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Phantasy [73]
1 year ago
6

3. Calculate the molarity of the HCI solution if 0.074 mol of HCl is dissolved in water to

Chemistry
1 answer:
mash [69]1 year ago
8 0

Answer:

Molarity = 0.037 M

pH = 1.43

Explanation:

First, you need to calculate the molarity. After converting mL to L, you can plug the values into the molarity equation and simplify.

2,000 mL / 1,000 = 2 L

Molarity (M) = moles / volume (L)

Molarity = 0.074 moles / 2 L

Molarity = 0.037 M

You can plug the molarity of the hydrogen ion into the pH equation to find the pH. Remember that H₃O⁺ and H⁺ can be considered the same thing. Since HCl dissociates into just one H⁺ and one Cl⁻, the molarity of the HCl solution is the same as the molarity of the H⁺.

pH = -log[H⁺]

pH = -log[0.037]

pH = 1.43

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What is the concentration of chloride ion when 16.8ml of a 0.554 M solution of barium chloride is combined with 16.8ml of a 0.59
ozzi

Answer:

The concentration of chloride ion is 2.887 M.

Explanation:

Here we have a combination BaCl₂  and AlCl₃

1 mole BaCl₂ can produce 2 moles chloride ions as follows;

BaCl₂ → Ba²⁺ + 2Cl⁻

Therefore;

0.554  mole BaCl₂ can produce 2×0.554  moles chloride ions;

1 mole AlCl₃ can produce 3 moles chloride ions as follows;

AlCl₃ → Al³⁺ + 3Cl⁻

Therefore;

0.593 mole AlCl₃ can produce 3×0.593 moles chloride ions

Total concentration of chloride ions = 2×0.554 + 3×0.593 = 2.887 moles of Cl⁻ in 2×16.8 ml solution

The concentration of Cl⁻ is 2.887 M

4 0
3 years ago
Humans exhale carbon dioxide when they breathe. If the concentration of carbon dioxide in the air that
Verizon [17]

Answer:

4 people

Explanation:

In order to know this, we need to gather the data.

First we have 3.5x10⁴ g of LiOH. This grams are to be last for the whole 9 days mission shuttle.

So, we first need to know how much of CO₂ will be neutralized by this mass. This can be done calculating the moles of LiOH, and then, the moles of CO₂

The molecular weight of LiOH is 23.94 g/mol and for CO₂ is 44 g/mol so:

CO₂ + 2LiOH -------> Li₂CO₃ + H₂O

moles of LiOH = 3.5x10⁴ / 23.94 = 1461.988 moles

We have a mole ratio of 1:2 so the moles of CO₂:

moles CO₂ = 1461.988 / 2 = 730.994 moles

The mass of CO₂ for the 9 days will be:

m CO₂ = 730.994 * 44 = 32163.74 g

Now, let's see how much it has to be used per day:

m CO₂/d = 32163.74 / 9 = 3573.75 g per day

This means that we can only take a few astronauts, and this number is:

n° astronaut = 3573.75 / 8.8x10²

<h2>n° astronaut = 4.06</h2><h2>In other words, only 4 astronaut</h2>
3 0
3 years ago
Jerome and a classmate are conducting an experiment. Which piece of safety equipment is least likely to be at Jerome’s lab stati
Kruka [31]

Answer:

A fire extinguisher

Explanation:

A lab station usually contains equipment for the people working at the station to use. However, most rooms will only have one fire extinguisher for the whole room meaning it would most likely be located somewhere that is easy to access by all and not just a singular lab station.

~Hope this Helps!~

5 0
3 years ago
Read 2 more answers
Calculate the oxidation number of the iodine (I) in each compound: HIO4 = I2 = NaI = HIO3 =
Makovka662 [10]
1) in periodic acid (HIO₄), iodine has oxidation number +7, hydrogen has oxidation number +1, oxygen has -2, compound has neutral charge:
+1 + x + 4 · (-2) = 0.
x = +7.

2) in molecule of iodine (I₂), iodine has oxidation number 0, because iodine is nonpolar molecule.

3) in sodium iodide (NaI), iodine has oxidation number -1, sodium has oxidation number +1:
+1 + x = 0.
x = -1.

4) in iodic acid (HIO₃), iodine has oxidation number +5, hydrogen has oxidation number +1, oxygen has -2, compound has neutral charge:
+1 + x + 3 · (-2) = 0.
x = +5.
8 0
3 years ago
Read 2 more answers
1. Magnesium chloride solution reacts with silver nitrate solution to form magnesium nitrate
Whitepunk [10]

a. 1,4332 g

b. 7.54~g

<h3>Further explanation</h3>

Given

Reaction

MgCl2 (s) + 2 AgNO3 (aq) → Mg(NO3)2 (aq) + 2 AgCl (s)

20 cm of 2.5 mol/dm^3 of MgCl2

20 cm of 2.5 g/dm^3 of MgCl2

Required

the mass of silver chloride - AgCl

Solution

a. mol MgCl2 :

\tt 20~cm^3=20\times 10^{-3}~dm^3\\\\mol=M\times V\\\\mol=2.5~mol/dm^3\times 20\times 10^{-3}DM^3=0.05

From equation, mol AgCl = 2 x mol MgCl2=2 x 0.05=0.1

mass AgCl(MW=143,32 g/mol)= 0.1 x 143,32=1,4332 g

b. mol MgCl2 (MW=95.211 /mol):

\tt mol=M\times V\\\\mol=\dfrac{2.5~g/dm^3}{95,211 g/mol}=0.0263~mol/dm^3

From equation, mol AgCl = 2 x mol MgCl2=2 x 0.0263=0.0526

mass AgCl(MW=143,32 g/mol)= 0.0526 x 143,32=7.54~g

6 0
2 years ago
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