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BartSMP [9]
3 years ago
15

A trial of this decomposition experiment, using different quantities of reactants than those listed in the question above produc

ed the following data: Volume of O2 produced at room conditions 280 mL Barometric pressure 740 torr Temperature of water 24°C Termperature of O2 25°C Vapor Pressure due to water at 25°C 22.4 torr For the conditions listed above, calculate the volume of O2(g) produced at standard conditions of temperature and pressure. (enter your answer in liters)
Chemistry
1 answer:
Paul [167]3 years ago
5 0

Answer : The volume of O_2(g) produced at standard conditions of temperature and pressure is 0.2422 L

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of O_2 gas = (740-22.4) torr = 717.6 torr

P_2 = final pressure of O_2 gas at STP= 760 torr

V_1 = initial volume of O_2 gas = 280 mL

V_2 = final volume of O_2 gas at STP = ?

T_1 = initial temperature of O_2 gas = 25^oC=273+25=298K

T_2 = final temperature of O_2 gas = 0^oC=273+0=273K

Now put all the given values in the above equation, we get:

\frac{717.6torr\times 280mL}{298K}=\frac{760torr\times V_2}{273K}

V_2=242.2mL=0.2422L

Therefore, the volume of O_2(g) produced at standard conditions of temperature and pressure is 0.2422 L

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The molecular weight of table salt, NaCl, is 58.5 g/mol. A tablespoon of salt weighs 6.37 grams. Calculate the number of moles o
cestrela7 [59]

Answer:

The number of moles of salt in one tablespoon is = <u>0.11 mole</u>

<u>Grams </u>cancel each other.

Explanation:

<u>Moles</u> : It is the unit of quantity . It is the mass of the substance present in exactly 12g of C-12.

moles=\frac{given\ mass}{Molar\ mass}

<u>Moles Calculation:</u>

Given mass = 6.37  gram

Molar mass = 58.5 g/mol

moles=\frac{6.37}{58.5}

= 0.1088

= 0.11 mole

<u>Units calculation</u>

moles=\frac{given\ mass}{Molar\ mass}

moles=\frac{g}{g/mole}

moles=\frac{g}{g}\times mole

<u>g ang g cancels each other </u>

moles = moles

<u>Hence unit = gram (g ) cancel each other.</u>

3 0
3 years ago
Read 2 more answers
A mixture containing nitrogen and hydrogen weighs 3.49 g and occupies a volume of 7.45 L at 305 K and 1.03 atm. Calculate the ma
-BARSIC- [3]

Answer:

Mass percent N₂ = 89%

Mass percent H₂ = 11%

Explanation:

First we <u>use PV=nRT to calculate n</u>, which is the total number of moles of nitrogen and hydrogen:

  • 1.03 atm * 7.45 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 305 K
  • n = 0.307 mol

So now we know that

  • MolH₂ + MolN₂ = 0.307 mol

and

  • MolH₂ * 2 g/mol + MolN₂ * 28 g/mol = 3.49 g

So we have a <u>system of two equations and two unknowns</u>. We use algebra to solve it:

Express MolH₂ in terms of MolN₂:

  • MolH₂ + MolN₂ = 0.307 mol
  • MolH₂ = 0.307 - MolN₂

Replace that value in the second equation:

  • MolH₂ * 2 g/mol + MolN₂ * 28 g/mol = 3.49
  • (0.307-MolN₂) * 2 + MolN₂ * 28 = 3.49
  • 0.614 - 2MolN₂ + 28molN₂ = 3.49
  • 0.614 + 26MolN₂ = 3.49
  • MolN₂ = 0.111 mol

Now we calculate MolH₂:

  • MolH₂ + MolN₂ = 0.307 mol
  • MolH₂ + 0.111 = 0.307
  • MolH₂ = 0.196 mol

Finally, we convert each of those mol numbers to mass, to <u>calculate the mass percent</u>:

  • N₂ ⇒ 0.111 mol * 28 g/mol = 3.108 g N₂
  • H₂ ⇒ 0.196 mol * 2 g/mol = 0.392 g H₂

Mass % N₂ = 3.108/3.49 * 100% = 89.05% ≅ 89%

Mass % H₂ = 0.392/3.49 * 100% = 11.15% ≅ 11%

5 0
3 years ago
If one jellybean weighs 0.350g how many jelly beans are in 504g of jellybeans?
allsm [11]

Answer:

1440g

Explanation:

Divide 504g by 0.350g

4 0
3 years ago
Another chemistry question i’m not good at this at all:( has me stressed
adoni [48]

Answer:

Average atomic mass = 15.86 amu.

Explanation:

Given data:

Number of atoms of Z-16.000 amu = 205

Number of atoms of Z-14.000 amu = 15

Average atomic mass  = ?

Solution:

Total number of atoms = 205 + 15 = 220

Percentage of Z-16.000 = 205/220 ×100 = 93.18%

Percentage of Z-14.000 = 15/220 ×100 = 6.82 %

Average atomic mass  = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass)  / 100

Average atomic mass  = (93.18×16.000)+(6.82×14.000) /100

Average atomic mass =  1490.88 + 95.48 / 100

Average atomic mass =  1586.36 / 100

Average atomic mass = 15.86 amu.

5 0
3 years ago
I need help with with number 1515. How many moles of NO2 are in a 37.058 L container at 101.28 kPa and -139.86
Andre45 [30]

Explanation:

To solve this question, we will use the Clayperon Equation:

P.V = n.R.T

where:

P = 101.28 kPa

1 atm = 101,325 Pa

x atm = 101,280 Pa

x = 1 atm

V = 37.058 L

n = we don't know

R = 0.082 atm.L/K.mol

T = -139.88 ºC = -139.88+273.15 = 133.27 K

1*37.058 = n*0.082*133.27

n = 0.29 moles

Answer: 0.29 moles

5 0
1 year ago
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