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nexus9112 [7]
3 years ago
7

How many valence electrons does an atom of any halogen have?

Chemistry
1 answer:
Makovka662 [10]3 years ago
7 0
All halogens are stable so they have 8 electrons in their last shell
You might be interested in
a compound containing only sulfur and nitrogen is by mass; the molar mass is g/mol. what are the empirical and molecular formula
kolezko [41]

The chemical compound's empirical formula is NS.

The chemical compound's molecular formula is N4S4.

<h3>What does a chemical empirical formula look like?</h3>
  • The empirical formula of a compound that gives the proportion (ratios) of the elements in the complex but not the precise number or arrangement of atoms is known as an empirical formula.
  • This would be the compound's element to whole number ratio with the lowest value.
<h3>What sort of empirical formula would that be?</h3>
  • The chemical structure of glucose is C6H12O6. Every mole of carbon and oxygen is accompanied by two moles of hydrogen.
  • Glucose has the empirical formula CH2O.
  • Ribose has the chemical formula C5H10O5, which can be simplified to the empirical formula CH2O.

learn more about empirical formula here

brainly.com/question/1603500

#SPJ4

the question you are looking for is

A compound containing only sulfur and nitrogen is 69.6% S by mass; the molar mass is 184 g/mol. What are the empirical and molecular formulas of the compound?

3 0
1 year ago
ANSWER FAST PLZZZ!!!!!!!!!!!!!
goldfiish [28.3K]

Answer:

1 and 2

Explanation:

1 is negative

2 is negative

3 is positive

5 0
3 years ago
Express 749 000 000 in scientific notation,
Dima020 [189]

Answer:

7.49 × 108

Explanation:

Scientific notation is a way to express numbers in a form that makes numbers that are too small or too large more convenient to write. It is commonly used in mathematics, engineering, and science, as it can help simplify arithmetic operations. In scientific notation, numbers are written as a base, b, referred to as the significant, multiplied by 10 raised to an integer exponent, n, which is referred to as the order of magnitude:

8 0
3 years ago
What is the molar out of a solution that contains 33.5g of CaCl2 in 600.0mL of water
omeli [17]

Answer:

Here's what I got.

Explanation:

Interestingly enough, I'm not getting

0.0341% w/v

either. Here's why.

Start by calculating the percent composition of chlorine,

Cl

, in calcium chloride, This will help you calculate the mass of chloride anions,

Cl

−

, present in your sample.

To do that, use the molar mass of calcium chloride, the molar mass of elemental chlorine, and the fact that

1

mole of calcium chloride contains

2

moles of chlorine atoms.

2

×

35.453

g mol

−

1

110.98

g mol

−

1

⋅

100

%

=

63.89% Cl

This means that for every

100 g

of calcium chloride, you get

63.89 g

of chlorine.

As you know, the mass of an ion is approximately equal to the mass of the neutral atom, so you can say that for every

100 g

of calcium chloride, you get

63.89 g

of chloride anions,

Cl

−

.

This implies that your sample contains

0.543

g CaCl

2

⋅

63.89 g Cl

−

100

g CaCl

2

=

0.3469 g Cl

−

Now, in order to find the mass by volume percent concentration of chloride anions in the resulting solution, you must determine the mass of chloride anions present in

100 mL

of this solution.

Since you know that

500 mL

of solution contain

0.3469 g

of chloride anions, you can say that

100 mL

of solution will contain

100

mL solution

⋅

0.3469 g Cl

−

500

mL solution

=

0.06938 g Cl

−

Therefore, you can say that the mass by volume percent concentration of chloride anions will be

% m/v = 0.069% Cl

−

−−−−−−−−−−−−−−−−−−−

I'll leave the answer rounded to two sig figs, but keep in mind that you have one significant figure for the volume of the solution.

.

ALTERNATIVE APPROACH

Alternatively, you can start by calculating the number of moles of calcium chloride present in your sample

0.543

g

⋅

1 mole CaCl

2

110.98

g

=

0.004893 moles CaCl

2

To find the molarity of this solution, calculate the number of moles of calcium chloride present in

1 L

=

10

3

mL

of solution by using the fact that you have

0.004893

moles present in

500 mL

of solution.

10

3

mL solution

⋅

0.004893 moles CaCl

2

500

mL solution

=

0.009786 moles CaCl

2

You can thus say your solution has

[

CaCl

2

]

=

0.009786 mol L

−

1

Since every mole of calcium chloride delivers

2

moles of chloride anions to the solution, you can say that you have

[

Cl

−

]

=

2

⋅

0.009786 mol L

−

1

[

Cl

−

]

=

0.01957 mol L

−

This implies that

100 mL

of this solution will contain

100

mL solution

⋅

0.01957 moles Cl

−

10

3

mL solution

=

0.001957 moles Cl

−

Finally, to convert this to grams, use the molar mass of elemental chlorine

0.001957

moles Cl

−

⋅

35.453 g

1

mole Cl

−

=

0.06938 g Cl

−

Once again, you have

% m/v = 0.069% Cl

−

−−−−−−−−−−−−−−−−−−−

In reference to the explanation you provided, you have

0.341 g L

−

1

=

0.0341 g/100 mL

=

0.0341% m/v

because you have

1 L

=

10

3

mL

.

However, this solution does not contain

0.341 g

of chloride anions in

1 L

. Using

[

Cl

−

]

=

0.01957 mol L

−

1

you have

n

=

c

⋅

V

so

n

=

0.01957 mol

⋅

10

−

3

mL

−

1

⋅

500

mL

n

=

0.009785 moles

This is how many moles of chloride anions you have in

500 mL

of solution. Consequently,

100 mL

of solution will contain

100

mL solution

⋅

0.009785 moles Cl

−

500

mL solution

=

0.001957 moles Cl

−

So once again, you have

0.06938 g

of chloride anions in

100 mL

of solution, the equivalent of

0.069% m/v

.

Explanation:

i think this is it

8 0
2 years ago
Pure magnesium metal is often found as ribbons and can easily burn in the presence of oxygen. when 2.59 g of magnesium ribbon bu
stepan [7]
Burning Mg in the air and reacting with O2 forming a white powder of MnO

So the equation is going to be:
Mn + O2 ⇒ MnO (this equation is not conserved)

to make it equilibrium:
1- First we should put 2Mno to equal the O2 on both sides.
So it will be:
Mg + O2⇒ 2MgO
2- Second we should put 2Mn to equal the Mn on both sides.
2Mg + O2⇒ 2MgO (this equation is conserved)
After putting the physical states the final equilibrium equation is going to be:
                        Δ
2Mg(s) + O2(g)⇒ 2MgO(s)

 

4 0
3 years ago
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