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3 years ago

Determine the mass (in grams) of 8.02 x 1024 atoms of mercury (Hg).

Chemistry

1 answer:

Virty [35]3 years ago

3 0

<h3>Answer:</h3>

2670 g Hg

<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

Reading a Periodic Table
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Stoichiometry

Using Dimensional Analysis

<h3>Explanation: </h3>

Step 1: Define

8.02 × 10²⁴ atoms Hg

Step 2: Identify Conversions

Avogadro's Number

Molar Mass of Hg - 200.59 g/mol

Step 3: Convert

Set up: $\displaystyle 8.02 \cdot 10^{24} \ atoms \ Hg(\frac{1 \ mol \ Hg}{6.022 \cdot 10^{23} \ atoms \ Hg})(\frac{200.59 \ g \ Hg}{1 \ mol \ Hg})$
Divide/Multiply: $\displaystyle 2671.42 \ g \ Hg$

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

2671.42 g Hg ≈ 2670 g Hg

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2 years ago

A mass of 100.0 g of NaCl is added to 100.0 mL of water, but not all of it dissolves. A mass of 59.5 grams of NaCl solid remains

Taya2010 [7]

Answer:

The molarity of the dissolved NaCl is 6.93 M

Explanation:

Step 1: Data given

Mass of NaCl = 100.0 grams

Volume of water = 100.0 mL = 0.1 L

Remaining mass NaCl = 59.5 grams

Molar mass NaCl= 58.44 g/mol

Step 2: Calculate the dissolved mass of NaCl

100 - 59. 5 = 40.5 grams

Step 3: Calculate moles

Moles NaCl = 40.5 grams / 58.44 g/mol

Moles NaCl = 0.693 moles

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2 years ago

Read 2 more answers

Suppose you have just added 100 ml of a solution containing 0.5 mol of acetic acid per liter to 400 ml of 0.5 m naoh. what is th

Tpy6a [65]

$pH = 13.5$

Explanation:

Sodium hydroxide completely ionizes in water to produce sodium ions and hydroxide ions. Hydroxide ions are in excess and neutralize all acetic acid added by the following ionic equation:

$\text{HAc} + \text{OH}^{-} \to \text{Ac}^{-} + \text{H}_2\text{O}$

The mixture would contain

$0.4 \times 0.5 - 0.1 \times 0.5 = 0.15 \; \text{mol}$ of $\text{OH}^{-}$ and
$0.1 \times 0.5 = 0.05 \; \text{mol}$ of $\text{Ac}^{-}$

if $\text{Ac}^{-}$ undergoes no hydrolysis; the solution is of volume $0.1 + 0.4 = 0.5 \; \text{L}$ after the mixing. The two species would thus be of concentration $0.30 \; \text{mol} \cdot \text{L}^{-1}$ and $0.10 \; \text{mol} \cdot \text{L}^{-1}$ , respectively.

Construct a RICE table for the hydrolysis of $\text{Ac}^{-}$ under a basic aqueous environment (with a negligible hydronium concentration.)

$\begin{array}{cccccccc} \text{R} & \text{Ac}^{-}(aq) &+ & \text{H}_2\text{O}(aq) & \leftrightharpoons & \text{HAc}(aq) & + & \text{OH}^{-} (aq)\\ \text{I} & 0.10 \; \text{M} & & & & & &0.30 \; \text{M}\\ \text{C} & -x \; \text{M}& & & & +x \; \text{M}& & +x \; \text{M} \\ \text{E} & (0.10 - x) \; \text{M} & & & & x \; \text{M} & & (0.30 +x) \; \text{M} \end{array}$

The question supplied the acid dissociation constant $pK_a$ for acetic acid $\text{HAc}$ ; however, calculating the hydrolysis equilibrium taking place in this basic mixture requires the base dissociation constant $pK_b$ for its conjugate base, $\text{Ac}^{-}$ . The following relationship relates the two quantities:

$pK_{b} (\text{Ac}^{-}) = pK_{w} - pK_{a}( \text{HAc})$

... where the water self-ionization constant $pK_w \approx 14$ under standard conditions. Thus $pK_{b} (\text{Ac}^{-}) = 14 - 4.7 = 9.3$ . By the definition of $pK_b$ :

$[\text{HAc} (aq)] \cdot [\text{OH}^{-} (aq)] / [\text{Ac}^{-} (aq) ] = K_b = 10^{-pK_{b}}$

$x \cdot (0.3 + x) / (0.1 - x) = 10^{-9.3}$

$x = 1.67 \times 10^{-10} \; \text{M} \approx 0 \; \text{M}$

$[\text{OH}^{-}] = 0.30 +x \approx 0.30 \; \text{M}$

$pH = pK_{w} - pOH = 14 + \text{log}_{10}[\text{OH}^{-}] = 14 + \text{log}_{10}{0.30} = 13.5$

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2 years ago

What is the average mass of a single chlorine atom in grams

Archy [21]

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Answer:

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