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2 years ago

Would you expect salts of the gold(I) ion, Au+, to be colored? Explain.

Chemistry

1 answer:

kolezko [41]2 years ago

8 0

Answer:

No. Au⁺ has a complete 5d sublevel.

Explanation:

The color of the transition metal/ion arises due to d-d transition which means that when energy of certain frequency is passed though the solution of the transition metal ions, the electrons of the d orbital gain some energy and goes to higher energy level. The energy absorbed lies in the visible region and thus they are appeared colored.

The electronic configuration of Au is:

1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶4f¹⁴5d¹⁰6s¹

If the gold atom looses one electron, then the electronic configuration is:

1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶4f¹⁴5d¹⁰

<u>Hence, in the configuration, the 5d orbital is full filled and there is no scope for d-d transition. </u>

<u>Thus, there will Au⁺ will not be colored.</u>

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If 36.9 mL of B2H6 reacted with excess oxygen gas, determine the actual yield of B2O3 if the percent yield of B2O3 was 75.7%. (T

zhuklara [117]

Answer: The actual yield of $B_2O_3$ is 60.0 g

Explanation:-

The balanced chemical reaction :

$B_2H_6(l)+3O_2(g)\rightarrow B_2O_3(s)+3H_2O(l)$

Mass of $B_2H_6$ = $Density\times Volume=1.131g/ml\times 36.9ml=41.7g$

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} B_2H_6=\frac{41.7g}{27.668g/mol}=1.51moles$

According to stoichiometry:

1 mole of $B_2H_6$ gives = 1 mole of $B_2O_3$

1.51 moles of $B_2H_6$ gives = $\frac{1}{1}\times 1.51=1.51$ moles of $B_2O_3$

Theoretical yield of $B_2O_3=moles\times {Molar mass}}=1.14mol\times 69.62g/mol=79.3g$

Percent yield of $B_2O_3$ = $75.7\%$

$\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100$

$75.7\%=\frac{\text{Actual yield}}{79.3}\times 100$

${\text{Actual yield}}=60.0g$

Thus the actual yield of $B_2O_3$ is 60.0 g

7 0

2 years ago

At what temperature would the volume of a gas be 0.550 L if it had a volume of 0.432 L at –20.0 o C?

Gnom [1K]

The temperature that would the volume of a gas be 0.550l if it had a volume of 0.432 L at -20.0 c is calculated using the Charles law formula

that is v1/T1=V2/T2
V1=0.550 l
t1=?
T2= -20 c +273 = 253 K
v2= 0.432 l

by making T1 the subject of the formula T1= V1T2/V2

T1= (0.55lL x253)/ 0.432 l = 322.11 K or 322.11-273 = 49.11 C

8 0

2 years ago

Read 2 more answers

A sample contains 2.2 g of the radioisotope niobium-91 and 15.4 g of its daughter isotope, zirconium-91. how many half-lives hav

dybincka [34]

Answer: 3

Explanation: This is a radioactive decay and all the radioactive process follows first order kinetics.

Equation for the reaction of decay of $_{19}^{40}\textrm{K}$ radioisotope follows: