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3 years ago

Why can't we say that acids react with all metals to produce hydrogen ?

Chemistry

2 answers:

chubhunter [2.5K]3 years ago

6 0

Answer:

Explanation:

Usually when an acid reacts with metal, salt and hydrogen are produced. Acid + metal = salt + hydrogen. Example calcium nitrate + hydrogen. Hope this helps

meriva3 years ago

4 0

Answer:

The more reactive the metal, the more rapid the reaction is. A metal below hydrogen in the reactivity series will not react with dilute acids.

Explanation:

You might be interested in

How many grams of precipitate will be formed when 20.5 mL of 0.800 M

Anton [14]

Answer:

There will be formed 1.84 grams of precipitate (NaNO3)

Explanation:

Step 1: The balanced equation

CO(NO3)2 (aq) + 2 NaOH (aq) → CO(OH)2 (s) + 2 NaNO3 (aq)

Step 2: Data given

Volume of 0.800 M CO(NO3)2 = 20.5 mL = 0.0205 L

Volume of 0.800 M NaOH = 27.0 mL = 0.027 L

Molar mass of NaNO3 = 84.99 g/mol

Step 3: Calculate moles of CO(NO3)2

Moles CO(NO3)2 = Molarity * volume

Moles CO(NO3)2 = 0.800 M * 0.0205

Moles CO(NO3)2 = 0.0164 moles

Step 4: Calculate moles NaOH

moles of NaOH = 0.800 M * 0.027 L

moles NaOH = 0.0216 moles

Step 5: Calculate limiting reactant

For 1 mole CO(NO3)2 consumed, we need 2 moles of NaOH to produce 1 mole of CO(OH)2 and 2 moles of NaNO3

NaOH is the limiting reactant. It will completely be consumed.

CO(NO3)2 is in excess. There willbe 0.0216 / 2 = 0.0108 moles of CO(NO3)2 consumed. There will remain 0.0164 - 0.0108 = 0.0056 moles of CO(OH)2

Step 6: Calculate moles of NaNO3

For 2 moles of NaOH consumed, we have 2 moles of NaNO3

For 0.0216 moles of NaOH, we have 0.0216 moles of NaNO3

Step 7: Calculate mass of NaNO3

mass of NaNO3 = moles of NaNO3 * Molar mass of NaNO3

mass of NaNO3 = 0.0216 moles * 84.99 g/mol = 1.84 grams

There will be formed 1.84 grams of precipitate (NaNO3)

5 0

3 years ago

Use the solubility rules from the Lab 4 introduction and your knowledge of qualitative separation schemes from the lab to answer

cluponka [151]

Answer:

a13+a13

Explanation:

4 0

3 years ago

A chemistry student is given 700. mL of a clear aqueous solution at 26.Â° C. He is told an unknown amount of a certain compound

Alex73 [517]

Answer:

The correct answer is - yes, 4.57 g of solute per 100 ml of solution

Explanation:

The correct answer is yes we can calculate the solubility of X in the water at 22.0Â°C. The salt will remain after the evaporate from the dissolved and cooled down at 26Â°C.

Then, the amount of solute dissolved in the 700 ml solution at 26°C is the weighed precipitate: 0.032 kg = 32 g.

Then solublity will be :

32. g solute / 700 ml solution = y / 100 ml solution

⇒ y = 32. g solute × 100 ml solution / 700 ml solution = 4.57 g.

Thus, the answer is 4.57 g of solute per 100 ml of solution.

5 0

3 years ago

If 12.6 grams of iron (III) oxide reacts with 9.65 grams of carbon monoxide to produce 7.23 g of pure iron, what are the theoret

Mariana [72]

The balanced equation is Fe₂O₃ + 3 CO = 2 Fe + 3 CO₂.
Next step is to convert everything to moles.
12.6g Fe₂O₃ x (1 mol Fe₂O₃ / 159.7g Fe₂O₃) = 0.07890 mol Fe₂O₃
9.65g CO x (1 mol CO / 28.01g CO) = 0.3445 mol CO
The third step is to determine the limiting and excess reactants.
0.07890 mol Fe₂O₃ x (3 mol CO/1 mol Fe₂O₃) = 0.2367 mol CO
Therefore Fe₂O₃ is the limiting reagent while CO is in excess.

0.07890 mol Fe x (2 mol Fe(s) / 1 mol Fe₂O₃) = 0.1578 mol Fe(s)
0.1578 mol Fe x (55.84g Fe / mole Fe) = 8.812g Fe is the theoretical yield
%yield = (7.23g / 8.812g) x 100% = 82.0% is the percent yield

3 0

3 years ago

Property of a substance

Scrat [10]

There are many properties to substances.

I'll list some examples below:

Mass
Volume
Density
Conductivity
Malleability
Boiling point
Melting point
Heat capacity

Hope this helps! :3

8 0

3 years ago