All categories

2 years ago

How many L are required to make 3.5 M hydrochloric acid using 1.1 moles?

Chemistry

1 answer:

ch4aika [34]2 years ago

4 0

<h3>Answer:</h3>

0.31 L

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

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS
Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Moles

<u>Aqueous Solutions</u>

Molarity = moles of solute / liters of solution

<u>Equilibrium - Acid/Base</u>

Strong Acids
Weak Acids
Writing Acids

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 3.5 M HCl

[Given] 1.1 moles

[Solve] Liters solution

<u>Step 2: Solve</u>

Substitute in variables [Molarity]: $\displaystyle 3.5 \ M = \frac{1.1 \ moles}{x \ L}$
[Solution] Cross-Multiply [Equality Property]: $\displaystyle x \ L = \frac{1.1 \ mol}{3.5 \ M}$
[Solution] Divide: $\displaystyle x = 0.314286 \ L$

<u>Step 3: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

0.314286 L ≈ 0.31 L

You might be interested in

What is the mass of 5 mole of ammonia . Calculate the number of NH₃ molecules, nitrogen atom and hydrogen atoms in it..

Aloiza [94]

Molar mass of NH_3

$\\ \sf\longmapsto 14u+3(1u)$

$\\ \sf\longmapsto 14u+3u$

$\\ \sf\longmapsto 17g/mol$

We know.

No of moles=Given mass/Molar mass

$\\ \sf\longmapsto Given\;Mass=17(5)$

$\\ \sf\longmapsto Given \:Mass\:of\:NH_3=85g$

Now

Lets write the balanced equation

$\\ \sf\longmapsto N_2+3H_2=2NH_3$

There is 2moles of Ammonia
3moles of H_2
1mole of N_2

Now

$\boxed{\sf No\:of\:Molecules =No\:of\;moles\times Avagadro\:no}$

For Hydrogen

$\\ \sf\longmapsto 3\times 6.023\times 10^{23}$

$\\ \sf\longmapsto 18.069\times 10^{23}$

$\\ \sf\longmapsto 1.8\times 10^{22}molecules$

For Ammonia

$\\ \sf\longmapsto 2\times 6.023\times 10^{23}$

$\\ \sf\longmapsto 12.046\times 10^{23}$

$\\ \sf\longmapsto 1.2\times 10^{22}molecules$

For Nitrogen

$\\ \sf\longmapsto 1\times 6.023\times 10^{23}$

$\\ \sf\longmapsto 6.023\times 10^{23}molecules$

3 0

2 years ago

1.34 milligrams is the same as _____ kg and _____g.

Karolina [17]

Answer C is for kg and but it's .00134 for grams

3 0

3 years ago

Read 2 more answers

Assignment Directions: Pick one of the following chemists and perform a bit of research on him/her. Answer the following questio

avanturin [10]

Don’t worry about it, let me do your homework!

Born February 27, 1869, Alice Hamilton was an American physician, research scientist, and author who is best known as a leading expert in the field of occupational health and a pioneer in the field of industrial toxicology. She was also the first woman appointed to the faculty of Harvard University.

8 0

3 years ago

URGENT

Vadim26 [7]

Answer:

Explanation:

Discussion

When Pressure increases equilibrium shifts to the side with the smallest number of moles. But which side is that?

N2(g) + 3H2(g) ⇌ 2NH3(g)

The left side has 1 mol of nitrogen (N2) and 3 moles of Hydrogen = 4 mols

on the left side.

The right side has 2 mols of NH3 = 2 mols on the right.

Conclusion: You tell the number of mols by the Balance numbers to the left of each chemical in an equation.

Since the left side N2 + 3H2 = 4 mols, the equilibrium does NOT shift left.

2NH3 is only two mols.

The equilibrium shifts Right

Answer

7 0

2 years ago

Identify the Bronsted-Lowry acid, the Bronsted-Lowry base, the conjugate acid and the conjugate base for each of the following r

Vlad1618 [11]

Answer:

Acids → H₂CO₃ from equilibrium 1 and water, from equilibrium 2.

Bases → Water from equilibrium 1 and ammonia from equilibrium 2.

In 1st equilibrium, H₃O⁺ is the conjugate acid and HCO₃⁻ the conjugate base.

In 2nd equilibrium, NH₄⁺ is the conjugate acid, and OH⁻, the conjugate base.

Explanation:

By the Bronsted-Lowry you know that acids are the one that release protons and base are the ones that catch them.

For the first equilibrium:

H₂CO₃(aq) + H₂O(l) ⇄ H₃O⁺(aq) + HCO₃⁻(aq)

Carbonic acid is the acid → It donates the proton to water, so the water becomes the base. As H₂CO₃ is the acid, the bicarbonate is the conjugate base (it can accept the proton from water to become carbonic acid, again) and the hydronium is the conjugate acid (it would release the proton to become water).

For the second equilibrium:

NH₃(aq) + H₂O(l) ⇄ NH₄⁺ (aq) + OH⁻(aq)

This is the opposite situation → Water relase the proton to ammonia, that's why water is the acid and NH₃, the base (it accepted to become ammonium). The NH₄⁺ is the conjugate acid (it can release the H⁺ to become ammonia) and the OH⁻ is the conjugate base (It can accept the proton to become water, again).

5 0

3 years ago