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2 years ago

For a hydrogen atom, which electronic transition would result in the emission of a photon with the highest energy?

Chemistry

1 answer:

BartSMP [9]2 years ago

6 0

Answer: A

Explanation:

Considering the order of filling electrons into orbitals, movement from a higher to a lower energy level results in the emission of a photon with energy equal to the energy difference between the two energy levels. However, the energies of different orbitals are close together for high values of n (principal quantum number). Their relative energies may change significantly when they form ions. This implies that energy levels are better separated and have high differences in energy for low values of n. Hence the answer. This means that photons transiting between these Lowe n levels will posses higher photon energy due to larger energy difference between levels.

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When 70. milliliter of 3.0-molar Na2CO3 is added to 30. milliliters of 1.0-molar NaHCO3 the resulting concentration of Na+ is 2

tiny-mole [99]

Answer : The resulting concentration of $Na^+$ ion is, 4.5 M

Explanation : Given,

Concentration of $Na_2CO_3$ = $M_1$ = 3.0 M = 3.0 mol/L

Volume of $Na_2CO_3$ = $V_1$ = 70 mL = 0.07 L

Concentration of $NaHCO_3$ = $M_2$ = 1.0 M = 1.0 mol/L

Volume of $NaHCO_3$ = $V_2$ = 30 mL = 0.03 L

First we have to calculate the moles of $Na_2CO_3$ and $NaHCO_3$

$\text{Moles of }Na_2CO_3=\text{Concentration of }Na_2CO_3\times \text{Volume of }Na_2CO_3=3.0mol/L\times 0.07L=0.21mol$

and,

$\text{Moles of }NaHCO_3=\text{Concentration of }NaHCO_3\times \text{Volume of }NaHCO_3=1.0mol/L\times 0.03L=0.03mol$

Now we have to calculate the moles of $Na^+$ ions.

As, 1 mole of $Na_2CO_3$ will give 2 moles of $Na^+$ ions

So, 0.21 moles of $Na_2CO_3$ will give $2\times 0.21=0.42$ moles of $Na^+$ ions

and,

As, 1 mole of $NaHCO_3$ will give 1 mole of $Na^+$ ions

So, 0.03 moles of $NaHCO_3$ will give 0.03 moles of $Na^+$ ions

So,

Total number of moles of $Na^+$ ions = 0.42 + 0.03 =0.45 mole

Total volume of both solution = 70 mL + 30 mL = 100 mL = 0.1 L

Now we have to calculate the concentration of $Na^+$ ions.

$\text{Concentration of }Na^+=\frac{\text{Moles of }Na^+}{\text{Volume of solution}}=\frac{0.45mol}{0.1L}=4.5mol/L=4.5M$

Therefore, the resulting concentration of $Na^+$ ion is, 4.5 M

6 0

2 years ago

A metallurgist reacts 320.0 grams of 75.0% by mass silver nitrate solution with an excess of copper metal. How many grams of sil

denpristay [2]

Answer:

= 152.40 g

Explanation:

The equation for the reaction is;

Cu(s) + AgNO3 → Ag(s) + Cu(NO3)2

Mass of silver nitrate = 320.0 g × 0.75

= 240.0 g

Molar mass of silver nitrate = 169.87 g/mol

Therefore;

Moles of silver nitrate = 240.0 g/169.87 g/mol

= 1.413 moles

Mole ratio of Silver nitrate to silver metal = 1 : 1

Therefore, moles of silver metal = 1.413 moles

Hence;

Mass of silver metal = 1.413 moles × 107.868 g/mol

= 152.40 g

4 0

2 years ago

For the reaction PCl5(g) <--> PCl3(g) Cl2(g) at equilibrium, which statement correctly describes the effects of increasing

xenn [34]

The given question is incomplete. The complete question is :

For the reaction $PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)$ at equilibrium, which statement correctly describes the effects of increasing pressure and adding $PCl_5$ , respectively

a) Increasing pressure causes shift to reactants, adding $PCl_5$ causes shift to products.

b) Increasing pressure causes shift to products ,adding $PCl_5$ causes shift to reactants.

c) Increasing pressure causes shift to products, adding $PCl_5$ causes shift to products.

d) Increasing pressure causes shift to reactants,adding $PCl_5$ causes shift to reactants

Answer: Increasing pressure causes shift to reactants, adding $PCl_5$ causes shift to products.

Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

For the given equation:

$PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)$

a) If the pressure is increased, the volume will decrease according to Boyle's Law. Now, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where decrease in pressure is taking place. As the number of moles of gas molecules is lesser at the reactant side. So, the equilibrium will shift in the left direction. i.e. towards reactants.

b) If $PCl_5$ is added, the equilibrium will shift in the direction where $PCl_5$ is decreasing. So, the equilibrium will shift in the right direction. i.e. towards products.

8 0

2 years ago

A 6.50-g sample of copper metal at 25.0 Â°c is heated by the addition of 84.0 j of energy. the final temperature of the copper i

anzhelika [568]

The final temperature of the copper is 59.0. The specific heat capacity of copper is 0.38 j/g -k

3 0

2 years ago

A solution contains cr3+ ions. the addition of 0.063 l of 1.50 m naf solution was needed to completely precipitate the chromium

steposvetlana [31]

Cr{3+} + 3 NaF → CrF3 + 3 Na{+} 

First calculate the total mols of NaF.

(0.063 L) x (1.50 mol/L NaF) = 0.0945 mol NaF total

Using stoichiometric ratio:

0.0945 mol NaF * (1 mol Cr3+ / 3 mol NaF) * (51.9961 g Cr3+/mol) = 1.6379 g Cr3+

6 0

2 years ago