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Ratling [72]
3 years ago
6

Answers for worksheet activity B

Chemistry
1 answer:
7nadin3 [17]3 years ago
5 0
Where is the picture??
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PLS HELP A gas has a volume of 250 mL at 96.5 kPa. What will the pressure of the gas be if the
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Answer:

Solve the following problems (assuming constant temperature). Assume all numbers are 3 sig figs. 1. A sample of oxygen gas occupies a volume of 250 mL at 740 torr pressure. ... the gas exert if the volume was decreased to 2.00 liters? ... A 175 mL sample of neon had its pressure changed from 75.0 kPa to 150 kPa.

Explanation:

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Select all of the ways collection areas use to gain water. precipitation falls direction into the collection area runoff from pr
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The answer will be A & B not only because it makes sense but evaporation is taking water away not adding to (gaining)
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3 years ago
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

7 0
3 years ago
A compound has the empirical formula . a 256-ml flask, at 373 k and 750. torr, contains 0.527 g of the gaseous compound. give th
CaHeK987 [17]
<span>Answer: From the ideal gas law, MM=mRTPV; where MM = molecular mass; m = mass; P = pressure in atmospheres; V= volume in litres; R = gas constant with appropriate units. So, 0.800â‹…gĂ—0.0821â‹…Lâ‹…atmâ‹…Kâ’1â‹…molâ’1Ă—373â‹…K0.256â‹…LĂ—0.987â‹…atm = 97.0 gâ‹…molâ’1. nĂ—(12.01+1.01+2Ă—35.45)â‹…gâ‹…molâ’1 = 97.0â‹…gâ‹…molâ’1. Clearly, n = 1. And molecular formula = C2H2Cl2. I seem to recall (but can't be bothered to look up) that vinylidene chloride, H2C=C(Cl)2 is a low boiling point gas, whereas the 1,2 dichloro species is a volatile liquid. At any rate we have supplied the molecular formula as required.</span>
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If an automobile air bag has a volume of 11.5 l , what mass of nan3 (in
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Moles of NaN3 at STP  = volume of gas / 22.4 = 11.5/22.4 = 0.5mole. Massof NaN3 = moles of NaN3 x molecular weight = 0.5 x 65 = 32.5 grams.
8 0
3 years ago
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