.0gNa*\frac{1molNa}{23.0gNa}=0.783molNa

And the moles of sodium consumed by 21.4 g of ammonia (2:2 mole ratio):

n_{Na}^{\ consumed}=21.4gNH_3*\frac{1molNH_3}{17gNH_3} *\frac{2molNa}{2molNH_3} =1.26molNa

In such a way, since less moles of sodium are available than consumed by ammonia, we can say, sodium is the limiting reactant. Furthermore, the mass of both sodium amide (39 g/mol) and hydrogen gas (2 g/mol) that are produced turn out:

m_{NaNH_2}=0.783molNa*\frac{2molNaNH_2}{2molNa}*\frac{39gNaNH_2}{1molNaNH_2}=30.42gNaNH_2

m_{H_2}=0.783molNa*\frac{1molH_2}{2molNa}*\frac{2gH_2}{1molH_2}=0.783gH_2

Best regards.

3 0
3 years ago
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