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ASHA 777 [7]
3 years ago
10

Please do question 9 and 10

Chemistry
1 answer:
DanielleElmas [232]3 years ago
8 0

Answer:

9) 60; 10) 0.2583

Explanation:

9) M(SiO₂)=28+16*2=60 (gr/mol);

10) if one mole is in 60gr., then for 15.5 gr.:

1/60=x/15.5; ⇒ x=15.5/60≈0.2583(mol).

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Charge of nucleus is always positive whether it is element or isotope.
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If different atoms can come together to form all living and nonliving things, why is there a limit to different combinations we
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Explanation:

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Select the correct answer.
givi [52]

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{1s^2 2s^2 2p^6} 3s^2 3p^4

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Explanation:

i didnt understand the rest of that but this is the e- configuration on top and the bottom is noble gas configuration

6 0
3 years ago
The teacher tells your group to make a stock solution of sodium chloride, and then diluting it to
sergejj [24]

The idea here is that you need to figure out how many moles of magnesium chloride,

MgCl

2

, you need to have in the target solution, then use this value to determine what volume of the stock solution would contain this many moles.

As you know, molarity is defined as the number of moles of solute, which in your case is magnesium chloride, divided by liters of solution.

c

=

n

V

So, how many moles of magnesium chloride must be present in the target solution?

c

=

n

V

⇒

n

=

c

⋅

V

n

=

0.158 M

⋅

250.0

⋅

10

−

3

L

=

0.0395 moles MgCl

2

Now determine what volume of the target solution would contain this many moles of magnesium chloride

c

=

n

V

⇒

V

=

n

c

V

=

0.0395

moles

3.15

moles

L

=

0.01254 L

Rounded to three sig figs and expressed in mililiters, the volume will be

V

=

12.5 mL

So, to prepare your target solution, use a

12.5-mL

sample of the stock solution and add enough water to make the volume of the total solution equal to

250.0 mL

.

This is equivalent to diluting the

12.5-mL

sample of the stock solution by a dilution factor of

20

.

3 0
2 years ago
How many milliliters of 0.125 M FeCl3 are needed to react with an excess of Na2S to produce 3.75 g of Fe2S3 if the percent yield
Katyanochek1 [597]

Answer:

0.912 mL

Explanation:

3 Na2S(aq) + 2 FeCl3(aq) → Fe2S3(s) + 6 NaCl(aq)

FeCl3 is the limiting reactant.

Number of moles of iron III sulphide produced= 3.75g/87.92 g/mol = 0.043 moles

Hence actual yield of Iron III sulphide = 0.043 moles

Theoretical yield of Iron III sulphide = actual yield ×100%/ %yield

Theoretical yield of iron III sulphide= 0.043 ×100/75 = 0.057 moles of Iron III sulphide

From the reaction equation,

2moles of iron III chloride produced 1 mole of iron III sulphide

x moles of iron III chloride, will produce 0.057 of iron III sulphide

x= 2× 0.057= 0.114 moles of iron III chloride

But

Volume= number of moles/ concentration

Volume= 0.114/0.125

Volume= 0.912 mL

4 0
3 years ago
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