Please helppp final quiz helppp!!!!

Please help with chemistry question... It would help me tremendously and if you can, please show work :)?

Alexxandr [17]

Answer:

5.06 atm

Explanation:

Step 1:

Data obtained from the question. This includes:

Mass of S2O = 175g

Volume (V) = 16600 mL

Temperature (T) = 195°C

Pressure (P)

Step 2:

Determination of the number of mole of S2O in 175g of S2O.

Mass of S2O = 175g

Molar Mass of S2O = (32x2) + 16 = 64 + 16 = 80g/mol

Number of mole of S2O =.?

Number of mole = Mass/Molar Mass

Number of mole of S2O = 175/80

Number of mole of S2O = 2.1875 moles

Step 3:

Conversion to appropriate units.

It is essential to always express the various variables in the right units of measurement in order to obtain the desired answer in the right units.

For volume:

1000mL = 1L

Therefore, 16600mL = 16600/1000 = 16.6L

For temperature:

Temperature (Kelvin) = temperature (celsius) + 273

Temperature (celsius) = 195°C

Temperature (Kelvin) = 195°C + 273 = 468K

Step 4:

Determination of the pressure.

The pressure can be obtained by the application of the ideal gas equation. This is illustrated below:

Volume (V) = 16.6L

Temperature (T) = 468K

Number of mole (n) = 2.1875 moles

Gas constant (R) = 0.082atm.L/Kmol

Pressure (P) =

PV = nRT

P x 16.6 = 2.1875 x 0.082 x 468

Divide both side by 16.6

P = (2.1875 x 0.082 x 468) /16.6

P = 5.06 atm

Therefore, the pressure is 5.06 atm

6 0

3 years ago

What is the name of the element with the atomic number 79

natima [27]

Look it up, it’s not that hard.

6 0

3 years ago

The theoretical yield of Cl2 from certain starting amounts of MnO2 and HCl was calculated as 60.25 g and 65.02 g, respectively.

user100 [1]

Answer:

c 43.38 g

Explanation:

The reaction between MnO2 and HCl can be represented by the following balanced equation:

MnO2 + HCl ---> Cl2 + MnCl2 + H2O

From the balanced equation, the theoretically required molar ratio of MnO2 to HCl is 1:1, therefore the yields would have been expected to be equal.

For the fact that HCl gives a higher yield (65.02g) than MnO2 (60.25g) according to the problem statement, HCl should be in excess, while the limiting reagent should be MnO2 .

Thus, the theoretical yield of Cl2 will be 60.25 g.

By definition, the percentage yield is given by

% Yield = (Actual Yield) / (Theoretical Yield),

This can be simplified to

Actual Yield = % Yield * Theoretical Yield

Plugging in the given values we have

Actual Yield = 72% * 60.25 = 43.38 g

5 0

3 years ago

Help whats the answer

vesna_86 [32]

Answer:

I would make the replacement parts with the 3D printer to help fix the well

Explanation:

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6 0

2 years ago

if 14.0 g of aluminium reacts with excess sulfuric acid to produce 75.26 g of aluminium sulfate, what is the percent yield?

vlada-n [284]

Taking into account definition of percent yield, the percent yield for the reaction is 84.88%.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 Al + 3 H₂SO₄ → Al₂(SO₄)₃ + 3 H₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Al: 2 moles
H₂SO₄: 3 moles
Al₂(SO₄)₃. 1 mole
H₂: 3 moles

The molar mass of the compounds is:

Al: 27 g/mole
H₂SO₄: 98 g/mole
Al₂(SO₄)₃: 342 g/mole
H₂: 2 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Al: 2 moles ×27 g/mole= 54 grams
H₂SO₄: 3 moles ×98 g/mole= 294 grams
Al₂(SO₄)₃: 1 mole ×342 g/mole= 342 grams
H₂: 3 moles ×2 g/mole= 6 grams

<h3>Mass of aluminium sulfate formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 54 grams of aluminium form 342 grams of aluminium sulfate, 14 grams of aluminium form how much mass of aluminium sulfate?

$mass of aluminium sulfate=\frac{14 grams of aluminium x342 grams of aluminium sulfate}{54 grams of aluminium}$

<u><em>mass of aluminium sulfate= 88.67 grams</em></u>

Then, 88.67 grams of aluminium sulfate can be produced if 14.0 g of aluminium reacts with excess sulfuric acid.

<h3>Percent yield</h3>

The percent yield is the ratio of the actual return to the theoretical return expressed as a percentage.

The percent yield is calculated as the experimental yield divided by the theoretical yield multiplied by 100%:

$percent yield=\frac{actual yield}{theorical yield} x100$

where the theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.

<h3>Percent yield for the reaction in this case</h3>

In this case, you know: