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slavikrds [6]
3 years ago
15

Which substance has Hf defined as 0 kJ/mol?

Chemistry
1 answer:
Elina [12.6K]3 years ago
5 0
Co i belive. what are the options?

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What are the 5 parts of an atom?
saul85 [17]
N the nucleus, there are protons and neutrons. That's two parts. 

<span>Outside the nucleus, there are electrons. That's a total of three parts and that is all that chemists discuss. Theoretical physicists talk about other sub nuclear "parts", but if you include them, the number goes way over 5. </span>

<span>The "shells", "sub-shells", "orbits", "orbitals", "probabilities" and the like are not physical parts, but only descriptions of energy</span>
3 0
3 years ago
Chemical equation for hydrogen chloride dissolving in water
irga5000 [103]

Answer:

The answer is HCl → H 2 O H+ + Cl.

7 0
4 years ago
Which would melt first, the element
balandron [24]

Explanation:

First, get both values into the same units

To covert from Kelvin to degrees C, subtract 273.15

1210 - 273.15 =936.85

Melting point of germanium=936.85 degrees C

Melting point of gold= 1064 degrees C

Therefore germanium would melt first because it has the lower melting point.

4 0
3 years ago
During the discussion of gaseous diffusion for enriching uranium, it was claimed that 235UF6 diffuses 0.4% faster than 238UF6. S
Kay [80]

<u>Answer:</u> The below calculations proves that the rate of diffusion of ^{235}UF_6 is 0.4 % faster than the rate of diffusion of ^{238}UF_6

<u>Explanation:</u>

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation:

\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}

We are given:

Molar mass of ^{235}UF_6=349.034348g/mol

Molar mass of ^{238}UF_6=352.041206g/mol

By taking their ratio, we get:

\frac{Rate_{(^{235}UF_6)}}{Rate_{(^{238}UF_6)}}=\sqrt{\frac{M_{(^{238}UF_6)}}{M_{(^{235}UF_6)}}}

\frac{Rate_{(^{235}UF_6)}}{Rate_{(^{238}UF_6)}}=\sqrt{\frac{352.041206}{349.034348}}\\\\\frac{Rate_{(^{235}UF_6)}}{Rate_{(^{238}UF_6)}}=\frac{1.00429816}{1}

From the above relation, it is clear that rate of effusion of ^{235}UF_6 is faster than ^{238}UF_6

Difference in the rate of both the gases, Rate_{(^{235}UF_6)}-Rate_{(^{238}UF_6)}=1.00429816-1=0.00429816

To calculate the percentage increase in the rate, we use the equation:

\%\text{ increase}=\frac{\Delta R}{Rate_{(^{235}UF_6)}}\times 100

Putting values in above equation, we get:

\%\text{ increase}=\frac{0.00429816}{1.00429816}\times 100\\\\\%\text{ increase}=0.4\%

The above calculations proves that the rate of diffusion of ^{235}UF_6 is 0.4 % faster than the rate of diffusion of ^{238}UF_6

3 0
3 years ago
Elements in the same period have the same______. Every element in the first period has______ shell/orbital for its______. Every
vodomira [7]

Answer:

1. Number of shell

2. 1 shell

3. Atom

4. 2 shells

5. Atom

Explanation:

5 0
3 years ago
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