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katen-ka-za [31]
2 years ago
5

A bulb is found to weigh 5.76 grams more when filled with air at 298 K and 1.00 atmosphere pressure than when it is evacuated. W

hat is the volume of the bulb? (Consider air to be 80% nitrogen and 20% oxygen by volume)
Chemistry
1 answer:
garik1379 [7]2 years ago
6 0

Answer:

Subtract water vapor pressure from total pressure to get partial pressure of gas A: PA=1.03 atm- 1 atm=0.03 atm.

What is the total pressure of the gases at 298 K?

98.8 kPa

A sample of nitrogen gas is bubbled through water at 298 K and the volume collected is 250 mL. The total pressure of the gas, which is saturated with water vapour, is found to be 98.8 kPa at 298 K.

The total pressure of a mixture of gases can be defined as the sum of the pressures of each individual gas: Ptotal=P1+P2+… +Pn. + P n . The partial pressure of an individual gas is equal to the total pressure multiplied by the mole fraction of that gas.

How do you find the partial pressure of water in air?

e is vapor pressure Rv = R∗/Mv = 461.5Jkg−1K−1 and Mv = 18.01gmol−1, ϵ = Mv/Md = 0.622. The vapor pressure is the partial pressure of the water vapor. where es is in Pascals and T is in Celsius.

ExpHow do you find the pressure of h2?

For the high pressures in which hydrogen gas is often stored, the van der Waals equation can be used. It is P+a(n/V)^2=nRT. For diatomic hydrogen gas, a=0.244atm L^2/mol^2 and b=0.0266L/mol.lanation:

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A compound has the percent composition 47.40% Pd, 28.50% O, 21.40% C, and 2.69% H. Based on this information, which molecular fo
Helga [31]

Answer:

Pd(O₂CCH₃)₂

Explanation

5 0
3 years ago
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Determine the approximate amount of potassium hydrogen phthalate, KHP, that you will need to neutralize 6.00 ml of 0.100 M NaOH.
Sveta_85 [38]

Answer:

potassium hydrogen phthalate KHP MOLAR MASS = 204.233 glmol

to get 1000 ml

Molar concentration = Mass concentration/Molar Mass

mass concentration = molar concentration x molar mass

mass concentration=0.1 M,

molar mass= 204.233 g/mol

so to get 1L

mass conc = 204.233 x 0.1

= 20.4233g  for 1L or 1000 ml

to get 6.00 ml

if 20.4233g is for 1000ml

then to 6.00 ml

= 20.4233 x 6 / 1000

= 0.123g for 6.00 ml

according to the equation below

NaOH(aq) + KHC8H4O4(aq) --> KNaC8H4O4(aq) + H2O(l)

number of moles of NaOH is equal to that of KHP

so the same amount will be needed too, which is

= 0.123g

6 0
3 years ago
True of false. A d-orbital lies in the x, y, and z area defined by the geometric axis of space.​
evablogger [386]

Answer:

true

Explanation:

3 0
3 years ago
How would granite be classified as
vredina [299]

Answer:

Granite is an intrusive igneous rock which means it is cooled slowly deep upper the Earth's crust. It is composed of 25% to 35% quartz and over 50% potassium- and sodium rich feldspars.

Explanation:

4 0
3 years ago
Read 2 more answers
WILL MARK BRANILEST
ahrayia [7]

Answer:

FeCl3 is the limiting reactant

O2 is in excess

Theoretical yield Cl2 = 9.84 grams

The % yield is 96.5 %

Explanation:

Step 1: Data given

Mass of FeCl3 = 15.0 grams

Moles O2 = 4.0 moles

Mass of Cl2 produced = 9.5 grams

Step 2: The balanced equation

4FeCl3 + 3O2 → 2Fe2O3 + 6Cl2

Step 3: Calculate moles FeCl3

Moles FeCl3 = mass FeCl3 / molar mass FeCl3

Moles FeCl3 = 15.0 grams / 162.2 g/mol

Moles FeCl3 = 0.0925 moles

Step 4: Calculate limiting reactant

FeCl3 is the limiting reactant. Because we have way more (more than ratio 3:4) moles O2 than FeCl3. It will completely be consumed (0.0925 moles). O2 is in excess. There will react = 0.069375 moles O2

There will remain 4.0 - 0.069375 = 3.930625 moles

Step 5: Calculate moles Cl2

For 4 moles FeCl3 we need 3 moles O2 to produce 2 moles Fe2O3 and 6 moles Cl2

For 0.0925 moles FeCl3 moles we'll have 6/4 * 0.0925 = 0.13875 moles Cl2.

Step 6: Calculate mass Cl2

Mass Cl2 = moles * molar mass

Mass Cl2 = 0.13875 moles * 70.9 g/mol

Mass Cl2 = 9.84 grams

Step 7: Calculate % yield

% yield = (actual yield / theoretical yield) * 100%

% yield = (9.5 grams / 9.84 grams ) * 100%

% yield = 96.5 %

The % yield is 96.5 %

4 0
4 years ago
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