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Serga [27]
3 years ago
15

How many moles of compound are there in the following?

Chemistry
1 answer:
slavikrds [6]3 years ago
8 0

a) (NH4)2SO4 --- 1 mole of it contains 2 moles of N, 8 moles of H, 1 mole of S, and 4 moles of O.

MM = (2 moles N x 14.0 g/mole) + (8 moles H x 1.01 g/mole) + (1 mole S x 32.1 g/mole) + (4 moles O x 16.0 g/mole) = 132 g/mole.

6.60 g (NH4)2SO4 x (1 mole (NH4)2SO4 / 132 g (NH4)2SO4) = 0.0500 moles (NH4)2SO4

b) The molar mass for Ca(OH)2 = 74.0 g/mole, calculated like (NH4)2SO4 above.

4.5 kg Ca(OH)2 x (1000 g / 1 kg) x (1 mole Ca(OH)2 / 74.0 g Ca(OH)2) = 60.8 moles Ca(OH)2

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Which of the following is necessary to increase the rate of a reaction?
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Rate of Reaction ca be increased by applying following changes,

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2) Decrease in Activation Energy:
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3) Concentration of Reactants:
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4) Increasing Surface Area of Reactant:
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6 0
3 years ago
What is a object that produces heat by producing light please?
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Consider a solution containing 0.100 M fluoride ions and 0.126 M hydrogen fluoride. The concentration of fluoride ions after the
Nataliya [291]

Answer: The concentration of fluoride ions after the addition of 9.00 mL of 0.0100 M HCl to 25.0 mL of this solution is 0.0709 M.

Explanation:

Given: Concentration of hydrogen fluoride = 0.126 M

Concentration of fluoride ions = 0.1 M

Volume of HCl = 9.0 mL

Concentration of HCl = 0.01 M

Volume of HCl = 25.0 mL

Moles of F^{-} ions are calculated as follows.

Moles of F^{-} = molarity \times volume\\= 0.1 M \times 0.025 L\\= 0.0025 mol

Moles of HF are as follows.

Moles of HF = Molarity \times Volume\\= 0.126 M \times 0.025 L\\= 0.00315 mol

Moles of HCl are as follows.

Moles of HCl = Molarity \times volume\\= 0.01 M \times 0.009 L\\= 0.00009 mol

Now, reaction equation with initial and final moles will be as follows.

                        H^{+}  + F^{-}  \rightarrow  HF

Initial:      0.00009  0.0025    0.00315

Equilibrium:      (0.0025 - 0.00009)    (0.00315 + 0.00009)

                                = 0.00241                    = 0.00324

Total volume = (9.00 mL + 25.0 mL) = 34.0 mL = 0.034 L

Hence, concentration of fluoride ions is calculated as follows.

Concentration = \frac{moles}{volume}\\= \frac{0.00241 mol}{0.034 L}\\= 0.0709 M

Thus, we can conclude that concentration of fluoride ions after the addition of 9.00 mL of 0.0100 M HCl to 25.0 mL of this solution is 0.0709 M.

7 0
3 years ago
The dumping of wastewater from this pipe is an example of what type of pollution?
Veseljchak [2.6K]

Answer:

dumping of wastewater is an example of water polution.

Explanation:

water polution is when an alienated foreign contaminent is put into a water source either intentionally or accidentally

3 0
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