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2 years ago

16. If the velocity of hydrogen molecule is 5 x 10^4cm sec-¹, then its de Broglie wavelength is

Chemistry

1 answer:

Aleks04 [339]2 years ago

5 0

Answer:

Correct option is B)

According to de-Broglie,

λ=mvh=6.023×10232×5×104cm/sec6.62×10−27ergsec=4×10−8cm=4Ao

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What is the energy of one photon of red light that has a wavelength of 678 nm?

Vitek1552 [10]

Answer:

E = 2.93×10⁻¹⁹ J

Explanation:

Given data:

Wavelength of photon = 678 nm (678×10⁻⁹ m)

Energy of photon = ?

Solution:

Formula:

E = h c/λ

h = 6.63×10⁻³⁴ Js

c = 3×10⁸ m/s

E = 6.63×10⁻³⁴ Js × 3×10⁸ m/s / 678×10⁻⁹ m

E = 19.89 ×10⁻²⁶ Jm / 678×10⁻⁹ m

E = 0.0293×10⁻¹⁷ J

E = 2.93×10⁻¹⁹ J

5 0

2 years ago

The least dense gas is neon fluorine krypton chlorine

DedPeter [7]

Answer:

Out of the options provided Neon is the least dense gas

Explanation:

6 0

2 years ago

Read 2 more answers

How could two objects have the same temperature but different thermal energies? One object could have more particles and greater

Tju [1.3M]

Answer:

One object could have more particles and greater total kinetic energy

Explanation:

The higher the temperature the more the particles. So, when we have high temperature, there is more particles interacting.

Temperature can simply be defined or gotten by taking the average of of the kinetic energy of the particles in the object that is the keywords here are TAKING THE AVERAGE KINETIC ENERGY

Thermal energy is the energy that can be gotten from adding up all the total kinetic energy of the particles in the object. So, the keywords here are; ADDING UP THE TOTAL KINETIC ENERGIES.

So, when two particles of the same temperature have the different thermal energy it means that One object could have more particles and thus having greater total kinetic energy.

6 0

2 years ago

A gas of 190 mL at a pressure of 74 atm can be expected to change its pressure when its volume changes to 30.0 mL. Express its n

kozerog [31]

Answer : The new pressure of the gas will be, 468.66 atm

Explanation :

Boyle's Law : This law states that pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

$P\propto \frac{1}{V}$ (At constant temperature and number of moles)

or,

$P_1V_1=P_2V_2$

where,

$P_1$ = initial pressure of the gas = 74 atm

$P_2$ = final pressure of the gas = ?

$V_1$ = initial volume of the gas = 190 ml

$V_2$ = final volume of the gas = 30 ml

Now we put all the given values in the above formula, we get the final or new pressure of the gas.

$74atm\times 190ml=P_2\times 30ml$

$P_2=468.66atm$

Therefore, the new pressure of the gas will be, 468.66 atm

4 0

2 years ago

Read 2 more answers

Balancing the reaction by oxidation number method k2cr2o7+sncl2+hcl

Mumz [18]

Answer:

$K_2Cr_2O_7 (aq) + 14 HCl (aq) + 3 SnCl_2 (aq)\rightarrow 2 CrCl_3 (aq) + 7 H_2O (l) + 3 SnCl_4 (aq) + 2 KCl (aq)$

Explanation:

The products of this reaction are given by:

$K_2Cr_2O_7 (aq) + SnCl_2 (aq) + HCl (aq)\rightarrow KCl (aq) + SnCl_4 (aq) + CrCl_3 (aq) + H_2O (l)$

Firstly, dichromate anion becomes chromium(III) cation, let's write this change:

$Cr_2O_7^{2-} (aq)\rightarrow Cr^{3+} (aq)$

The following steps should be taken:

balance the main element, chromium: multiply the right side by 2 to get 2 chromium species on both side:

$Cr_2O_7^{2-} (aq)\rightarrow 2 Cr^{3+} (aq)$

balance oxygen atoms by adding 7 water molecules on the right:

$Cr_2O_7^{2-} (aq)\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)$

balance the hydrogen atoms by adding 14 protons on the left:

$Cr_2O_7^{2-} (aq) + 14 H^+ (aq)\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)$

balance the charge (the total net charge on the left is 12+, on the right we have 6+, so 6 electrons are needed on the left):

$Cr_2O_7^{2-} (aq) + 14 H^+ (aq) + 6e^-\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)$

Similarly, tin(II) cation becomes tin(IV) cation:

$Sn^{2+} (aq)\rightarrow Sn^{4+} (aq) + 2e^-$

Now that we have the two half-equations, multiply the second one by 3, so that it also has 6 electrons that will be cancelled out upon addition of the two half-equations:

$Cr_2O_7^{2-} (aq) + 14 H^+ (aq) + 6e^-\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)$

$3 Sn^{2+} (aq)\rightarrow 3 Sn^{4+} (aq) + 6e^-$

Add them together:

$Cr_2O_7^{2-} (aq) + 14 H^+ (aq) + 3 Sn^{2+} (aq)\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l) + 3 Sn^{4+} (aq)$

Adding the ions spectators:

$K_2Cr_2O_7 (aq) + 14 HCl (aq) + 3 SnCl_2 (aq)\rightarrow 2 CrCl_3 (aq) + 7 H_2O (l) + 3 SnCl_4 (aq) + 2 KCl (aq)$

7 0

3 years ago