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stepan [7]
3 years ago
9

What is the electron configuration of a potassium ion when it has formed an ionic bond with a bromine ion?

Chemistry
2 answers:
garri49 [273]3 years ago
6 0

Answer:

A

Explanation:

devlian [24]3 years ago
5 0
The answer is A right
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1. A 5.04 mL sample of Cu has a mass of 45 grams. What is the density of Cu? Show units in final answer
babunello [35]
I think the answer is number D…. I think
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2 years ago
If the compound has a molar mass of 156 g/mol, what is its molecular formula?
oksano4ka [1.4K]
I believe that the choices for this question are:
C2H4O2, C4H8O4 CH2O, C6H12O6 C3H6O3, C6H12O6 C2H4O2, C6H12O6 
 
The answer to this based on the molar masses given is:
C2H4O2, C6H12O6 
 
To prove calculate the molar mass:
C2H4O2 = 2*12 + 4*1 + 2*16 = 60
C6H12O6 = 6*12 + 12*1 + 6*16 = 180
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2 years ago
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vaieri [72.5K]
I believe this is it

7 0
3 years ago
a sealed container filled with argon gas at 35 c has a pressure of 832 torr. if the volume of the container is decreased by a fa
lesya692 [45]

Answer:

If the volume of the container is decreased by a factor of 2 the pressure is is increased by the same factor to 1664 torr.

Explanation:

Here we have Boyle's law which states that, at constant temperature, the volume of a given mass of gas is inversely proportional to its pressure

V ∝ 1/P or V₁·P₁ = V₂·P₂

Where:

V₁ = Initial volume

V₂ = Final volume = V₁/2

P₁ = Initial pressure = 832 torr

P₂ = Final pressure  = Required

From V₁·P₁ = V₂·P₂ we have,

P₂ = V₁·P₁/V₂ = V₁·P₁/(V₁/2)

P₂  = 2·V₁·P₁/V₁ = 2·P₁ = 2× 832 torr = 1664 torr

6 0
3 years ago
Read 2 more answers
34g aluminum are combined with 39g chlorine gas which is the limiting reactant
vampirchik [111]

The limiting reactant is chlorine (Cl2).

<u>Explanation</u>:

Limiting reactant is the amount of product formed which gets limited by the reagent without continuing it.

        2 Al + 3 Cl2 ==> 2 AlCl3  represents the balanced equation.

Number of moles Al present = 34 g Al x 1 mole Al / 26.98 g

                                                = 1.260 g moles of Al

Number of moles Cl2 present = 39 g Cl2 x 1 mole Cl2 / 35.45 g

                                                  = 1.10 g moles of Cl2

Dividing each reactant by it's coefficient in the balanced equation obtains:

1.260 moles Al / 2 = 0.63 g moles of Al

1.11 moles Cl2 / 3 = 0.36 g moles of Cl2

The reactant which produces a lesser amount of product is called as limiting reactant.

Here the Limiting reactant is Cl2.

6 0
3 years ago
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