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Dafna1 [17]
3 years ago
11

Consider a solution that contains 0.274 M potassium chloride and 0.155 M magnesium chloride.

Chemistry
1 answer:
solong [7]3 years ago
3 0

Answer:

Concentration of chloride ions = 0.584M

Explanation:

The step by step calculations is shown as attached below.

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A student sets up the following equation to convert a measurement.
dybincka [34]

Answer:

\frac{1 m}{100 cm}

Explanation:

The final answer has a different set of units. In particular, meters (m) changes to centimeters (cm). To make this change, you need to multiply the first value by proportions.

When writing these proportions, it is important that they are arranged in a way that allows for the cancellation of units. For instance, since m is located in the denominator, it must be located in the numerator of the conversion.

<u>Proportion:</u>

1 m = 100 cm

The full expression:

<h3>-1.7*10^5\frac{V}{m}  ·  \frac{1 m}{100 cm}  =  ? \frac{V}{cm}</h3><h2>                 ^</h2>

As you can see, the old unit (m) cancels out and you are left with cm in the denominator.

7 0
2 years ago
3. Given 20g of Barium Hydroxide, how many grams of
anastassius [24]

The number of grams of ammonium nitrate (NH₄NO₃) it would take for all the barium hydroxide to react is 18.7g

First, we will write the balanced chemical equation for the reaction

The balanced chemical equation for the reaction is

Ba(OH)₂ + 2NH₄NO₃ → 2NH₄OH + Ba(NO₃)₂

This means, 1 mole of barium hydroxide is required to react with 2 moles of ammonium nitrate

Now, we will calculate the number of moles of barium hydroxide present.

Mass of barium hydroxide (Ba(OH)₂) = 20 g

Using the formula

Number\ of\ moles = \frac{Mass}{Molar\ mass}

Molar mass of Ba(OH)₂ = 171.34 g/mol

∴ Number of moles of Ba(OH)₂ present =\frac{20}{171.34}

Number of moles of Ba(OH)₂ present = 0.116727 mole

Now,

Since 1 mole of barium hydroxide is required to react with 2 moles of ammonium nitrate

Then,

0.116727 mole of barium hydroxide will react with 2 × 0.116727 mole of ammonium nitrate

2 × 0.116727 = 0.233454 mole

∴ Number of moles of NH₄NO₃ required is 0.233454 mole

Now, for the mass of ammonium nitrate (NH₄NO₃) required

From the formula

Mass = Number of moles × Molar mass

Molar mass of NH₄NO₃ = 80.043 g/mol

∴ Mass of NH₄NO₃ required = 0.233454 × 80.043

Mass of NH₄NO₃ required = 18.68636 g

Mass of NH₄NO₃ required ≅ 18.7g

Hence, the number of grams of ammonium nitrate (NH₄NO₃) it would take for all the barium hydroxide to react is 18.7g

Learn more on determining mass of reactant required here: brainly.com/question/11232389

6 0
2 years ago
Question 5 of 5
rjkz [21]
Probably Fresh vegetables, it can rot out, that’d be my guess, it’s not canned
8 0
3 years ago
Read 2 more answers
I need chemistry help text me at (951) 897-8325 PLEASE
user100 [1]
Just post the question on here
7 0
2 years ago
Name the 3 seperate metals group
umka21 [38]

Akali Metals

Akali-Earth Metals

and Other Metals

8 0
3 years ago
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