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quester [9]
3 years ago
8

During a flame test, a lithium salt produces a characteristic red flame. This red color is produced when electrons in excited li

thium atoms
1. are lost by the atoms
2. are gained by the atoms
3. return to lower energy states within the atoms
4. move to higher energy states
Chemistry
1 answer:
hoa [83]3 years ago
6 0
During a flame test, a lithium salt produces a characteristic red flame. This red color is produced when electrons in excited lithium atoms <span>return to lower energy states within the atoms.

Hope I've Helped!
</span>
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One of relatively few reactions that takes place directly between two solids at room temperature is In this equation, the in Ba(
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Answer:

3.14 grams of ammonium thiocyanate must be used to react completely with 6.5 g barium hydroxide octahydrate.

Explanation:

Ba(OH)_2.8H_2O(s)+NH_4SCN(s)\rightarrow Ba(SCN)_2(s)+8H_2O(l)+NH_3(g)

The balance chemical equation is :

Ba(OH)_2.8H_2O(s)+2NH_4SCN(s)\rightarrow Ba(SCN)_2(s)+10H_2O(l)+2NH_3(g)

Mass of barium hydroxide octahydrate = 6.5 g

Moles of  barium hydroxide octahydrate = \frac{6.5 g}{315 g/mol}=0.020635 mol

According to reaction, 2 moles of ammonium thiocyanate reacts with1 mole of  barium hydroxide octahydrate. The 0.020635 moles of barium hydroxide octahydrate will react with:

\frac{2}{1}\times 0.020635 mol=0.04127 mol

Mass of 0.04127 moles of ammonium thiocyanate;

0.04127 mol\times 76 g/mol=3.136 g\approx 3.14 g

3.14 grams of ammonium thiocyanate must be used to react completely with 6.5 g barium hydroxide octahydrate

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Fluorine has 7 valence electrons what type of bond is likely to form between two atoms?
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g A laboratory analysis of an unknown compound found the following composition: C 75.68% ; H 8.80% ; O 15.52%. What is the empir
sammy [17]

Answer:

THE EMPIRICAL FORMULA FOR THE UNKNOWN COMPOUND IS C7H9O

Explanation:

The empirical formula for the unknown compound can be obtained by following the processes below:

1 . Write out the percentage composition of the individual elements in the compound

C = 75.68 %

H = 8.80 %

O = 15.52 %

2. Divide the percentage composition by the atomic masses of the elements

C = 75 .68 / 12 = 6.3066

H = 8.80 / 1 = 8.8000

O = 15.52 / 16 = 0.9700

3. Divide the individual results by the lowest values

C = 6.3066 / 0.9700 = 6.5016

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4. Round up the values to the whole number

C = 7

H = 9

O = 1

5 Write out the empirical formula for the compound

C7H90

In conclusion, the empirical formula for the unknown compound is therefore C7H9O

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