I am a metal that has 3 valence electrons and 123 neutrons

If 100 ml of a 0.5 M HCI Solution is diluted with water to 1000ml, what is the new concetration?

Vesna [10]

Answer:

$0.05\ \text{M HCl}$

Explanation:

$V_1$ = Initial volume = 100 mL

$V_2$ = Final volume = 1000 mL

$M_1$ = Initial concentration = 0.5 M

$M_2$ = Final concentration

We have the relation

$\dfrac{M_1}{M_2}=\dfrac{V_2}{V_1}\\\Rightarrow M_2=M_1\dfrac{V_1}{V_2}\\\Rightarrow M_2=0.5\times \dfrac{100}{1000}\\\Rightarrow M_2=0.05\ \text{M HCl}$

The new concentration is $0.05\ \text{M HCl}$ .

3 0

3 years ago

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

H = 8.8000 / 0.9700 = 9.0722

O = 0.9700 / 0.9700 = 1

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

7 0

3 years ago

Which one would it be 1,2,3,4

sesenic [268]

I think it's 1. PCl5= PCl3 + Cl2. I can't tell the exact reason but it's something to do with Atomic Redox Reaction

6 0

3 years ago

In a reaction involving the iodination of acetone, the following volumes were used to make up the reaction mixture: 5 mL 4.0M ac

adell [148]

Explanation:

Below is an attachment containing the solution.

7 0

4 years ago

A student titrated a solution containing 3.7066 g of an unknown Diprotic acid to the end point using 28.94 ml of 0.3021 M KOH so

uranmaximum [27]

Answer:

Molar mass of the diprotic acid is 424 grams

Explanation:

[hint: diprotic acid only contains 2 hydrogen protons]

Ionic equation:

${ \bf{2OH { }^{ - } _{(aq)} + 2H { }^{ + } _{(aq)}→ 2H _{2} O _{(l)} }}$

first, we get moles of potassium hydroxide in 28.94 ml :

${ \sf{1 \: l \: of \: KOH \: contains \: 0.3021 \: moles}} \\ { \sf{0.02894 \: l \: of \: KOH \: contain \: (0.02894 \times 0.3021) \: moles}} \\ { \underline{ = 0.008743 \: moles}}$

since mole ratio of diprotic acid : base is 2 : 2, moles are the same.

Therefore, moles of acid that reacted are 0.008743 moles.

${ \sf{0.008743 \: moles \: of \: acid \: weigh \: 3.7066 \: g}} \\ { \sf{1 \: mole \: of \: acid \: weighs \: ( \frac{1 \times 3.7066}{0.008743}) \: g }} \\ = { \underline{423.95 \: g \approx424 \: grams}}$

for the molar mass:

8 0

3 years ago