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adell [148]
3 years ago
10

a chemist uses hot hydrogen gas to convert chromium (iii) oxide to pure chromium. how many grams of hydrogen are needed to produ

ce 90 grams of water h2o?
Chemistry
1 answer:
MA_775_DIABLO [31]3 years ago
6 0

Answer:

Explanation:

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The answer is letter A definitively .
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What is the smallest division on 4 balance
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Answer:0.1g

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At a certain temperature the vapor pressure of pure thiophene is measured to be . Suppose a solution is prepared by mixing of th
Lesechka [4]

Answer:

0.35 atm

Explanation:

It seems the question is incomplete. But an internet search shows me these values for the question:

" At a certain temperature the vapor pressure of pure thiophene (C₄H₄S) is measured to be 0.60 atm. Suppose a solution is prepared by mixing 137. g of thiophene and 111. g of heptane (C₇H₁₆). Calculate the partial pressure of thiophene vapor above this solution. Be sure your answer has the correct number of significant digits. Note for advanced students: you may assume the solution is ideal."

Keep in mind that if the values in your question are different, your answer will be different too. <em>However the methodology will remain the same.</em>

First we <u>calculate the moles of thiophene and heptane</u>, using their molar mass:

  • 137 g thiophene ÷ 84.14 g/mol = 1.63 moles thiophene
  • 111 g heptane ÷ 100 g/mol = 1.11 moles heptane

Total number of moles = 1.63 + 1.11 = 2.74 moles

The<u> mole fraction of thiophene</u> is:

  • 1.63 / 2.74 = 0.59

Finally, the <u>partial pressure of thiophene vapor is</u>:

Partial pressure = Mole Fraction * Vapor pressure of Pure Thiophene

  • Partial Pressure = 0.59 * 0.60 atm
  • Pp = 0.35 atm

3 0
3 years ago
The partial pressure of O2 in air at sea level is 0.21atm. The solubility of O2 in water at 20∘C, with 1 atm O2 pressure is 1.38
adell [148]

Answer:

1.21x10^{-3} M

Explanation:

Henry's law relational the partial pressure and the concentration of a gas, which is its solubility. So, at the sea level, the total pressure of the air is 1 atm, and the partial pressure of O2 is 0.21 atm. So 21% of the air is O2.

Partial pressure = Henry's constant x molar concentration

0.21 = Hx1.38x10^{-3}

H = \frac{0.21}{1.38x10^{-3} }

H = 152.17 atm/M

For a pressure of 665 torr, knowing that 1 atm = 760 torr, so 665 tor = 0.875 atm, the ar concentration is the same, so 21% is O2, and the partial pressure of O2 must be:

P = 0.21*0.875 = 0.1837 atm

Then, the molar concentration [O2], will be:

P = Hx[O2]

0.1837 = 152.17x[O2]

[O2] = 0.1837/15.17

[O2] = 1.21x10^{-3} M

7 0
3 years ago
1. Silver nitrate will react with aluminum metal, yielding aluminum nitrate and silver metal. If you start with 0.223 moles of a
Archy [21]

Answer:

Explanation:

1)

Given data:

Number of moles of aluminium = 0.223 mol

Mass of silver produced = ?

Solution:

Chemical equation:

3AgNO₃  +   Al  →  3Ag + Al(NO₃)₃

Now we will compare the moles of Al with silver.

                               Al           :            Ag

                                1            :             3

                                0.223   :         3×0.223= 0.669 mol

Grams of silver:

Mass = number of moles × molar mass

Mass = 0.669 mol × 107.87 g/mol

Mass = 72.2 g

2)

Given data:

Number of moles of mercury(II) oxide produced = 3.12 mol

Mass of mercury = ?

Solution:

Chemical equation:

2Hg + O₂  →   2HgO

Now we will compare the moles of mercury with mercury(II) oxide.

                         HgO         :         Hg

                            2            :          2

                          3.12          :       3.12

Mass of Hg:

Mass = number of moles × molar mass

Mass = 3.12 mol × 200.59 g/mol

Mass = 625.84 g

3)

Given data:

Number of moles of dinitrogen pentoxide = 12.99 mol

Mass of oxygen = ?

Solution:

Chemical equation:

2N₂  + 5O₂   →  2N₂O₅

Now we will compare the moles of N₂O₅ with oxygen.

                 N₂O₅          :           O₂

                     2            :             5

                    12.99      :         5/2×12.99 = 32.48 mol

Mass of oxygen:

Mass = number of moles × molar mass

Mass = 32.48 mol × 32 g/mol

Mass = 1039.36 g

4)

Given data:

Number of moles of benzene = 0.103 mol

Mass of carbon dioxide = ?

Solution:

Chemical equation:

2C₆H₆  + 15O₂   →  12CO₂ + 6H₂O

Now we will compare the moles of N₂O₅ with oxygen.

                  C₆H₆         :           CO₂

                     2            :             12

                    0.103      :         12/2×0.103 = 0.618 mol

Mass of carbon dioxide:

Mass = number of moles × molar mass

Mass = 0.618 mol × 44 g/mol

Mass = 27.192 g

3 0
3 years ago
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