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nikdorinn [45]
3 years ago
8

Passing an electric current through a sample of water (H2O) can cause the water to decompose into hydrogen gas (H2) and oxygen g

as (O2) according to the following equation.. . 2H2O= 2H2 + O2 . . The molar mass of H2O is 18.01 g/mol. The molar mass of O2 is 32.00 g/mol. What mass of H2O, in grams, must react to produce 50.00 g of O2?. 14.07. 23.05. 28.14. 56.28
Chemistry
2 answers:
Setler [38]3 years ago
8 0
The balanced chemical reaction is:

<span>2H2O= 2H2 + O2
</span>
We are given the amount of oxygen to be produced in the reaction. The starting point for the calculations will be this amount.

50 g ( 1 mol O2 / 32 g O2 ) ( 2 mol H2O / 1 mol O2 ) ( 18.01 g H2O / 1 mol H2O) = 56.28 g of H2O is needed.

Therefore, the correct answer is the last option.
Katyanochek1 [597]3 years ago
3 0

Answer: In more simpler terms the answer is 56.28

Explanation:

The other dude explained it above ^^^

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When 6.040 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 18.95 grams of CO2 and 7.759 grams of H
algol [13]

Answer: The empirical formula is CH_2 and molecular formula is C_4H_8

Explanation:

We are given:

Mass of CO_2 = 18.95 g

Mass of H_2O= 7.759 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 18.59 g of carbon dioxide, =\frac{12}{44}\times 18.59=5.07g of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 7.759 g of water, =\frac{2}{18}\times 7.759=0.862g of hydrogen will be contained.

Mass of C = 5.07 g

Mass of H = 0.862 g

Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{5.07g}{12g/mole}=0.422moles

Moles of H=\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.862g}{1g/mole}=0.862moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =\frac{0.422}{0.422}=1

For H =\frac{0.862}{0.422}=2

The ratio of C : H = 1: 2

Hence the empirical formula is CH_2.

The empirical weight of CH_2 = 1(12)+2(1)= 14 g.

The molecular weight = 56.1 g/mole

Now we have to calculate the molecular formula.

n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{56.1}{14}=4

The molecular formula will be=4\times CH_2=C_4H_8

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3 years ago
Meteorologists use weather balloons to carry weather instruments high into the atmosphere. When it is first released at Earth’s
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The pressure of the gas used in the weather balloon increases to expand the balloon.

Explanation:

  • Weather balloons contain the boxes where the weather measurement instruments are present that is attached to the large balloon.
  • Weather balloon uses gases like Hydrogen or Helium. When the weather balloon rises to the atmosphere, the air pressure decreases. This leads to the increase in the pressure of hydrogen or Helium gas used in the weather balloon. This expands the balloon.
  • The gas particles hits the balloon container and generates the pressure. The increase of pressure thus helps the weather balloon to move in a constant speed through the atmosphere.
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3 years ago
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user100 [1]
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What must be satisfied for a hypothesis to be useful?
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Answer:

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A phosphate buffer is involved in the formation of urine. The developing urine contains H2PO4 and HPO42- in the same concentrati
Katen [24]

Answer:

The ionization equation is

H_{2}PO_{4}^{-}  +H_{2}O ⇄HPO_{4}^{-2}  +H_{3}O^{+} (1)

Explanation:

The ionization equation is

H_{2}PO_{4}^{-}  +H_{2}O ⇄HPO_{4}^{-2}  +H_{3}O^{+} (1)

As the Bronsted definition sais, an acid is a substance with the ability to give protons thus, H2PO4 is the acid and HPO42- is the conjugate base.

The Ka expression is the ratio between the concentration of products and reactants of the equilibrium reaction so,

Ka = \frac{[HPO_{4}^{-2}] [H_{3}O^{+}]}{[H_{2}PO_{4}^{-}] [H_{2}O]} = 6.2x10^{-8}

The pKa is

-Log (Ka) = -Log (6.2x10^{-8}) = 7.2

The pKa of H2CO3 is 6,35, thus this a stronger acid than H2PO4. The higher the pKa of an acid greater the capacity to donate protons.

In the body H2CO3 is a more optimal buffer for regulating pH due to the combination of the two acid-base equilibriums and the two pKa.

If the urine is acidified, according to Le Chatlier's Principle the equilibrium (1)  moves to the left neutralizing the excess proton concentration.

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3 years ago
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