What happens to the equilibrium when the temperature is reduced

A 15.00-ml sample of a naoh solution of unknown concentration requires 17.88 ml of a 0.1053 m h2so4 solution to reach the equiva

Sergeu [11.5K]

Answer:

0.1255 M

Explanation:

1) Data:

Base: NaOH

Vb = 15.00 ml = 15.00 / 1,000 liter

Mb = ?

Acid: H₂SO₄

Va = 17.88 ml = 17.88 / 1,000 liter

Ma = 0.1053

2) Chemical reaction:

The titration is an acid-base (neutralization) reaction to yield a salt and water:

Acid + Base → Salt + Water

H₂SO₄ (aq) + NaOH(aq) → Na₂SO₄ (aq) + H₂O (l)

3) Balanced chemical equation:

H₂SO₄ (aq) + 2 NaOH(aq) → Na₂SO₄ (aq) + 2H₂O (l)

Placing coefficient 2 in front of NaOH and H₂O balances the equation

4) Stoichiometric mole ratio:

The coefficients of the balanced chemical equation show that 1 mole of H₂SO₄ react with 2 moles of NaOH. Hence, the mole ratio is:

1 mole H₂SO₄ : 2 mole NaOH

5) Calculations:

a) Molarity formula: M = n / V (in liter)

⇒ n = M × V

b) Nunber of moles of acid:

nₐ = Ma × Va = 0.1053 (17.88 / 1,000)

c) Number of moles of base, nb:

nb = Mb × Vb = Mb × (15.00 / 1,000)

d) At equivalence point number of moles of acid = number of moles of base

nₐ = nb

0.1053 × (17.88 / 1,000) = Mb × (15.00 / 1,000)

Mb = 0.1053 × 17.88 / 15.00 = 0.1255 mole/liter = 0.1255 M

3 0

3 years ago

Read 2 more answers

How many moles of iron is 6.022 x 10^22 atoms of iron? (Report answer as a number rounded to one place past the decimal.) *

yuradex [85]

<h3>Answer:</h3>

$\displaystyle 0.1 \ mol \ Fe$

<h3>General Formulas and Concepts:</h3>

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Chemistry

Atomic Structure

Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Stoichiometry

Using Dimensional Analysis

<h3>Explanation:</h3>

Step 1: Define

6.022 × 10²² atoms Fe (iron)

Step 2: Identify Conversions

Avogadro's Number

Step 3: Convert

Set up: $\displaystyle 6.022 \cdot 10^{22} \ atoms \ Fe(\frac{1 \ mol \ Fe}{6.022 \cdot 10^{23} \ atoms \ Fe})$
Divide: $\displaystyle 0.1 \ mol \ Fe$

7 0

3 years ago

Which element has the strongest attraction for electrons?

NemiM [27]

Answer:

The answer is Flourine

3 0

3 years ago

The first step in industrial nitric acid production is the catalyzed oxidation of ammonia. Without a catalyst, a different react

murzikaleks [220]

Answer: The value of $K_{eq}$ is $4.84\times 10^{-5}$

Explanation:

We are given:

Initial moles of ammonia = 0.0280 moles

Initial moles of oxygen gas = 0.0120 moles

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

$\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}$

So, concentration of ammonia = $\frac{0.0280}{1.00}=0.0280M$

Concentration of oxygen gas = $\frac{0.0120}{1.00}=0.0120M$

The given chemical equation follows:

$4NH_3(g)+3O_2(g)\rightleftharpoons 2N_2(g)+6H_2O(g)$

Initial: 0.0280 0.0120

At eqllm: 0.0280-4x 0.0120-3x 2x 6x

We are given:

Equilibrium concentration of nitrogen gas = $3.00\times 10^{-3}M=0.003$

Evaluating the value of 'x', we get:

$\Rightarrow 2x=0.003\\\\\Rightarrow x=0.0015M$

Now, equilibrium concentration of ammonia = $0.0280-4x=[0.0280-(4\times 0.0015)]=0.022M$

Equilibrium concentration of oxygen gas = $0.0120-3x=[0.0120-(3\times 0.0015)]=0.0075M$

Equilibrium concentration of water = $6x=(6\times 0.0015)]=0.009M$

The expression of $K_{eq}$ for the above reaction follows:

$K_{eq}=\frac{[H_2O]^6\times [N_2]^2}{[NH_3]^4\times [O_2]^3}$

Putting values in above expression, we get:

$K_{eq}=\frac{(0.009)^6\times (0.003)^2}{(0.022)^4\times (0.0075)^3}\\\\K_{eq}=4.84\times 10^{-5}$

Hence, the value of $K_{eq}$ is $4.84\times 10^{-5}$

4 0

3 years ago

Exactly 5000 mL of air at 223K is warmed and has a new volume of 8.36 liters. What is the new temperature?

34kurt

Answer:

The new temperature is 373 K

Explanation:

Step 1: Data given

Volume air = 5000 mL = 5.0 L

Temperature = 223K

New volume = 8.36 L

Step 2: Calculate the new temperature

V1/T1 = V2/T2

⇒V1 = the initial volume = 5.0 L

⇒T1 = the initial temperature = 223 K

⇒V2 = the new volume = 8.36 L

⇒T2 = the new temperature

5.0/223 = 8.36 /T2

T2 = 373 K

The new temperature is 373 K

7 0

3 years ago