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Enter your answer in the provided box. an aqueous solution containing 10 g of an optically pure substance was diluted to 500 ml

sladkih [1.3K]

Answer:

[α] = -77.5° / $\frac{\textup{dm-g}}{\textup{mL}}$

Explanation:

Given;

Mass of optically pure substance in the solution = 10 g

Volume of water = 500 mL

Length of the polarimeter, l = 20 cm = 20 × 0.1 dm = 2 dm

measured rotation = - 3.10°

Now,

The specific rotation ( [α] ) is given as:

[α] = $\frac{\alpha}{c\times l}$

here,

α is the measured rotation = -3.10°

c is the concentration

or

c = $\frac{\textup{Mass of optically pure substance in the solution}}{\textup{Volume of water}}$

or

c = $\frac{10}{500}$

or

c = 0.02 g/mL

on substituting the values, we get

[α] = $\frac{-3.10^o}{0.02\times2}$

or

[α] = -77.5° / $\frac{\textup{dm-g}}{\textup{mL}}$

7 0

3 years ago

Percent concentration is one of the most common and basic concentration measurement used by general public. true or false?

Aleksandr [31]

Answer:

<u>TRUE:</u> <em>Percent concentration is one of the most common and basic concentration measurement used by general public</em>

Explanation:

In chemistry there are many <em>concentration measurements</em> used to describe the mixtures. Some of them are, percent, molarity, molality, and molar fraction, among others.

Percent concentration is a popular one because it is commonly understood and used by the non specialist people, i.e. general public.

The percent concentration of a component is defined as: (amount of component in the mixture / amount of mixture) × 100.

The amounts may be measured in mass units (e.g grams) or volume units (e.g. mililiters).

For solutions, mass percent concentration is:

% = (mass of solute / mass of solution) × 100.

And voluem percen contration is:

% = (volume of solute / volume of solution) × 100

Since percentage is used in many profesional and personal activities, most persons use it.

For example, rubbing alcohol, that everyone buys in pharmacies, is 70%; vinager, used in the food, is acetic acid at 5% - 8%.

5 0

3 years ago

What greek letter are used in the precise definition of a limit?

Ivan

Epsilon-delta would be the green letter that is used in the precision definition of a limit. (ε, δ).

5 0

3 years ago

How many moles of gold, Au, are in 3.60 x 10^-5 g of gold?

zlopas [31]

<h3>Answer:</h3>

1.83 × 10⁻⁷ mol Au

<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>

<u>Atomic Structure</u>

Reading a Periodic Table
Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.60 × 10⁻⁵ g Au (Gold)

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

Molar Mass of Au - 196.97 g/mol

<u>Step 3: Convert</u>

Set up: $\displaystyle 3.60 \cdot 10^{-5} \ g \ Au(\frac{1 \ mol \ Au}{196.97 \ g \ Au})$
Multiply: $\displaystyle 1.82769 \cdot 10^{-7} \ mol \ Au$

<u>Step 4: Check</u>

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

1.82769 × 10⁻⁷ mol Au ≈ 1.83 × 10⁻⁷ mol Au

4 0

2 years ago

A bottle of wine contains 9.81 grams of C2H5OH, dissolved in 87.5 grams of water. The final volume of the solution is 100.0 mL.

dimulka [17.4K]

Answer:

[EtOH] = 2.2M and Wt% EtOH = 10.1% (w/w)

Explanation:

1. Molarity = moles solute / Volume solution in Liters

=> moles solute = mass solute / formula weight of solute = 9.8g/46g·mol⁻¹ = 0.213mol EtOH

=> volume of solution (assuming density of final solution is 1.0g/ml) ...

volume solution = 9.81gEtOH + 87.5gH₂O = 97.31g solution x 1g/ml = 97.31ml = 0.09731 Liter solution

Concentration (Molarity) = moles/Liters = 0.213mol/0.09731L = 2.2M in EtOH

2. Weight Percent EtOH in solution (assuming density of final solution is 1.0g/ml)

From part 1 => [EtOH] = 2.2M in EtOH = 2.2moles EtOH/1.0L soln

= {(2.2mol)(46g/mol)]/1000g soln] x 100% = 10.1% (w/w) in EtOH.

3 0

3 years ago