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2 years ago

How many moles in 4.93 x 10E23 atoms of silver?

Chemistry

1 answer:

leva [86]2 years ago

8 0

<h3>Answer:</h3>

0.819 mol Ag

<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> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

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

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

4.93 × 10²³ atoms Ag

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

Avogadro's Number

<u>Step 3: Convert</u>

Set up: $\displaystyle 4.93 \cdot 10^{23} \ atoms \ Ag(\frac{1 \ mol \ Ag}{6.022 \cdot 10^{23} \ atoms \ Ag})$
Divide: $\displaystyle 0.818665 \ mol \ Ag$

<u>Step 4: Check</u>

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

0.818665 mol Ag ≈ 0.819 mol Ag

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How many sulfur atoms are present in 100 grams of this compound? Report your answer to three significant figures.

kap26 [50]

Answer:

1.88 × 10²⁴ atoms

Explanation:

Step 1: Given data

Mass of sulfur: 100 g

Step 2: Calculate the moles corresponding to 100 g of sulfur

The molar mass of sulfur is 32.07 g/mol. The moles corresponding to 100 g of sulfur are:

100 g × (1 mol/32.07 g) = 3.12 mol

Step 3: Calculate the number of atoms in 3.12 moles of sulfur

We will use Avogadro's number: there are 6.02 × 10²³ atoms of sulfur in 1 mole of sulfur.

3.12 mol × (6.02 × 10²³ atoms/1 mol) = 1.88 × 10²⁴ atoms

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3 years ago

An alkane with the formula C6H14 can be prepared by hydrogenation of either of two precursor alkenes having the formula C6H12.

ankoles [38]

Two precursor alkenes

H₃C CH₃
I I
H₂C=C-CH-CH₃ 2,3-dimethyl-1-butene

H₃C   CH₃
I   I
H₃C-CH=CH-CH₃ 2,3-dimethyl-2-butene

alkane

H₃C CH₃
  I I
H₃C-CH-CH-CH₃ 2,3-dimethylbutane

H₃C  CH₃ H₃C CH₃
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H₂C=C-CH-CH₃ + H₂ → H₃C-CH-CH-CH₃

H₃C CH₃ H₃C   CH₃
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H₂C-C=CH-CH₃ + H₂ → H₃C-CH-CH-CH₃

6 0

3 years ago

Read 2 more answers

Type the correct answer in the box.

tatiyna

the mass of 1 mole of phosphorus atoms is approximately 30 g

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3 years ago

Read 2 more answers

When taking a measurement with a pH meter, keep the instrument in the ______________ until it is needed. Rinse the pH meter with

nydimaria [60]

Answer:

Storage solution; deionized water; stabilizes.

Explanation:

The pH scale measures the concentration of hydrogen ions in acidic and alkaline solutions.

In chemistry, it literally means power of hydrogen ions and it is a measure of the molar concentration of hydrogen ions in a particular solution, thus specifying the acidity, neutrality or basicity of chemical solutions.

Mathematically, the pH of a solution is given by;

$pH = -log_{10}(H^{+})$

Hence, a solution with a pH of 7 is neutral. Also, a solution with a pH below 7 is acidic but basic (alkaline) if it's pH is above 7.

A pH meter can be defined as a scientific instrument or device designed and developed for the measurement of the hydrogen-ion concentration in water-based solutions, in order to determine their level of acidity or alkanility.

When using a pH meter to take a measurement, you should keep it in a storage solution until it is needed. Also, a deionized water should be used to rinse the pH meter and gently pat dry.

Furthermore, the pH meter should be placed in a given sample solution and a reading of the measurement taken when the pH of the solution stabilizes.

8 0

3 years ago

Be sure to answer all parts. Calculate the pH during the titration of 30.00 mL of 0.1000 M KOH with 0.1000 M HBr solution after

Fantom [35]

Answer:

(a) pH = 12.73

(b) pH = 10.52

Explanation:

The net balanced reaction equation is:

KOH + HBr ⇒ H₂O + KBr

The amount of KOH present is:

n = CV = (0.1000 molL⁻¹)(30.00 mL) = 3.000 mmol

(a) The amount of HBr added in 9.00 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(9.00 mL) = 0.900 mmol

This amount of HBr will neutralize an equivalent amount of KOH (0.900 mmol), leaving the following amount of KOH:

(3.000 mmol) - (0.900 mmol) = 2.100 mmol KOH

After the addition of HBr, the volume of the KOH solution is 39.00 mL. The concentration of KOH is calculated as follows:

C = n/V = (2.100 mmol) / (39.00 mL) = 0.0538461 M KOH

The pOH and pH of the solution can then be calculated:

pOH = -log[OH⁻] = -log(0.0538461) = 1.2688

pH = 14 - pOH = 14 - 1.2688 = 12.73

(b) The amount of HBr added in 29.80 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(29.80 mL) = 2.980 mmol

This amount of HBr will neutralize an equivalent amount of KOH, leaving the following amount of KOH:

(3.000 mmol) - (2.980 mmol) = 0.0200 mmol KOH

After the addition of HBr, the volume of the KOH solution is 59.80 mL. The concentration of KOH is calculated as follows:

C = n/V = (0.0200 mmol) / (59.80 mL) = 0.0003344 M KOH

The pOH and pH of the solution can then be calculated:

pOH = -log[OH⁻] = -log(0.0003344) = 3.476

pH = 14 - pOH = 14 - 3.476 = 10.52

(0.1000 molL⁻¹)(38.00 mL) = 3.800 mmol

This amount of HBr will neutralize all of the KOH present. The amount of HBr in excess is:

(3.800 mmol) - (3.000 mmol) = 0.800 mmol HBr

After the addition of HBr, the volume of the analyte solution is 68.00 mL. The concentration of HBr is calculated as follows:

C = n/V = (0.800 mmol) / (68.00 mL) = 0.01176 M HBr

The pH of the solution can then be calculated:

pH = -log[H⁺] = -log(0.01176) = 1.93

4 0

3 years ago