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FinnZ [79.3K]
4 years ago
8

Help please! I'd appreciate it 

Chemistry
1 answer:
babymother [125]4 years ago
6 0

Answer:

16.56 g

Explanation:

Mass is the production of Volume and Density.

m = V. d = 6 × 2.76 = 16.56 g

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g What are the relative rates of diffusion of the three naturally occurring isotopes of krypton, 80Kr80Kr , 82Kr82Kr, and 83Kr83
emmasim [6.3K]

Solution :

According to the Graham's law of diffusion, we know that, the rate of the diffusion varies inversely to the molar mass of the gas, i.e.

Rate of diffusion, $r_d = \frac{a}{\sqrt M}$

where, the 'M' is the molar mass of the gas.

Now in the case of the isotopes of the Krypton,

Atomic mass of $^{80}Kr$ = 80 AMU

Atomic mass of $^{82}Kr$ = 82 AMU

Atomic mass of $^{83}Kr$ = 83 AMU

So the ratio of the rate of diffusion of the three isotopes are :

$M_{d,^{80}Kr}:M_{d,^{82}Kr}:M_{d,^{83}Kr}$

$=\frac{1}{\sqrt{80}}:\frac{1}{\sqrt{82}}:\frac{1}{\sqrt{83}}$

$=0.1118:0.1104:0.10976$

Dividing the above three with the smallest number among the three i.e. 0.10976, we get the relative rates of diffusion.

∴ $M_{d,^{80}Kr}:M_{d,^{82}Kr}:M_{d,^{83}Kr}$

 = 1.02 : 1.01 : 1

Hence the relative rate of diffusion are :

$^{80}Kr(1.02)>^{82}Kr(1.01)>^{83}Kr(1.00)$

3 0
3 years ago
The ____________________ metals are all of the elements located in groups 3-12 on the Periodic Table.
S_A_V [24]

Answer:

Transition Metals are all of the elements kn group 3-12.

6 0
4 years ago
Determine the end (final) value of n in a hydrogen atom transition, if the electron starts in and the atom emits a photon of lig
Usimov [2.4K]

Complete Question

Determine the end (final) value of n in a hydrogen atom transition, if the electron starts in n=4 and the atom emits a photon of light with a wavelength of 486 nm. Group of answer choices

Answer:

n=2

Explanation:

From the question we are told that:

Wavelength \lambda=486nm=>486*10^{-9}

Generally the equation for Atom Transition is mathematically given by

\frac{1}{\lambda}=R_{\infty }(\frac{1}{n_1^2}-\frac{1}{n_2^2})

Where

Rydberg constant R_{\infty}=1.097*10^7

Therefore

\frac{1}{486*10^{-9}}=1.097*10^7*(\frac{1}{n_1^2}-\frac{1}{4^2})

(\frac{1}{n_1^2}-\frac{1}{4^2})=\frac{1}{486*10^{-9}*1.097*10^7}

n_1^2=3.98

n=1.99

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6 0
3 years ago
The solution set of axequalsb is the set of all vectors of the form wequalspplusbold v subscript h​, where bold v subscript h is
ELEN [110]
Answer : Incorrect

Explanation : In the given solution set of Ax = b;
is set of all vectors to form w= p + V_{h}
here V_{h} is is any solution of the equation Ax = 0;

this stands to be incorrect, as the correct would only exists when some vector p is found to be like this Ap=b
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4 years ago
The mass of a neutron is 1.67 x 10^-24 g. Approximately what number of neutrons would equal a mass of one gram?
Lelechka [254]
1. a sulfur-32 atom contains 16 protons, 16 neutrons, and 16 electrons. what is the mass (in grams) of a sulfur-32 atom?

The mass in grams of a sulfur-32 atom is 32.

That is how the isotopes are identified, showing the mass number of the isotope, which is the sum of protons and neutrons: 16 protons + 16 neutrons = 32 mass number = mass in grams of the isotope.

2. the mass of a neutron is 1.67 x 10-24 g. approximately what number of neutrons would equal a mass of one gram?

divide 1 gram by the number of neutrons per gram:

number of neutrons = 1 g / 1.67 * 10 ^ - 24 g / neutrons = 0.5998 * 10 ^ 24 neutrons = 5.998 * 10 ^23 neutrons.

3 0
4 years ago
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