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3 years ago

A 4.00-gram sample of sodium chloride was added to a beaker containing 40.0 mL 0.033 M lead(II)

Chemistry

1 answer:

melamori03 [73]3 years ago

6 0

The answer is going to be 2.5

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Binigyan ang isang liham ng pasasalamat at gagawin ko rin siya ng kabutihan sa ibang pagkakataon

anzhelika [568]

Answer:

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2 years ago

How many moles is 1.25 x 1024 atoms of bromine?

Irina-Kira [14]

Answer:

2.08 moles (3 s.f.)

Explanation:

number of moles

= number of atoms ÷ Avogadro's constant

Avogadro's constant= 6.022 ×10²³

Thus, number of moles

= 1.25×10²⁴ ÷ (6.022 ×10²³)

= 2.08 moles (3 s.f.)

8 0

2 years ago

How did the surface area of the metal strip expose to the solution affect the

frez [133]

Answer:

See explanation below

Explanation:

In an electrochemical cell, electricity is obtained by the gradual deterioration of the anode.

Hence, surface area of the metal will affect the length of time within which the electrochemical cell works.

The greater the surface area of the metal, the longer the electrochemical cell can function and the greater the quantity of electricity produced, hence the answer above.

3 0

2 years ago

What are the limitations of using chemical indicators in determining the pH of a given solution?

tino4ka555 [31]

Answer:

1) chemical indicators won't work above it's pH range so therefore it probably won't change colour.

2) the solution should be clear and colourless to see colour change.

3) indicators tend to be of low accuracy so it's not 100% reliable.

5 0

3 years ago

Calculate the mass, in grams, of Ag2CrO4 that will precipitate when 50.0mL of 0.20M AgNO3 solution is mixed with 40.0mL of 0.10M

Darina [25.2K]

Answer:

1.327 g Ag₂CrO₄

Explanation:

The reaction that takes place is:

2AgNO₃(aq) + K₂CrO₄(aq) → Ag₂CrO₄(s) + 2KNO₃(aq)

First we need to identify the limiting reactant:

We have:

0.20 M * 50.0 mL = 10 mmol of AgNO₃

0.10 M * 40.0 mL = 4 mmol of K₂CrO₄

If 4 mmol of K₂CrO₄ were to react completely, it would require (4*2) 8 mmol of AgNO₃. There's more than 8 mmol of AgNO₃ so AgNO₃ is the excess reactant. That makes K₂CrO₄ the limiting reactant.

Now we calculate the mass of Ag₂CrO₄ formed, using the limiting reactant:

4 mmol K₂CrO₄ * $\frac{1mmolAg_2CrO_4}{1mmolK_2CrO_4} *\frac{331.73mg}{1mmolAg_2CrO_4}$ = 1326.92 mg Ag₂CrO₄

1326.92 mg / 1000 = 1.327 g Ag₂CrO₄

7 0

2 years ago