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stiks02 [169]
3 years ago
7

g Select the statement that best answers the following question What effect does the cation of an ionic compound have on the app

earance of the solution? The cation affects the intensity of the color more than the color of the solution. The cation affects the color of the solution more than the intensity of the color. The cation does not affect the color or color intensity of the solution. The cation only affects the intensity of the color in a solution.
Chemistry
1 answer:
liberstina [14]3 years ago
7 0

Answer:

The cation affects the intensity of the color more than the color of the solution.

Explanation:

According to Beer Lambert law, the intensity of the colour of the solution depends on the concentration of the specie responsible for the colour in the solution.

Let us recall that transition metal compounds are coloured in solution due to electronic transitions.

Therefore, the cation affects the intensity of the color more than the color of the solution.

You might be interested in
If you began a reaction with the following ions in solution, what would be the net ionic equation?
Mumz [18]

Answer:

2 PO₄³⁻(aq) + 3 Fe²⁺(aq) ⇒ Fe₃(PO₄)₂

Explanation:

Let's consider the complete ionic equation between the ions present. It includes all the ions and the insoluble compounds (Fe₃(PO₄)₂ is insoluble).

Na⁺(aq) + 2 PO₄³⁻(aq) + 3 Fe²⁺(aq) + NO₃⁻(aq) ⇒ Fe₃(PO₄)₂ + Na⁺(aq) + NO₃⁻(aq)

The net ionic equation includes only the ions that participate in the reaction (not spectator ions) and the insoluble compounds.

2 PO₄³⁻(aq) + 3 Fe²⁺(aq) ⇒ Fe₃(PO₄)₂

6 0
3 years ago
What is the conecntration of Fe3 and the concentration of No3- present in the solution that result when 30.0 ml of 1.75M Fe(No3)
zvonat [6]

Answer:

[Fe^{+3}]=0.700 M

[NO_{3}^{-}]=2.10 M

Explanation:

Here, a solution of Fe(NO₃)₃ is diluted, as the total volume of the solution has increased. The formula for dilution of the compound is mathematically expressed as:

C_{1}. V_{1}= C_{2}.V_{2}

Here, C and V are the concentration and volume respectively. The numbers at the subscript denote the initial and final values. The concentration of Fe(NO₃)₃ is 1.75 M. As ferric nitrate dissociates completely in water, the initial concentration of ferric is also 1.75 M.

Solving for [Fe],

[Fe^{+3}]=\frac{C_{1}.V_{1}}{V_{2} }

[Fe^{+3}]=\frac{(1.75).(30.0)}{45.0+30.0 }

[Fe^{+3}]=0.700 M

For [NO₃⁻],

There are three moles of nitrate is 1 mole of Fe(NO₃)₃. This means that the initial concentration of nitrate ions will be three times the concentration of ferric nitrate i.e., it will be 5.25 M.

[NO_{3}^{-}]=\frac{C_{1}.V_{1}}{V_{2} }

[NO_{3}^{-}]=\frac{(5.25)(30.0)}{30.0+45.0 }

[NO_{3}^{-}]=2.10 M

7 0
3 years ago
What is the total pressure in atmospheres in a 10.0L vessel containing 2.50 x 10-3 mol H2, 1.00 x 10-3
grigory [225]

Answer:

Total pressure = 16.42× 10⁻⁹atm

Explanation:

Given data:

Moles of H₂ = 2.50 × 10⁻³ mol

Moles of He = 1.00 × 10⁻³ mol

Mass of Ne = 3 × 10⁻⁴ mol

Volume = 10 L

Temperature = 35°C

Total pressure = ?

Solution:

Pressure of hydrogen:

P = nRT / V

P = 2.50 × 10⁻³ mol× 0.0821 atm. L.mol⁻¹ .k⁻¹ × 308 K / 10 L

p = 63.22× 10⁻³  atm. L /10 L

P = 6.3 × 10⁻³atm

Pressure of helium:

P = nRT / V

P = 1.00 × 10⁻³ mol× 0.0821 atm. L.mol⁻¹ .k⁻¹ × 308 K / 10 L

p = 25.29 × 10⁻³ atm. L /10 L

P = 2.53× 10⁻³ atm

Pressure of neon:

P = nRT / V

P = 3 × 10⁻⁴ mol× 0.0821 atm. L.mol⁻¹ .k⁻¹ × 308 K / 10 L

p = 75.86× 10⁻³ atm. L /10 L

P = 7.59× 10⁻³ atm

Total pressure of mixture:

P(mixture)  = pressure of hydrogen + pressure of helium+ pressure of neon

P(mixture)  = 6.3 × 10⁻³atm + 2.53× 10⁻³ atm + 7.59× 10⁻³ atm

P(mixture) = 16.42× 10⁻⁹atm

8 0
3 years ago
Please solve quickly
slavikrds [6]

Answer:

Explanation:

mass of one virus = 9.0 x 10⁻¹² mg

mass of one mole = 6.02 x 10²³ x mass of one virus

= 6.02 x 10²³ x 9.0 x 10⁻¹²

= 54.18 x 10¹¹ mg

= 54 x 10⁸ g .

= 54 x 10⁵ kg .

b )

let n be no of moles of virus that will be equal to weight of oil tanker

n x 54 x 10⁵ = 3 x 10⁷

n = 5.5555  

rounding off to 2 significant figure

5.6 moles Ans .

6 0
4 years ago
A 20.0–milliliter sample of 0.200–molar K2CO3 so­lution is added to 30.0 milliliters of 0.400–mo­lar Ba(NO3)2 solution. Barium c
iragen [17]

Answer:

[Ba^2+] = 0.160 M

Explanation:

First, let's calculate the moles of each reactant with the following expression:

n = M * V

moles of K2CO3 = 0.02 x 0.200 = 0.004 moles

moles of Ba(NO3)2 = 0.03 x 0.400 = 0.012 moles

Now, let's write the equation that it's taking place. If it's neccesary, we will balance that.

Ba(NO3)2 + K2CO3 --> BaCO3 + 2KNO3

As you can see, 0.04 moles of  K2CO3 will react with only 0.004 moles of Ba(NO3) because is the limiting reactant. Therefore, you'll have a remanent of

0.012 - 0.004 = 0.008 moles of Ba(NO3)2

These moles are in total volume of 50 mL (30 + 20 = 50)

So finally, the concentration of Ba in solution will be:

[Ba] = 0.008 / 0.050 = 0.160 M

6 0
4 years ago
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