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3 years ago

Of an 8 M solution of HCI, 1 L is diluted to 2 M. What is the volume of the diluted solution?

Chemistry

1 answer:

jok3333 [9.3K]3 years ago

4 0

Answer: The volume of diluted solution is 4 L

Explanation:

According to the dilution law,

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock $HCl$ solution = 8 M

$V_1$ = volume of stock $HCl$ solution = 1 L

$M_1$ = molarity of diluted $HCl$ solution = 2 M

$V_1$ = volume of diluted $HCl$ solution = ?

Putting in the values we get:

$8M\times 1L=2M\times V_2$

$V_2=4L$

Therefore, volume of diluted solution is 4 L

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A solution is prepared by dissolving ammonium sulfate in enough water to make of stock solution. A sample of this stock solution

Pani-rosa [81]

Answer: molarity of ammonium ions = 0.274mol/L

molarity of sulfate ions = 0.137mol/L

Note: The complete question is given below

A solution is prepared by dissolving 10.8 g ammonium sulfate in enough water to make 100.0 mL of stock solution. A 10.00-mL sample of this stock solution is added to 50.00 mL of water. Calculate the concentration of ammonium ions and sulfate ions in the final solution.

Explanation:

Molar concentration = no of moles/volume in liters

no of moles = mass/molar mass

mass of ammonium sulfate = 10.8g, molar mass of ammonium sulfate, (NH₄)₂SO₄ = (14+4)*2 + 32+ (16)*4 = 132g/mol

no of moles = 10.8g/132g/mol = 0.0820moles

Molarity of stock solution = 0.0820mol/(100ml/1000ml* 1L) = 0.0820mol/0.1L Molarity of stock solution = 0.820mol/L

Concentration of final solution is obtained from the dilution formula,

C1V1 = C2V2

C1 = 0.820M, V1 = 10mL, C2 = ?, V2 = 60mL

C2 = C1V1/V2

C2 = 0.820*10/60 = 0.137mol/L

molar concentration of ions = molarity of solution * no of ions

molarity of ammonium ions = 0.137mol/L * 2 = 0.274mol/L

molarity of sulfate ions = 0.137 mol/L * 1 = 0.137mol/L

4 0

3 years ago

Used oil needs to be kept in what kind of containers

Paul [167]

Answer:

Never store used oil in anything other than tanks and storage containers. Used oil may also be stored in units that are permitted to store regulated hazardous waste. Tanks and containers storing used oil do not need to be RCRA permitted, however, as long as they are labeled and in good condition.

8 0

3 years ago

2H20(g) → 2H2(g) + 02 (8)

fiasKO [112]

Answer:

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5 0

3 years ago

Pls help

Irina-Kira [14]

Answer:

CrO₂ --------------------> Cr⁴⁺ and O²⁻

VCO₃ -------------------> V²⁺ and CO₃²⁻

Cr₂(SO₄)₃ -------------> Cr³⁺ and SO₄²⁻

(NH₄)₂S ----------------> NH₄⁺ and S²⁻

Explanation:

Within ionic compounds, the cation is listed first, followed by the anion. Some of the ions are polyatomic, meaning they are covalently bonded to other elements. Polyatomic ions always have a specific charge.

All of these ionic compounds have an overall charge of 0. As such, the charges of the cations and anions must balance out. In order to do so, there are some compounds which have more than one atom of each ion.

2.) CrO₂

------> Oxygen (O) always forms the anion, O²⁻.

------> Therefore, if there are 2 oxygen anions, the chromium (Cr) must have the cationic form of Cr⁴⁺.

------> +4 + (-2) + (-2) = 0

3.) VCO₃

------> Carbonate (CO₃), a polyatomic ion, always has the state CO₃²⁻.

------> If there is only one atom of each ion, the charges must perfectly balance, making vanadium (V) be the cation V²⁺.

------> +2 + (-2) = 0

4.) Cr₂(SO₄)₃

------> Sulfate (SO₄), a polyatomic ion, always has the state SO₄²⁻.

-------> The only way the charges could balance out is if the chromium (Cr) is in the cationic form Cr³⁺.

------> +3 + 3 + (-2) + (-2) + (-2) = 0

5.) (NH₄)₂S

------> Ammonium (NH₄), a polyatomic ion, always has the state NH₄⁺.

------> Sulfur (S) always forms the anion S²⁻.

------> +1 + 1 + (-2) = 0

3 0

2 years ago

For the following reaction, find the value of Q and predict the direction of change, given that a 1L flask initially contains 2

Tresset [83]

Answer:

C) Q < K, reaction will make more products

Explanation:

1/8 S8(s) + 3 F2(g) ↔ SF6(g)

∴ Kc = 0.425 = [ SF6 ] / [ F2 ]³

∴ Q = [ SF6 ] / [ F2 ]³

∴ [ SF6 ] = 2 mol/L

∴ [ F2 ] = 2 mol/L

⇒ Q = ( 2 ) / ( 2³)

⇒ Q = 0.25

⇒ Q < K, reaction will make more products

5 0

3 years ago