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MrRa [10]
3 years ago
7

In part A of the experiment you will need to make a 4.80 x 10-4 M KSCN solution starting with a 2.00 x 10-3M KSCN solution. You

will be making up the solution in a 25.00 mL volumetric flask and using 0.5 M HNO3 as the diluent. What volume of 2.00 x 10-3 M KSCN will you need
Chemistry
1 answer:
ryzh [129]3 years ago
7 0

Answer:

Answer to you need to make a 6.00 x 10-4 M KSCN solution starting with a 2.00 x 10-3 M KSCN solution. You will be making up the so. ... X 10-3 M KSCN Solution. You Will Be Making Up The Solution In A 25 ML Volumetric Flask And Using 0.5 M HNO3 As The Diluent. What Volume Of 2.00 X 10-3 M KSCN Will You Need?

Explanation:

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That unknown substance is water

Explanation:

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What is the product of the reaction of (S)-2-bromobutane with sodium methoxide in acetone?
Serggg [28]

Answer:

2-methoxybutane

Explanation:

This reaction is an example of Nucleophilic substitution reaction. Also, the reaction of (S)-2-bromobutane with sodium methoxide in acetone, is bimolecular nucleophilic substitution (SN2). The reaction equation is given below.

(S)-2-bromobutane + sodium methoxide (in acetone) → 2-methoxybutane

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What does lithium 6 and lithium 7 look like
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Lithium 6 would have 6 valence electrons in the outer orbital, while lithium 7 would have 7 in the outer orbital. 
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3 years ago
If 63.8 grams of aluminum metal (Al) react with 72.3 grams of sulfur (S) in a synthesis reaction, how many grams of the excess r
Jobisdone [24]

Answer:

23.2 g of Al will be left over when the reaction is complete

Explanation:

2Al  +  3S  → Al₂S₃

1 mol of Al = 26.98 g

1 mol of S = 32.06 g

Mole = Mass / Molar mass

63.8 g/ 26.98 g/m = 2.36 mole of Al

72.3 g / 32.06 g/m = 2.25 mole of S

2 mole of Aluminun react with 3 mole of sulfur

2.36 mole of Al react with (2.36 .3)/2 = 3.54 m of S

As I have 2.25 mole of S, and I need 3.54 S, is my limiting reagent so the limiting in excess is the Al.

3 mole of S react with 2 mole of Al

2.25 mole of S react with (2.25 m . 2)/3 = 1.50 mole

I need 1.50 mole of Al and I have 2.36, that's why the Al is in excess.

2.36 mole of Al - 1.50 mole of Al = 0.86 mole

This is the quantity of Al without reaction.

Molar mass . mole = Mass →  26.98 g/m . 0.86 m = 23.2 g

7 0
3 years ago
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