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3 years ago

Convert 25.4 grams of barium phosphate, Ba3(PO4)2 to formula units.

1 answer:

7 0

To solve this questions you first need to find the number of moles of barium phosphate you have. The molar mass of barium phosphate is 601.93g/mol.
24.4/601.83 = 0.0402 moles barium phosphate
Then you need to use avagadro’s number, 6.022 x 10^23, which is the number of molecules or formula units in a mole.

6.022 x 10^23 * 0.0402 = 2.42 x 10^22 formula units

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What type of reaction is most likely to occur when barium reacts with fluorine? combustion synthesis decomposition single replac

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Given the ion C2O4-2, what species would you expect to form with each of the following ions?

Ksivusya [100]

Answer:

A. K₂C₂O₄ Potassium oxalate

B. CuC₂O₄ Copper oxalate

C. Bi₂(C₂O₄)₃ Bismuth (III) oxalate

D. Pb(C₂O₄)₂ Lead (IV) oxalate

E. (NH₄)₂C₂O₄ Ammonium oxalate

F. HC₂O₄⁻ Acid oxalate

Explanation:

C₂O₄⁻² → oxalate anion

This is the conjugate base from the H₂C₂O₄ which is the oxalic acid. A weak dyprotic acid that can release 2 protons.

A. 2K⁺ + C₂O₄⁻² → K₂C₂O₄ Potassium oxalate

It can be formed by the neutralization of the acid with the base

H₂C₂O₄ + 2KOH → K₂C₂O₄ + 2H₂O

B. Cu²⁺ + C₂O₄⁻² ⇄ CuC₂O₄ ↓

This is a precipitate.

C. 2Bi³⁺ + 3C₂O₄⁻² ⇄ Bi₂(C₂O₄)₃ ↓

This is a precipitate.

D. Pb⁴⁺ + 2C₂O₄⁻² ⇄ Pb(C₂O₄)₂ ↓

This is a precipitate.

E. 2NH₄⁺ + C₂O₄⁻² ⇄ (NH₄)₂C₂O₄ ↓

This is a precipitate.

F. This is the conjugate strong base, for the weak acid

H⁺ + C₂O₄⁻² ⇄ HC₂O₄⁻

HC₂O₄⁻ + H₂O ⇄ C₂O₄⁻² + H₃O⁺ Ka

HC₂O₄⁻ + H₂O ⇄ H₂C₂O₄ + OH⁻ Kb

HC₂O₄⁻ is an amphoteric compound

6 0

3 years ago

what is the specific heat of a substance if 300 j are required to raise the temperature of a 267-g sample by 12 degrees c

slava [35]

Answer : The specific heat of the substance is 0.0936 J/g °C

Explanation :

The amount of heat Q can be calculated using following formula.

$Q = m \times C \times \bigtriangleup T$

Where Q is the amount of heat required = 300 J

m is the mass of the substance = 267 g

ΔT is the change in temperature = 12°C

C is the specific heat of the substance.

We want to solve for C, so the equation for Q is modified as follows.

$C = \frac{Q}{m \times \bigtriangleup T}$

Let us plug in the values in above equation.

$C = \frac{300J}{267g \times 12 C}$

$C = \frac{300J}{3204 g C}$

C = 0.0936 J/g °C

The specific heat of the substance is 0.0936 J/g°C

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3 years ago