4. *

How many formula units are there in 50.3 moles potassium chloride?

Alona [7]

<h3>Answer:</h3>

3.03 × 10²⁵ formula units KCl

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

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

Reading a Periodic Table
Writing Compounds

<u>Atomic Structure</u>

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

<h3>Explanation:</h3>

<u>Step 1: Define</u>

50.3 mol KCl (Potassium chloride)

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

<u /> $\displaystyle 50.3 \ mol \ KCl(\frac{6.022 \cdot 10^{23} \ formula \ units \ KCl}{1 \ mol \ KCl} )$ = 3.02907 × 10²⁵ formula units KCl

<u>Step 4: Check</u>

<em>We are given 3 sig figs. Follow sig fig rules and round.</em>

3.02907 × 10²⁵ formula units KCl ≈ 3.03 × 10²⁵ formula units KCl

6 0

3 years ago

How can you obtain hydrogen from a mixture of ethene, hydrogen and ammonia gases?

irga5000 [103]

Answer:

We normally separate unreacted hydrogen from ammonia (product) in Haber process. The reaction mixture contains some ammonia, plus a lot of unreacted hydrogen and nitrogen. The mixture is cooled and compressed, causing the ammonia gas to condense into a liquid.

3 0

3 years ago

When solutions of silver nitrate and sodium chloride are mixed, a precipitation reaction occurs. What mass of precipitate can be

Bas_tet [7]

Explanation:

The given precipitation reaction will be as follows.

$AgNO_{3}(aq) + NaCl(aq) \rightarrow AgCl(s) + NaNO_{3}(aq)$

Here, AgCl is the precipitate which is formed.

It is known that molarity is the number of moles present in a liter of solution.

Mathematically, Molarity = $\frac{\text{no. of moles}}{\text{volume in liter}}$

It is given that volume is 1.14 L and molarity is 0.269 M. Therefore, calculate number of moles as follows.

Molarity = $\frac{\text{no. of moles}}{\text{volume in liter}}$

0.269 M = $\frac{\text{no. of moles}}{1.14 L}$

no. of moles = 0.306 mol

As molar mass of AgCl is 143.32 g/mol. Also, relation between number of moles and mass is as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

0.307 mol = $\frac{mass}{143.32 g/mol}$

mass = 43.99 g

Thus, we can conclude that mass of precipitate produced is 43.99 g.

8 0

2 years ago

Determine whether the following hydroxide ion concentrations ([OH−]) correspond to acidic, basic, or neutral solutions by estima

Rzqust [24]

Answer:

See explanation below

Explanation:

To do this, we will use the following expression to calculate the [H⁺]:

[H⁺] = Kw / [OH⁻]

[H⁺] is the same as [H₃O⁺]. So we have the [OH⁻] so, let's replace every value into the above expression to calculate the hydronium concentration and say if it's acidic, basic or neutral. This can be known because if the [H⁺] > 1x10⁻⁷ M the solution is acidic. If it's [H⁺] < 1x10⁻⁷ M the solution is basic, and if it's [H⁺] = 1x10⁻⁷ M the solution is neutral.

a) [H⁺] = 1x10⁻¹⁴ / 6x10⁻¹² = 1.67x10⁻³ M. Acidic.

b) [H⁺] = 1x10⁻¹⁴ / 9x10⁻⁹ = 1.11x10⁻⁶ M. Acidic.

c) [H⁺] = 1x10⁻¹⁴ / 8x10⁻¹⁰ = 1.25x10⁻⁵ M. Acidic.

d) [H⁺] = 1x10⁻¹⁴ / 7x10⁻¹³ = 0.0143 M. Acidic.

e) [H⁺] = 1x10⁻¹⁴ / 2x10⁻² = 5x10⁻¹³ M. Basic.

f) [H⁺] = 1x10⁻¹⁴ / 9x10⁻⁴ = 1.11x10⁻¹¹ M. Basic.

g) [H⁺] = 1x10⁻¹⁴ / 5x10⁻⁵ = 2x10⁻¹⁰ M. Basic.

h) [H⁺] = 1x10⁻¹⁴ / 1x10⁻⁷ = 1x10⁻⁷ M. Neutral.

Part B.

In this part, we'll use the following expression and replace the given values:

[OH⁻] = Kw / [H⁺]

Replacing the values:

[OH⁻] = 1x10⁻¹⁴ / 5.2x10⁻⁵

[OH⁻] = 1.92x10⁻¹⁰ M

PArt C:

In this case, we will use expression of part A, and replace the given values:

[H⁺] = 1x10⁻¹⁴ / 2.7x10⁻²

[H⁺] = 3.7x10⁻¹³ M

8 0

2 years ago

You are comparing a reaction that produces a chemical

MaRussiya [10]

If the reaction is a chemical change, new substances with different properties and identities are formed. This may be indicated by the production of an odor, a change in color or energy, or the formation of a solid.

7 0

3 years ago

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