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cluponka [151]
2 years ago
14

1s22s22p63s23p3 what element does this represent

Chemistry
1 answer:
s344n2d4d5 [400]2 years ago
3 0

Answer:

Its phosphorus (P)

Explanation:

In writing the electron configuration for Phosphorus the first two electrons will go in the 1s orbital. Since 1s can only hold two electrons the next 2 electrons for Phosphorous go in the 2s orbital. The next six electrons will go in the 2p orbital. The p orbital can hold up to six electrons. We'll put six in the 2p orbital and then put the next two electrons in the 3s. Since the 3s if now full we'll move to the 3p where we'll place the remaining three electrons. Therefore the Phosphorus electron configuration will be 1s22s22p63s23p3.

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What is Bose Einstein state of matter and their examples
Valentin [98]

Answer:

A BEC ( Bose - Einstein condensate ) is a state of matter of a dilute gas of bosons cooled to temperatures very close to absolute zero is called BEC.

Examples - Superconductors and superfluids are the two examples of BEC.

Explanation:

8 0
3 years ago
What are the parts of a calorimeter and the function of each part?
chubhunter [2.5K]

A calorimeter is a device used to measure the amount of heat involved in a chemical or physical process. For example, when an exothermic reaction occurs in solution in a calorimeter, the heat produced by the reaction is absorbed by the solution, which increases its temperature.

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
How many atoms of Ca are present in 80.156 grams of Ca? (4 points) 1.2044 × 1024 2.4088 × 1024 4.8270 × 1025 1.9346 × 1027
Alika [10]

Answer: 1.2044 × 10^24

Explanation:

1 mole of calcium is 40 gram

Based on Avogadro's law:

1 mole of any substance has 6.02 x 10^23 atoms, also 1 mole of calcium is 40gram

So, 1 mole of calcium = 6.02 x 10^23 atoms

Also, 1 mole of calcium is 40gram

40 grams of calcium = 6.02 x 10^23 atoms

80.156 grams of calcium = Z atoms

To get the value of Z, cross multiply:

Z atoms x 40 grams = 6.02 x 10^23 atoms x 80.156 grams

40Z = 482.54 x 10^23

Z = (482.54 x 10^23/40)

Z = 12.06 x 10^23

Put result in standard form

Z = 1.206 x 10^24 atoms

Thus, 1.206 x 10^24 atoms of Ca are present in 80.156 grams of Ca

6 0
4 years ago
Read 2 more answers
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viktelen [127]
The water vapor will begin to contract. The particles will begin to bond with each other to seek heat so they're contracting.
3 0
4 years ago
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