All categories

3 years ago

As changes in energy levels of electrons increase, the frequencies of atomic line spectra they emit ___?

1 answer:

6 0

Ans: As changes in energy levels of electrons increase, the frequencies of atomic line spectra they emit will increase.

The energy (E) is related to the frequency (ν) by the following equation:

E = hν

where h = planck's constant

The change in energy i between levels is:

ΔΕ = h(Δν) -----(1)

Based on the above equation, as the changes in energy levels increase, the frequency of emitted radiation will also increase.

You might be interested in

Need a quick answer!!!

Anna11 [10]

N: 60 mol
p: 20 atm = 20×1013hPa = 20260 hPa
T: 50°C = 323 K
R: 83,14 hPa·dm³/mol·K
V: ?
.................

pV = nRT
V = nRT/p
V = (60×83,14×323)/20260
V = 79,529 dm³ ≈ 79,6 dm³

79,6 dm³ = 79,6 L

:•)

7 0

3 years ago

What promotes greater hardness in minerals?

Mashutka [201]

Smaller atoms and stronger bonds promotes greater hardness in minerals.

7 0

3 years ago

Read 2 more answers

The solubility of water in diethyl ether has been reported to be 1.468 % by mass.' Assuming that 23.0 mL of diethyl ether were a

melomori [17]

Explanation:

As it is given that solubility of water in diethyl ether is 1.468 %. This means that in 100 ml saturated solution water present is 1.468 ml.

Hence, amount of diethyl ether present will be calculated as follows.

(100ml - 1.468 ml)

= 98.532 ml

So, it means that 98.532 ml of diethyl ether can dissolve 1.468 ml of water.

Hence, 23 ml of diethyl ether can dissolve the amount of water will be calculated as follows.

Amount of water = $\frac{1.468 ml \times 23 ml}{98.532 ml}$

= 0.3427 ml

Now, when magnesium dissolves in water then the reaction will be as follows.

$Mg + H_{2}O \rightarrow Mg(OH)_{2}$

Molar mass of Mg = 24.305 g

Molar mass of $H_{2}O$ = 18 g

Therefore, amount of magnesium present in 0.3427 ml of water is calculated as follows.

Amount of Mg = $\frac{24.305 g \times 0.3427 ml}{18 g}$

= 0.462 g

6 0

3 years ago

Metals posses varying reduction potentials. Iron can reduce Cu+2 (aq) to copper metal. Silver can reduce Au+ (aq) to gold metal.

natta225 [31]

Answer: The true statement is iron can reduce $Au^+(aq)$ to gold metal

Explanation:

Single displacement reaction is defined as the reaction in which more reactive element displaces a less reactive element from its chemical reaction.

The reactivity of metal is determined by a series known as reactivity series. The metals lying above in the series are more reactive than the metals which lie below in the series.

$A+BC\rightarrow AC+B$

Metal A is more reactive than metal B.

We are given:

Iron can reduce copper, silver can reduce gold, sodium can reduce iron and copper can reduce silver metal.

The increasing order of reactivity thus follows:

$Au$

where, sodium is most reactive and gold is least reactive

For the given options:

Option 1: Copper cannot easily reduce sodium ion to sodium metal because it is less reactive.

$Cu(s)+Na^+(aq.)\rightarrow \text{ No reaction}$

Option 2: Iron cant easily reduce gold ion to gold metal because it is more reactive.

$Fe(s)+3Au^+(aq.)\rightarrow Fe^{3+}(aq.)+3Au(s)$

Option 3: Silver cannot easily reduce iron ion to iron metal because it is less reactive.

$Ag(s)+Fe^{3+}(aq.)\rightarrow \text{ No reaction}$

Hence, the true statement is iron can reduce $Au^+(aq)$ to gold metal

7 0

3 years ago

If m = 45g and V = 15ml, what is D in g/ml?

ch4aika [34]

Answer:

3 g/mL

Explanation:

We know that the density of an object can be measured by dividing its mass (g) to its volume (mL).

Formula

D=m/v

Given data:

Mass= 45 g

Volume= 15 mL

Now we will put the values in formula:

D=45 g/ 15 mL= 3 g/mL

6 0

3 years ago