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3 years ago

Rank the following steps involved in preparing a dilute solution from a stock solution in order from first to last.

Chemistry

1 answer:

Serggg [28]3 years ago

5 0

Answer: C - B - A - D

Explanation:

-Obtain a flask that is thoroughly cleaned

-Add the desired amount of stock solution

-Add deionized water until the volume reaches the mark on the flask.

- Cap the flask and invert several times to thoroughly mix the solution.

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If the bright lines in the spectrum of oxygen are matched by lines in the dark-line spectrum of the sun, what does that indicate

lbvjy [14]

Hello young fellow friend I think the anwser is (C)

7 0

3 years ago

Read 2 more answers

Which pair shares the same empirical formula?

ZanzabumX [31]

Answer:- 3. $CH_3$ and $C_2H_6$

Explanations:- An empirical formula is the simplest whole number ratio of atoms of each element present in the molecule/compound.

For example, the molecular formula of benzene is $C_6H_6$ . The ratio of C to H in it is 6:6 that could be simplified to 1:1. So, an empirical formula of benzene is CH.

In the first pair, the ratio of C to H in first molecule is 2:4 that could be simplified to 1:2 and the empirical formula is $CH_2$ . In second molecule the ratio of C to H is 6:6 and it could be simplified to 1:1. and the empirical formula is CH. Empirical formulas are different for both the molecules of first pair and so it is not the right choice.

In second pair, C to H ratio in first molecule is 1:2, so the empirical formula is $CH_2$ . The C to H ratio for second molecule is 1:4, so the empirical formula is $CH_4$ . Here also, the empirical formulas are not same and hence it is also not the right choice.

In third pair, C to H ratio in first molecule is 1:3, so the empirical formula is $CH_3$ . In second molecule the C to H ratio is 2:6 and it is simplified to 1:3. So, the empirical formula for this one is also $CH_3$ . Hence. this is the correct choice.

In fourth pair, first molecule empirical formula is CH. Second molecule has 2:4 that is 1:2 mole ratio of C to H and so its empirical formula is $CH_2$ . As the empirical formulas are different, it is not the right choice.

So, the only and only correct pair is the third one. 3. $CH_3$ and $C_2H_6$

4 0

3 years ago

Read 2 more answers

The reaction H2(g)+F2(g)?2HF(g) is spontaneous at all temperatures. A classmate calculates the entropy change for this reaction

Pachacha [2.7K]

Answer:

The correct answer is there is a mistake in the calculation. The second law of thermodynamics state that in any spontaneous process there is an increase in the entropy of the universe.

Explanation:

According to the second law a reaction will occur in a system spontaneously if the total entropy of both system and surrounding increases during the reaction.That means in case of spontaneous reaction entropy change is always positive.

But according to the question the reaction H2+F2=2HF is spontaneous in all temperature.So according to the second law of thermodynamics i can say that my classmate made a mistake in calculation that"s why his result for entropy change comes negative.

6 0

3 years ago

What is the mass, in grams, of 9.01 x 1024 molecules of methanol (CH, OH) ?

Zina [86]

Answer:

480.6 g

Explanation:

Given data:

Number of molecules of methanol = 9.01 ×10²⁴

Mass in gram = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ molecules

9.01 ×10²⁴molecules ×1 mol /6.022 × 10²³ molecules

1.5 ×10¹ mol

15 mol

Mass in gram:

Mass = number of moles × molar mass

Mass = 15 mol × 32.04 g/mol

Mass = 480.6 g

3 0

3 years ago

A 50.00-mL solution of 0.0350 M aniline ( Kb = 3.8 × 10–10) is titrated with a 0.0113 M solution of hydrochloric acid as the tit

nexus9112 [7]

Answer:

pH = 3.70

Explanation:

Moles of aniline in solution are:

0.0500L × (0.0350mol / 1L) = 1.750x10⁻³ mol aniline

Aniline is in equilibrium with water, thus:

C₆H₅NH₂ + H₂O ⇄ C₆H₅NH₃⁺ + OH⁻ Kb = 3.8x10⁻¹⁰

HCl reacts with aniline thus:

HCl + C₆H₅NH₂ → C₆H₅NH₃⁺ + Cl⁻

At equivalence point, all aniline reacts producing C₆H₅NH₃⁺. C₆H₅NH₃⁺ has its own equilibrium with water thus:

C₆H₅NH₃⁺(aq) + H₂O(l) ⇄ C₆H₅NH₂(aq) + H₃O⁺(aq)

Where Ka is defined as:

Ka = [C₆H₅NH₂] [H₃O⁺] / [C₆H₅NH₃⁺] = Kw / Kb = 1.0x10⁻¹⁴ / 3.8x10⁻¹⁰ = 2.63x10⁻⁵ (1)

As all aniline reacts producing C₆H₅NH₃⁺, moles in equilibrium are:

[C₆H₅NH₃⁺] = 1.750x10⁻³ mol - X

[C₆H₅NH₂] = X

[H₃O⁺] = X

Replacing in (1):

2.63x10⁻⁵ = [X] [X] / [1.750x10⁻³ mol - X]

4.6x10⁻⁸ - 2.63x10⁻⁵X = X²

0 = X² + 2.63x10⁻⁵X - 4.6x10⁻⁸

Solving for X:

X = -2.3x10⁻⁴ → False answer, there is no negative concentrations

X = 2.0x10⁻⁴ → Right answer

As [H₃O⁺] = X, [H₃O⁺] = 2.0x10⁻⁴.

Now, pH = -log[H₃O⁺]. Thus, pH at equivalence point is:

pH = 3.70

7 0

3 years ago