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pickupchik [31]

4 years ago

6

4.80 x 10^25 formula units pf calcium iodide (Cal2) will have what mass?

1 answer:

maw [93]4 years ago

7 0

Answer:

mass CaI2 = 23.424 Kg

Explanation:

From the periodic table we obtain for CaI2:

⇒ molecular mass CaI2: 40.078 + ((2)(126.90)) = 293.878 g/mol

∴ mol CaI2 = (4.80 E25 units )×(mol/6.022 E23 units) = 79.708 mol CaI2

⇒ mass CaI2 = (79.708 mol CaI2)×(293.878 g/mol) = 23424.43 g

⇒ mass CaI2 = 23.424 Kg

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pishuonlain [190]

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In the laboratory, a volume of 100 mL of sulfuric acid (H2SO4) is recorded. How many g are there of the liquid if its density is

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Explanation:

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4 years ago

HELP ASAP 10 POINTS

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3 years ago

How do we know that K2CrO4 and Ba(NO3)2 are not solids?

navik [9.2K]

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7 0

3 years ago

Calculate the standard reaction Gibbs free energy for the following cell reactions: (a) 2 Ce41(aq) 1 3 I2(aq) S 2 Ce31(aq) 1 I32

Law Incorporation [45]

<u>Answer:</u>

<u>For a:</u> The standard Gibbs free energy of the reaction is -347.4 kJ

<u>For b:</u> The standard Gibbs free energy of the reaction is 746.91 kJ

<u>Explanation:</u>

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$ ............(1)

<u>For a:</u>

The given chemical equation follows:

$2Ce^{4+}(aq.)+3I^{-}(aq.)\rightarrow 2Ce^{3+}(aq.)+I_3^-(aq.)$

<u>Oxidation half reaction:</u> $Ce^{4+}(aq.)\rightarrow Ce^{3+}(aq.)+e^-$ ( × 2)

<u>Reduction half reaction:</u> $3I^_(aq.)+2e^-\rightarrow I_3^-(aq.)$

We are given:

$n=2\\E^o_{cell}=+1.08V\\F=96500$

Putting values in equation 1, we get:

$\Delta G^o=-2\times 96500\times (+1.80)=-347,400J=-347.4kJ$

Hence, the standard Gibbs free energy of the reaction is -347.4 kJ

<u>For b:</u>

The given chemical equation follows:

$6Fe^{3+}(aq.)+2Cr^{3+}+7H_2O(l)(aq.)\rightarrow 6Fe^{2+}(aq.)+Cr_2O_7^{2-}(aq.)+14H^+(aq.)$

<u>Oxidation half reaction:</u> $Fe^{3+}(aq.)\rightarrow Fe^{2+}(aq.)+e^-$ ( × 6)

<u>Reduction half reaction:</u> $2Cr^{2+}(aq.)+7H_2O(l)+6e^-\rightarrow Cr_2O_7^{2-}(aq.)+14H^+(aq.)$

We are given:

$n=6\\E^o_{cell}=-1.29V\\F=96500$

Putting values in equation 1, we get:

$\Delta G^o=-6\times 96500\times (-1.29)=746,910J=746.91kJ$

Hence, the standard Gibbs free energy of the reaction is 746.91 kJ

7 0

3 years ago