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2 years ago

15

A 40.0L tank of ammonia has a pressure of 12.7kPa. Calculate the volume of the ammonia if its pressure is changed to 8.4 kPa whi

le its temperature remains constant

1 answer:

Ne4ueva [31]2 years ago

5 0

Answer:

60.5 L

Explanation:

If the gas temperature is constant, we can use Boyle's law:

P1V1 = P2V2

V2 = P1V1 / P2

P1 = 12.7 kPa

V1 = 40.0 L

P2 = 8.4 kPa

V2 = (12.7 kPa) (40.0 L) / (8.4 kPa) = 60.5 L

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A 36.165 mg36.165 mg sample of a chemical known to contain only carbon, hydrogen, sulfur, and oxygen is put into a combustion an

erma4kov [3.2K]

Answer:

The empirical formula for the compound is C6H12SO2.

Explanation:

We'll begin by writing out what was given from the question. This is shown:

Let us consider the First experiment:

Mass of the compound = 36.165 mg

Mass of CO2 = 64.425 mg

Mass of H2O = 26.373 mg

Data obtained from the Second experiment:

Mass of compound = 47.029 mg

Mass of SO2 = 20.32 mg

Next, we'll determine the mass of C, H and S. This is illustrated below:

Molar Mass of CO2 = 12 + (2x16) = 44g/mol

Mass of C in CO2 = 12/44 x 64.425 Mass of C = 17.57 mg

Molar Mass of H2O = (2x1) + 16 = 18g/mol

Mass of H in H2O = 2/18 x 26.373

Mass of H = 2.93 mg

Molar Mass of SO2 = 32 + (16x2) = 64g/mol

Mass of S in SO2 = 32/64 x 20.32

Mass of S = 10.16 mg

At this stage, it is important we determine the percentage composition of C, H, S and O. This is illustrated below:

% of C = 17.57/36.165 x 100 = 48.58%

% of H = 2.93/36.165 x 100 = 8.10%

% of S = 10.16/47.029 x 100 = 21.60%

% of O = 100 - (48.58 + 8.1 + 21.6)

% of O = 21.72%

Now we can easily obtain the empirical formula for the compound by doing the following.

Step 1:

Divide by their molar mass

C = 48.58/12 = 4.0483

H = 8.10/1 = 8.1

S = 21.60/32 = 0.675

O = 21.72/16 = 1.3575

Step 2:

Divide by the smallest:

C = 4.0483/0.675 = 6

H = 8.1/0.675 = 12

S = 0.675/0.675 = 1

O = 1.3575/0.675 = 2

From the calculations made above, empirical formula for the compound is C6H12SO2

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3 years ago

What are the steps for scientific method in order

Anna007 [38]

Make an observation.

Conduct research.

Form hypothesis.

Test hypothesis.

Record data.

Draw conclusion.

Replicate.

One thing that is designed to change in the set up of the experiment. ( The things that I can change) Independent Variable.

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3 years ago

For the reaction of hydrogen with iodine

fenix001 [56]

Answer:

$r_{H_2} = \frac{-1}{2} r_{HI}$

Explanation:

Hello!

In this case, considering the given chemical reaction:

$H_2(g) + I_2(g) \rightarrow 2HI(g)$

Thus, by applying the law of rate proportions, we can write:

$\frac{1}{-1} r_{H_2} = \frac{1}{-1}r_{i_2} = \frac{1}{2} r_{HI}$

Whereas the stoichiometric coefficients of reactants are negative due their disappearance and that of the product is positive due to its appearance. In such a way, when we relate the rate of disappearance of hydrogen gas to the rate of formation of hydrogen iodide, we obtain:

$r_{H_2} = \frac{-1}{2} r_{HI}$

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2 years ago

If a bar magnet's neutral region is broken into two, what will most likely occur? A. Neither segment will have a north or south

grigory [225]

The answer is D I JUST TOOKM THE TEST

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3 years ago

Read 2 more answers

Classify the following substances as a Bronsted-Lowry acid, Bronsted-Lowry base, Lewis acid, and/or Lewis base.

Margarita [4]

Answer :

HCl : Bronsted-Lowry acid

$BF_3\text{ and }^{+}CCl_3$ : Lewis-acid

$^-H, CH_2O,^-OCH_3\text{ and }NH_3$ : Bronsted-Lowry base

$^-H, CH_2O,^-OCH_3,CH_3Cl\text{ and }NH_3$ : Lewis-base

Explanation :

According to the Bronsted Lowry conjugate acid-base theory, an acid is defined as a substance which donates protons and a base is defined as a substance which accepts protons.

According to the Lewis concept, an acid is defined as a substance that accepts electron pairs and base is defined as a substance which donates electron pairs.

HCl :

It is a Bronsted-Lowry acid that donates protons.

$BF_3\text{ and }^{+}CCl_3$ :

It is a Lewis-acid that accepts electron pairs.

$^-H, CH_2O,^-OCH_3\text{ and }NH_3$ :

It is a Bronsted-Lowry base that accepts protons.

$^-H, CH_2O,^-OCH_3,CH_3Cl\text{ and }NH_3$ :

It is a Lewis-base that donates electron pairs.

7 0

3 years ago