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2 years ago

Mn2+

Chemistry

1 answer:

snow_lady [41]2 years ago

5 0

The molarity of the stock Mn²⁺ ions is 0.0288 M

Based on the dilution formula;

The molarity of A is 0.00144 M
The molarity of B is 0.0000576 M
The molarity of C is 0.000001152 M

<h3>What is the molarity of a solution?</h3>

The molarity of a solution is the number of moles of a solute dissolved in a given volume of solution in liters.

Molarity = number of moles/volume

The molarity of the stock solution is:

moles of Mn²⁺ ions = mass / molar mass

molar mass of Mn²⁺ ions = 55.0 g/mol

moles of Mn²⁺ ions = 1.584 / 55

moles of Mn²⁺ ions = 0.0288 moles

molarity of Mn²⁺ ions = 0.0288 / 1

molarity of Mn²⁺ ions = 0.0288 M

The dilution formula is used to determine the molarities of A, B, and C.

C₁V₁ = C₂V₂

C₂ = C₁V₁ / V₂

Where;

C₁ = initial molarity
V₁ = initial volume
C₂ = final molarity
V₂ = final volume

Molarity of A = 50 * 0.0288 / 1000

Molarity of A = 0.00144 M

Molarity of B = 10 * 0.00144 / 250

Molarity of B = 0.0000576 M

Molarity of C = 10 * 0.0000576 / 500

Molarity of C = 0.000001152 M

Learn more about molarity at: brainly.com/question/17138838

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Answer is: B) The kinetic energy is increasing proportional to the mass of the bowling ball.

Kinetic energy has to do with the speed of an object and how much mass it has; basically how the object is moving.

Ek = 1/2 · m(bowling ball) · v²(bowling ball).

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3 0

3 years ago

Can anyone help with this? please

Sloan [31]

Answer:

Volume of solution = 80.5 mL

Explanation:

Given data:

Molarity of solution = 4.50 mol/L

Mass of ethanol = 16.7 g

Volume of solution = ?

Solution:

Volume will be calculated from molarity formula.

Molarity = number of moles / volume in L

Number of moles:

Number of moles = mass/molar mass

Number of moles = 16.7 g/ 46.07 g/mol

Number of moles = 0.3625 mol

Volume of solution:

Molarity = number of moles / volume in L

4.50 mol/L = 0.3625 mol / volume in L

Volume in L = 0.3625 mol /4.50 mol/L

Volume in L = 0.0805 L

Volume in mL:

0.0805 L ×1000 mL/1 L

80.5 mL

3 0

3 years ago

) B5H9(l) is a colorless liquid that will explode when exposed to oxygen. How much heat is released when 0.211 mol of B5H9 react

Tom [10]

<u>Answer:</u> The amount of heat released when 0.211 moles of $B_5H_9(l)$ reacts is 554.8 kJ

<u>Explanation:</u>

The chemical equation for the reaction of $B_5H_9$ with oxygen gas follows:

$2B_5H_9(l)+12O_2(g)\rightarrow 5B_2O_3(s)+9H_2O(l)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(5\times \Delta H_f_{(B_2O_3(s))})+(9\times \Delta H_f_{(H_2O(l))})]-[(2\times \Delta H_f_{(B_5H_9(l))})+(12\times \Delta H_f_{(O_2(g))})]$

We are given:

$\Delta H_f_{(H_2O(l))}=-285.4kJ/mol\\\Delta H_f_{(B_2O_3(s))}=-1272kJ/mol\\\Delta H_f_{(B_5H_9(l))}=73.2kJ/mol\\\Delta H_f_{(O_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(2\times (-1272))+(9\times (-285.4))]-[(2\times (73.2))+(12\times (0))]\\\\\Delta H_{rxn}=-5259kJ$

To calculate the amount of heat released for the given amount of $B_5H_9(l)$ , we use unitary method, we get:

When 2 moles of $B_5H_9(l)$ reacts, the amount of heat released is 5259 kJ

So, when 0.211 moles of $B_5H_9(l)$ will react, the amount of heat released will be = $\frac{5259}{2}\times 0.211=554.8kJ$

Hence, the amount of heat released when 0.211 moles of $B_5H_9(l)$ reacts is 554.8 kJ

7 0

3 years ago

Which is always the first step in dealing with an accident in the lab?

marin [14]

The answer is the letter C

3 0

3 years ago

Read 2 more answers

I need some help with a long chemistry problem. Anything is appreciated!

Mkey [24]

Answer:

The CSI is wrong.

Explanation:

1. Find the volume of the pool

The formula for the volume of a cylinder is V = πr²h .

D = 12 m; h = 10 m

r = D/2 = (12 m/2) = 6.0 m

V = πr²h = π × (6.0 m)² × 10 m = π × 36 m²× 10 m = 360π m³ = 1100 m³

= 1.1. × 10⁶ L

2. Calculate the moles of OH⁻

n = cV = 1.0 × 10⁻² mol·L⁻¹ × 1.3 × 10⁶ L = 11 000 mol of OH⁻

3. Calculate the moles of acetic acid needed for neutralization

HA + OH⁻ ⟶ A⁻ + H₂O

The molar ratio of is 1 mol HA:1 mol OH⁻, so you need 11 000 moL of acetic acid.

4. Calculate the actual moles of acetic acid

You have four 5 L jugs of acetic acid pH 2 .

Volume = 20 L

[H⁺] = 10⁻² mol·L⁻¹ = 0.01 mol·L⁻¹

(a) Set up an ICE table

HA + H₂O ⇌ A⁻ + H₃O⁺

I/mol·L⁻¹: c 0 0

I/mol·L⁻¹: - 0.01 +0.01 +0.01

I/mol·L⁻¹: c - 0.01 0.01 0.01

$K_{\text{a}} = \dfrac{\text{[A]}^{-}\text{[H$_{3}$O$^{+}$]}}{\text{[HA]}} = 1.76 \times 10^{-5}$

(b) Calculate the concentration of acetic acid

$\begin{array}{rcl}\dfrac{\text{[A]}^{-}\text{[H$_{3}$O$^{+}$]}}{\text{[HA]}}& = & 1.76 \times 10^{-5}\\\\\dfrac{0.01\times 0.01}{c}& = & 1.76 \times 10^{-5}\\\\1 \times 10^{-4} & = & 1.76 \times 10^{-5}c\\c & = & \dfrac{1 \times 10^{-4}}{1.76 \times 10^{-5}}\\\\ & = & \text{6 mol/L}\\\end{array}$

The concentration of the acetic acid is 6 mol·L⁻¹

(c) Calculate the moles of acetic acid

$n = \text{20 L} \times \dfrac{\text{6 mol}}{\text{1 L}} = \textbf{100 mol}$

You have 100 mol of acetic acid.

The CSI is wrong.

You don't have enough acetic acid to neutralize the pool.

8 0

4 years ago