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Bad White [126]
2 years ago
14

A student is performing a titration to determine the concentration of a 20.0 mL sample of hydrochloric acid. What did the studen

t determine the concentration of the HCl to be if the end point was reached after 50.0 mL of 0.1 M NaOH was added.
Chemistry
1 answer:
irakobra [83]2 years ago
8 0

Answer:

0.25M HCl

Explanation:

The reaction of HCl with NaOH is:

HCl + NaOH ⇄ H₂O + NaCl

<em>Where 1 mole of HCl reacts per mole of NaOH</em>

The end point was reached when the student added:

0.0500L × (0.1mol / L) = 0.00500 moles of NaOH

As 1 mole of HCl reacted per mole of NaOH, moles of HCl present are:

<em>0.00500 moles HCl</em>

The volume of the sample of hydrochloric acid was 20.0mL = 0.0200L, and concentration of the sample is:

0.00500 mol HCl / 0.0200L = <em>0.25M HCl</em>

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O is what should go in the blank. O stands for Oxygen.
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Which of the following is not a catalyst? A. Nickel B. Platinum C. Enzymes D. Phosphorus
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The answer would be D because from my research it's the only one that didn't have a catalyst 
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b) If 1.5 mole of oxygen reacted, how many mole of iron (III) oxide would be formed from the reaction?
lyudmila [28]

Answer:

If there reacted 1.5 moles of O2, there will be produced 1.0 mol of Fe2O3

Explanation:

Step 1: Data given

Number of moles oxygen reacted = 1.5 moles

Step 2: The balanced equation

4Fe + 3O2 → 2Fe2O3

Step 3: Calculate moles of Fe2O3

For 4 moles Fe consumed, we need 3 moles of O2 to produce 2 moles of Fe2O3

For 1.5 moles O2 consumed, we'll have 2/3 * 1.5 = 1.0 mol of Fe2O3

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7 0
3 years ago
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gizmo_the_mogwai [7]

Answer:

It is because the object that is charged attracts the uncharged because it is giving electrons and passing it on to the object. As shown in the image the balloon is attracting the paper that is not charged. Actually telling,the charged object will cause the uncharged object to become charged, with a positive charge on one side and a negative charge on the other side. This process is called induction.

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2 years ago
Use the ideal gas law to calculate the concentrations of nitrogen and oxygen present in air at a pressure of 1.0 atm and a tempe
notka56 [123]

Answer:

[N₂] = 0.032 M

[O₂] = 0.0086 M

Explanation:

Ideal Gas Law → P . V = n .  R . T

We assume that the mixture of air occupies a volume of 1 L

78% N₂ → Mole fraction of N₂ = 0.78

21% O₂  → Mole fraction of O₂ = 0.21

1% another gases  → Mole fraction of another gases = 0.01

In a mixture, the total pressure of the system refers to total moles of the mixture

1 atm . 1L = n . 0.082L.atm/mol.K . 298K

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We apply the mole fraction to determine the moles

N₂ moles / Total moles = 0.78 → 0.78 . 0.0409 mol = 0.032 moles N₂

O₂ moles / Total moles = 0.21 → 0.21 . 0.0409 mol = 0.0086 moles O₂

4 0
3 years ago
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