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2 years ago

How does distillation separate a solvent from a solute?

Chemistry

1 answer:

Illusion [34]2 years ago

8 0

Answer:

The solution is heated to the the boiling point , and the solvent is collected as its gas condenses

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Pls answer rn due tmr pls pls fast

Lemur [1.5K]

Answer:

From bottom to top, Troposphere, Stratosphere, Mesosphere, Thermosphere, & at the top, Exosphere.

Explanation:

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2 years ago

Sulfuric acid dissolves and completely dissociates in water and a solvation sphere of water molecules forms around the ions. Thi

gizmo_the_mogwai [7]

The answer is D

Hope this help;)

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3 years ago

Read 2 more answers

The flask contains 10.0 mL of HCl and a few drops of phenolphthalein indicator. The buret contains 0.160 M NaOH. It requires 18.

olchik [2.2K]

Answer:

Approximately $0.291\; \rm M$ (rounded to two significant figures.)

Explanation:

The unit of concentration $\rm M$ is the same as $\rm mol \cdot L^{-1}$ (moles per liter.) On the other hand, the volume of both the $\rm NaOH$ solution and the original $\rm HCl$ solution here are in milliliters. Convert these two volumes to liters:

$V(\mathrm{NaOH}) = 18.2\; \rm mL = 18.2 \times 10^{-3}\; \rm L = 0.0182\; \rm L$ .
$V(\text{$\mathrm{HCl}$, original}) = 10.0\; \rm mL = 10.0\times 10^{-3}\; \rm L = 0.0100\; \rm L$ .

Calculate the number of moles of $\rm NaOH$ in that $0.0182\; \rm L$ of $0.160\; \rm M$ solution:

$\begin{aligned} n(\mathrm{NaOH}) &= c(\mathrm{NaOH})\cdot V(\mathrm{NaOH})\\ &= 0.160\; \rm mol \cdot L^{-1} \times 0.0182\; \rm L \approx 0.00291\; \rm mol\end{aligned}$ .

$\rm HCl$ reacts with $\rm NaOH$ at a one-to-one ratio:

$\rm HCl\; (aq) + NaOH\; (aq) \to NaCl\; (aq) + H_2O\; (l)$ .

Coefficient ratio:

$\displaystyle \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} = 1$ .

In other words, one mole of $\rm NaOH$ would neutralize exactly one mole of $\rm HCl$ . In this titration, $0.291\; \rm mol$ of $\rm NaOH\!$ was required. Therefore, the same amount of $\rm HC$ should be present in the original solution:

$\begin{aligned}&n(\text{$\mathrm{HCl}$, original})\\ &= n(\mathrm{NaOH})\cdot \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} \\ &\approx 0.00291\; \rm mol \times 1 = 0.00291\; \rm mol\end{aligned}$ .

Calculate the concentration of the original $\rm HCl$ solution:

$\displaystyle c(\text{$\mathrm{HCl}$, original}) = \frac{n(\text{$\mathrm{HCl}$, original})}{V(\text{$\mathrm{HCl}$, original})} \approx \frac{0.00291\; \rm mol}{0.0100\; \rm L} \approx 0.291\; \rm M$ .

5 0

2 years ago

Which of the following is an example of an abiotic factor? A wind factor B Human C bacterium D duck

solmaris [256]

Abiotic factors, also called abiotic components, are non-living parts of the environment which affects the living organisms and their function in the ecosystem. I think the correct answer is A. Wind factor is the only non-living part from the choices.

7 0

2 years ago

Read 2 more answers

What is the mass of 1.58 moles of CH4

HACTEHA [7]

<h3>Answer:</h3>

25.4 g CH₄

<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>

Reading a Periodic Table

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

1.58 mol CH₄

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

[PT] Molar Mass of C - 12.01 g/mol

[PT] Molar Mass of H - 1.01 g/mol

Molar Mass of CH₄ - 12.01 + 4(1.01) = 16.05 g/mol

<u>Step 3: Convert</u>

Set up: $\displaystyle 1.58 \ mol \ CH_4(\frac{16.05 \ g \ CH_4}{1 \ mol \ CH_4})$
Multiply/Divide: $\displaystyle 25.359 \ g \ CH_4$

<u>Step 4: Check</u>

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

25.359 g CH₄ ≈ 25.4 g CH₄

4 0

2 years ago