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melamori03 [73]

2 years ago

5

What is the wavelength of light with 4.01 x 10^-19 J of energy? (The speed of light in a vacuum is 3.00 x 10^8 m/s, and Planck's

constant is 6.626 x 10^-34 Jos.)
A. 1820 nm
B. 496 nm
C. 202 nm
D. 551 nm

1 answer:

Vesna [10]2 years ago

4 0

Answer:

D

hope it helps:)))!!!!!!

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Acetylene gas, C2H2, releases large quantities of heat when it burns, making it ideal for welding applications. This gas is gene

Karo-lina-s [1.5K]

Answer:

64 kg

Explanation:

The computation of the number of kg need to ordered is shown below:

As we know that

$CaC_2 + 2H_2O \rightarrow Ca(OH)^2 + C_2H_2$

Now as we know that

1 mole of $C_2H_2$ generated from 1 mole of $CaC_2$

Now

26g of $C_2H_2$ generated from 64g of $CaC_2$

And,

26kg of $C_2H_2$ generated from

= $\frac{64g\times26Kg}{26g}$

= 64kg $CaC_2$

Hence, the number of kg that required for ordering the calcium carbide is 64 kg

5 0

2 years ago

I need the 3 questions !!! About Rubidium (Rb) <br><br> ASAP PLEASE DUE IN LIKE 20 min

hjlf

Answer:

419kJ/mol
5,0,0,+12
That catches fire spontaneously

Explanation:

1. Topic: Chemistry

ElementFirst Ionization Energy (kJ/mol) Lithium520Sodium496Rubidium403Cesium376According to the above table, which is most likely to be the first ionization energy for potassium?

536kJ/mol

504kJ/mol

419kJ/mol

391kJ/mol

358kJ/mol

2. Topic: Chemistry, Atom

The correct set of four quantum numbers for the valence electrons of the rubidium atom (Z=37) is:

5,1,1,+12

5,0,1,+12

5,0,0,+12

5,1,0,+12

3. Rubidium and cesium are pyrophoric. Here the term pyrophoric means:

That hardly catches fire

That does not catch fire at all

That catches fire spontaneously

None of these

5 0

2 years ago

Using the following reaction:

Romashka-Z-Leto [24]

Answer:

Mn is the oxidizing agent.

N is the reducing agent.

Explanation:

Hello!

In this case, according to the undergoing chemical reaction, it is seen that the manganese in KMnO4 has an oxidation state of 7+, in MnSO4 of 2+ and nitrogen in KNO2 is 3+ and in KNO3 is 5+; thus we have the following half-reactions:

$Mn^{7+}+5e^-\rightarrow Mn^{2+}\\\\N^{3+}\rightarrow N^{5+}+2e^-$

Thus, since manganese is undergoing a decrease in the oxidation state, we infer it is the oxidizing agent whereas nitrogen, undergoing an increase in the oxidation state is the reducing agent.

Best regards!

5 0

2 years ago

A chemist must prepare of 800.0 ml potassium hydroxide solution with a pH of 13.00 at 25Â°.

ArbitrLikvidat [17]

Answer:

4.48 grams is the mass of potassium hydroxide that the chemist must weigh out in the second step.

Explanation:

The pH of the solution = 13.00

pH + pOH = 14

pOH = 14 - pH = 14 - 13.00 = 1.00

$pOH=-\log[OH^-]$

$1.00=-\log[OH^-]$

$[OH^-]=10^{-1.00} M=0.100 M$

$KOH(aq)\rightarrow K^+(aq)+OH^-(aq)$

$[KOH]=[OH^-]=[K^+]=0.100 M$

Molariy of the KOH = 0.100 M

Volume of the KOH solution = 800 mL= 0.800 L

1 mL = 0.001 L

Moles of KOH = n

$Molarity=\frac{Moles}{Volume(L)}$

$0.100 M=\frac{n}{0.800 L}$

n = 0.0800 mol

Mass of 0.0800 moles of KOH :

0.0800 mol × 56 g/mol = 4.48 g

4.48 grams is the mass of potassium hydroxide that the chemist must weigh out in the second step.

4 0

2 years ago

A 46.9 gram sample of a substance has a volume of about 3.5 centimeters3. It is solid at a room temperature of 23ºC. Out of the

horrorfan [7]

Answer : (C) Hafnium is the most likely identity of the given substance.

Solution : Given,

Mass of given substance (m) = 46.9 g

Volume of given substance (V) = 3.5 $Cm^{3}$

First, find the Density of given substance.

Formula used :

$Density=\frac{\text{Mass of given substance}}{\text{Voume of given substance}}$

Now,put all the values in this formula, we get

$Density=\frac{46.9 g}{3.5 Cm^{3} }$ = 13.4 g/ $Cm^{3}$

So, we conclude that the density of given substance (13.4 g/ $Cm^{3}$ ) is approximately equal to the density of Mercury and Hafnium (13.53 and 13.31 g/ $Cm^{3}$ respectively).

According to the question the substance is solid at room temperature but Mercury is liquid at room temperature. So, Mercury is not identical to the given substance.

Another element i.e, Hafnium is the element whose density is approximately equal to the given substance and also solid at room temperature. And we know that the melting point of solid is high.

So, Hafnium is the most likely element which is the identity of the given substance.

3 0

2 years ago

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