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3 years ago

Which formula represents lead(I I) chromate?

Chemistry

1 answer:

Ymorist [56]3 years ago

7 0

The formula for lead (ii) chromate is: PbCrO4

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A chemist has a small amount of compound ( b.p. = 65 oC) that must be fractionally distilled. Yet, the Chemist does not want to

koban [17]

Answer:

The chemist can either:

a. Use a small fractionation apparatus.

b. Add a compound with a much higher boiling point.

Explanation:

Using a smaller fractionation apparatus or Vigreux column will help to minimize loss of the distillate.

If a compound with a higher boiling point is added, the vapors of this liquid will displace the vapors of this small amount of compound with a lower boiling point. This compound with a higher boiling point is known as a Chaser.

6 0

2 years ago

Some water released 7800 J of energy when cooled from 78°C to 33°C. What was the mass of the water?

Karo-lina-s [1.5K]

Answer:

-41. 47

Explanation:

m = q / Cp x T

m = Mass

q = Energy (or joules)

Cp = Heat Capacity

T = Change in Temperature

Water's heat capacity is always 4.18.

This is the formula you'll need for change in temperature:

Final - Initial

So, 33 - 78 = -45

m = 7800 / 4.18 x -45

= -41.47

4 0

1 year ago

How much water will be required to dissolve 0.3g of benzoic acid at 95oC given that the solubility of BA at 95oC is 68.0g/L?

Ainat [17]

Write as a proportion, showing the relationship of both given information:
68.0g 0.3g
---------- = -----------
1L x ( your answer)

Cross multiply: 68.0g× X = 0.3g × 1L
68.0g (X)= 0.3g/L
Solve for X by dividing both sides by 68.0 g
68.0g (X) = 0.3g/L
------------- ------------------
68.0g 68.0g

Then enter into calculator 0.3/68 and that will be your solution. Make sure you round up.

6 0

2 years ago

If 50% of a radioactive element remains after 4000 years, what is the half-life?

ANTONII [103]

Hey there!

A half-life means after a certain amount of time, half of that substance will be gone/changed after that time.

If 50%, or half, of the element remains after 4000 years, that means the half life must be 4000 years.

Hope this helps!

4 0

3 years ago

According to the following reaction, how many grams of potassium phosphate will be formed upon the complete reaction of 29.6 gra

Ugo [173]

Answer:

There is 37.36 grams of K3PO4 produced

Explanation:

Step 1: Data given

Mass of H3PO4 = 29.6 grams

KOH is in excess

Molar mass of KOH = 56.11 g/mol

Molar mass of H3PO4 = 97.99 g/mol

Step 2: The balanced equation

3KOH(aq) + H3PO4(aq) ⇔ K3PO4(aq)+3H2O(l)

Step 3: Calculate mass of KOH

Mass KOH = mass KOH / molar mass KOH

Mass KOH = 29.6 grams / 56.11 g/mol

Mass KOH = 0.528 moles

Step 4: Calculate moles of K3PO4

Since KOH is the limiting reactant, We need 3 moles of KOH for each moles of H3PO4, to produce 1 mole of K3PO4 and 3 moles of H2O

For 0.528 moles of KOH we'll have 0.528/3 = 0.176 moles of K3PO4

Step 5: Calculate mass of K3PO4

Mass K3PO4 = moles K3PO4 * molar mass K3PO4

Mass K3PO4 = 0.176 moles * 212.27 g/mol

Mass K3PO4 = 37.36 grams

There is 37.36 grams of K3PO4 produced

8 0

3 years ago