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77julia77 [94]
2 years ago
9

What is the anion of the compound Cu(CH3COO)2?

Chemistry
1 answer:
olya-2409 [2.1K]2 years ago
4 0

Answer:

The anion of the compound is b CH3COO-

Explanation:

The compound cupric acetate Cu(CH3COO)2 undergo dissociation to form cupric ion Cu2+ and Acetate anion(CH3COO-)

           Cu(CH3COO)2⇒ Cu2+ + CH3COO-

From the above equation it can be stated that the anion of Cu(CH3COO)2 is CH3COO-.

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How many moles of water contain 1.3 x 10^24 molecules?
Vika [28.1K]
C is the correct answer
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3 years ago
A 100.0-mL sample of 1.00 M NaOH is mixed with 50.0 mL of 1.00 M H2SO4 in a large Styrofoam coffee cup calorimeter fitted with a
worty [1.4K]

Answer:

THE ENTHALPY CHANGE IN KJ/MOLE IS +114 KJ/MOLE.

Explanation:

Heat = mass * specific heat capacity * temperature rise

Total volume = 100 + 50 = 150 mL

Total mass = density * volume

Total mass = 1 * 150 mL = 150 g

So therefore, the heat evolved during the reaction is:

Heat = 150 * 4.18 * ( 31.4 - 22.3)

Heat = 150 * 4.18 * 9.1

Heat = 5705.7 J

Equation for the reaction:

2 NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2 H2O(l)  

From the equation, 2 moles of NaOH reacts with 1 mole of H2SO4 to produce 1 mole of Na2SO4 and 2 moles of water

50 mL of 1 M of H2SO4 contains

50 * 1 / 1000 mole of acid

= 0.05 mole of acid

The production of 1 mole of water evolved 5705.7 J of heat and hence the enthalpy changein kJ per mole will be:

0.05 mole of H2SO4 produces 5705.7 J of heat

1 mole of H2SO4 will produce 5705.7 / 0.05 J

= 114,114 J / mole

In kj/mole = 114 kJ/mole.

Hence, the enthalpy change of the reaction in kJ /mole is +114 kJ/mole.

5 0
3 years ago
Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb(NO3)2(aq)+2
Setler79 [48]

Answer:

a) volume of ammonium iodide required =349 mL

b) the moles of lead iodide formed = 0.0436 mol

Explanation:

The reaction is:

Pb(NO_{3})_{2}+2NH_{4}I -->PbI_{2}+2NH_{4}NO_{3}

It shows that one mole of lead nitrate will react with two moles of ammonium iodide to give one mole of lead iodide.

Let us calculate the moles of lead nitrate taken in the solution.

Moles=molarityX volume (L)

Moles of lead nitrate = 0.360 X 0.121 =0.0436 mol

the moles of ammonium iodide required = 2 X0.0436 = 0.0872 mol

The volume of ammonium iodide required will be:

volume=\frac{moles}{molarity}=\frac{0.0872}{0.250}=0.349L=349mL

the moles of lead iodide formed = moles of lead nitrate taken = 0.0436 mol

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the moon is full or

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