Convert 15.0 inches to meters. Show detailed work

A stock solution of HNO3 is prepared and found to contain 13.5 M of HNO3. If 25.0 mL of the stock solution is diluted to a final

salantis [7]

Answer : The correct option is, (C) 0.675 M

Explanation :

Using neutralization law,

$M_1V_1=M_2V_2$

where,

$M_1$ = concentration of $HNO_3$ = 13.5 M

$M_2$ = concentration of diluted solution = ?

$V_1$ = volume of $HNO_3$ = 25.0 ml = 0.0250 L

conversion used : (1 L = 1000 mL)

$V_2$ = volume of diluted solution = 0.500 L

Now put all the given values in the above law, we get the concentration of the diluted solution.

$13.5M\times 0.0250L=M_2\times 0.500L$

$M_2=0.675M$

Therefore, the concentration of the diluted solution is 0.675 M

8 0

3 years ago

Potassium cyanide is a toxic substance, and the median lethal dose depends on the mass of the person or animal that ingests it.

Artemon [7]

Answer:

88,7 mL of solution

Explanation:

Molarity (Represented as M) is an unit of chemical concentration that is defined as the ratio between moles of solute per liters of solution, that is:

Molarity = moles of solute / Liters of solution

If molarity of KCN solution is 0,0820M and moles of KCN are 7,27x10⁻³ moles:

0,0820M = 7,27x10⁻³ moles / Liters of solution

Liters of solution = 0,0887L = <em>88,7 mL of solution</em>

I hope it helps!

4 0

3 years ago

At a given temperature, the elementary reaction A --->B in the forward direction is first order in A with a rate constant of

Pani-rosa [81]

Answer:

The value of the equilibrium constant for the reaction A ⇒ B is Kc = 1.72 × 10³.

The value of the equilibrium constant for the reaction B ⇒ A is K'c = 5.81 × 10⁻⁴.

Explanation:

For the reaction A ⇒ B, the equilibrium constant (Kc) is equal to the forward rate constant (kf) divided by the reverse rate constant (ki).

$Kc=\frac{kf}{ki} =\frac{1.60 \times 10^{2} s^{-1} }{ 9.30 \times 10^{-2} s^{-1}} =1.72 \times 10^{3}$

If we consider the inverse reaction B ⇒ A, its equilibrium constant (K'c) is the inverse of the forward reaction equilibrium constant.

$K'c=\frac{1}{Kc} =\frac{1}{1.72 \times 10^{3} } =5.81 \times 10^{-4}$

4 0

3 years ago

Predict which of the following pairs of solutions, when mixed together, will cause a precipitate to form. (Select all that apply

Kay [80]

Answer:

25 mL of 1 × 10–5 M Co(NO₃)₂ and 75 mL of 5 × 10–4 M Na₂S

500 mL of 7.5 × 10–4 M AlCl₃ and 100 mL of 1.7 × 10–5 M Hg₂(NO₃)₂.

650 mL of 0.0080 M K₂SO₄ and 175 mL of 0.15 M AgNO₃

Explanation:

When 2 compounds that produce an insoluble substance are mixed together, <em>A precipitate will be formed if Q of reaction > Ksp</em>

For the solutions:

1.5 L of 0.025 M BaCl₂ and 1.25L of 0.014 M Pb(NO₃)₂.

Ksp is:

PbCl₂(s) ⇄ Pb²⁺(aq) + 2Cl⁻(aq)

Ksp = 2.4x10⁻⁴ = [Pb²⁺][Cl⁻]²

Molar concentration of each ion is:

[Pb²⁺] = 1.25L ₓ (0.014mol / L) = 0.0175mol / 2.75L = 6.36x10⁻³M

[Cl⁻] = 2 ₓ 1.5L ₓ (0.025mol / L) = 0.075mol / 2.75L = 0.0273M

Replacing in Ksp expression to find Q:

Q = [6.36x10⁻³M][0.0273M]² = 4.73x10⁻⁶

As Q < Ksp, the mixture will not produce a precipitate.

25 mL of 1 × 10–5 M Co(NO₃)₂ and 75 mL of 5 × 10–4 M Na₂S

Ksp is:

CoS(s) ⇄ Co²⁺(aq) + S²⁻(aq)

Ksp = 4.0x10⁻²¹ = [Co²⁺][S²⁻]

Molar concentration of each ion is:

[Co²⁺] = 0.025L ₓ (1x10⁻⁵mol / L) = 2.5x10⁻⁷mol / 0.1L = 2.5x10⁻⁶M

[S²⁻] = 0.075L ₓ (5x10⁻⁴mol / L) = 3.75x10⁻⁵mol / 0.1L = 3.75x10⁻⁴M

Replacing in Ksp expression to find Q:

Q = [2.5x10⁻⁶M][3.75x10⁻⁴M] = 9.38x10⁻⁶

As Q > Ksp, the mixture will produce a precipitate.

500 mL of 7.5 × 10–4 M AlCl₃ and 100 mL of 1.7 × 10–5 M Hg₂(NO₃)₂.

Ksp is:

Hg₂Cl₂(s) ⇄ 2Hg⁺(aq) + 2Cl⁻(aq)

Ksp = 3.5x10⁻¹⁸ = [Hg⁺]²[Cl⁻]²

Molar concentration of each ion is:

[Hg⁺] = 2ₓ0.100L ₓ (1.7x10⁻⁵mol / L) = 3.4x10⁻⁶mol / 0.6L = 5.67x10⁻⁶M

[Cl⁻] = 3 ₓ 0.500L ₓ (7.5x10⁻⁴mol / L) = 1.125x10⁻³mol / 0.6L = 1.88x10⁻³M

Replacing in Ksp expression to find Q:

Q = [5.67x10⁻⁶M]²[1.88x10⁻³M]² = 1.14x10⁻⁶

As Q > Ksp, the reaction will produce a precipitate.

650 mL of 0.0080 M K₂SO₄ and 175 mL of 0.15 M AgNO₃

Ksp is:

Ag₂SO₄(s) ⇄ 2Ag⁺(aq) + SO₄²⁻(aq)

Ksp = 1.5x10⁻⁵ = [Ag⁺]²[SO₄²⁻]

Molar concentration of each ion is:

[Ag⁺] = 0.175L ₓ (0.15mol / L) = 0.02625mol / 0.825L = 0.0318M

[SO₄²⁻] = 0.650L ₓ (0.080mol / L) = 0.052mol / 0.825L = 0.0630M

Replacing in Ksp expression to find Q:

Q = [0.0318M]²[0.0630M] = 6.37x10⁻⁵

As Q > Ksp, the reaction will produce a precipitate.

6 0

3 years ago

What is the molarity (M) of .5 liter of a solution that has 205 g of NaCl dissolved in it?

jeka94

Answer:

<h2>Molarity = 7 mol / L</h2>

Explanation:

Since the mass of NaCl and it's volume has been given we can find the molarity by using the formula

<h3> $C = \frac{m}{M \times v}$ </h3>

where

C is the molarity

m is the mass

M is the molar mass

v is the volume

From the question

v = 0.5 L

m = 205 g

We must first find the molar mass and then substitute the values into the above formula

M( Na) = 23 , M( Cl) = 35.5

Molar mass of NaCl = 23 + 35.5 =

58.5 g/mol

So the molarity of NaCl is

$C = \frac{205}{0.5 \times 58.5} \\ C = \frac{205}{29.25}$

C = 7.00854

We have the final answer as

<h3>Molarity = 7 mol / L</h3>

Hope this helps you

8 0

3 years ago