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3 years ago

The visible spectrum of the Cr(acac)3

Chemistry

1 answer:

Sati [7]3 years ago

3 0

Explanation:

The observable visible spectrum of Cr(acac)3 complex is different from that of

[Cr(en)3]Br3 due to strength of bonded ligand in the coordination sphere.

en is a strong field ligand compared to acac thus pairing occurs. In both the complexes

Cr is in +3 state, configuration [Ar] 3d^3. There are 3 unpaired e in Cr(acac)3 whereas 1 unpaired e in [Cr(en)3]Br3 due to pairing.

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How many grams of aluminum are required to react with 35 mL of 2.0 M hydrochloric acid, HCl?__ HCl + __ Al --> __ AlCl3 + __

kiruha [24]

Answer:

0.6258 g

Explanation:

To determine the number grams of aluminum in the above reaction;

determine the number of moles of HCl
determine the mole ratio,
use the mole ratio to calculate the number of moles of aluminum.
use RFM of Aluminum to determine the grams required.

Moles of HCl

35 mL of 2.0 M HCl

2 moles of HCl is contained in 1000 mL

x moles of HCl is contained in 35 mL

$x \: mol \: = \: \frac{2 \: \times \: 35}{1000} \\ = 0.07 \: moles \:$

We have 0.07 moles of HCl.

Mole ratio

Balanced equation

6HCl(aq) + 2Al(s) --> 2AlCl3(aq) + 3H2(g)

Hence mole ratio = 6 : 2 (HCl : Al

but moles of HCl is 0.07, therefore the moles of Al;

$= \frac{2}{6} \times 0.07 \\ \: = 0.0233333 \: moles$

Therefore we have 0.0233333 moles of aluminum.

Grams of Aluminum

We use the formula;

$grams \: = moles \: \times \: rfm$

The RFM (Relative formula mass) of aluminum is 26.982g/mol.

Substitute values into the formula;

$= 0.0233333 \: moles \: \times \: 26.982 \: \frac{g}{mol} \\ = 0.625799 \: grams$

The number of grams of aluminum required to react with HCl is 0.6258 g.

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2 years ago

125 ml of nitrogen gas is collected at 70.0 degrees Celsius. The pressure

dybincka [34]

Answer: Volume of the gas at STP is 22.53 L.

Explanation:

Given : Volume = 125 mL (as 1 mL = 0.001 L) = 0.125 L

Temperature = $70^{o}C = (70 + 273) K = 343 K$

Pressure = $125 kPa = 125 kPa \times \frac{0.01 atm}{1 kPa} = 1.25 atm$

According to the ideal gas equation, the volume of given nitrogen gas is calculated as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

$1.25 atm \times V = 1 mol \times 0.0821 L atm/mol K \times 343 K\\V = \frac{1 mol \times 0.0821 L atm/mol K \times 343 K}{1.25 atm}\\= \frac{28.1603}{1.25} L\\= 22.53 L$

Hence, volume of the gas at STP is 22.53 L.

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2 years ago

Think of a resource and predict What happens to the resource when competition for it increases

MatroZZZ [7]

Lets say foxes hunt rabbits. If their are too many foxes then they will compete for the rabbits. The rabbits get outnumbered by the amount of foxes so the foxes kill all the rabbits and the rabbits could eventually go extinct.

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seropon [69]

Answer:

Explanation:

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