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RUDIKE [14]
3 years ago
9

Write in exponential form: 34⋅34⋅34⋅34⋅34⋅34⋅34⋅34.

Chemistry
1 answer:
kvv77 [185]3 years ago
7 0

Answer:

34^8

Explanation:

You might be interested in
(Not drawn to scale) 32 What does this model represent? (1) a single atom (3) a mixture (2) a cell (4) a compound Open notes nav
Mademuasel [1]

UwU have a good day bye!

3 0
3 years ago
Propose a plausible mechanism for the reaction f2 + 2clo2 → 2fclo2 given that the rate law for the reaction is rate = k[f2][clo2
shepuryov [24]

<u>The given reaction is:</u>

F2 + ClO2 → 2FClO2

Rate = k[F2][ClO2]

<u>Explanation:</u>

The possible mechanism for this reaction can be broken down into two steps with the slow step being the rate determining step

Step 1:       F2 + ClO2 → FClO2 + F ----------- Slow

Step 2:      F + ClO2 → FClO2           ----------- Fast

-----------------------------------------------------------

Overall:  F2 + 2ClO2 → 2FClO2

Rate = k[F2][ClO2]

 


8 0
3 years ago
Read 2 more answers
In Universe L, recently discovered by an intrepid team of chemists who also happen to have studied interdimensional travel, quan
Advocard [28]

Answer:

Manganese, Fifth transition element

[X] 3d⁶ 4s¹

Iron, Sixth transition element

[X] 3d⁶ 4s²

Explanation:

Complete Question

In Universe L, recently discovered by an intrepid team of chemists who also happen to have studied interdimensional travel, quantum mechanics works as it does in our universe, except that there are six d orbitals instead of the usual number we observe here. Use these facts to write the ground-state electron configurations of the sixth and seventh elements in the first transition series in Universe L. Note; you may use [X] to stand for the electron configuration of the noble gas at the end of the row before the first transition series.

Solution

In our universe, there are 5 d orbitals.

And according to Aufbau's principles that electrons fill the lower energy orbitals before they fill higher energy orbitals and Hund's Rule that states that electrons are fed singly to all the orbitals of a subshell before pairing occurs.

The fifth and sixth transition elements in our universe is then Manganese and Iron respectively.

Manganese - [Ar] 3d⁵ 4s²

Iron - [Ar] 3d⁶ 4s²

So, in the new universe L, where there are six d orbitals, for manganese, the fifth transition metal, because half filled orbitals are more stable than partially filled orbitals (that woukd have been rhe case if we leave 5 electrons on the 3d orbital), the 4s orbital is filled to half of its capacity and the one electron removed from the 4s is used to fill the six 3d orbital to half of its capacity too.

For the sixth transition element, the new extra electron just fills the lower energy 4s orbital, leaving the six 3d orbitals all half-filled.

Hence, they both have ground state configurations of

- Manganese, Fifth transition element

[X] 3d⁶ 4s¹

- Iron, Sixth transition element

[X] 3d⁶ 4s²

Hope this Helps!!!

7 0
3 years ago
A mixture of H2 and water vapor is present in a closed vessel at 20. 00°C. The total pressure of the system is 755. 0 mmHg. The
MAVERICK [17]

The partial stress of H2 is 737.47 mmHg Let's observe the Ideal Gas Law to find out the whole mols.

We count on that the closed vessel has 1L of volume

  • P.V=n.R.T
  • We must convert mmHg to atm. 760 mmHg.
  • 1 atm
  • 755 mmHg (755/760) = 0.993 atm
  • 0.993 m.1L=n.0.082 L.atm/mol.K .
  • 293 K(0.993 atm 1.1L)/(0.082mol.K /L.atm).
  • 293K = n
  • 0.0413mols = n

These are the whole moles. Now we are able to know the moles of water vapor, to discover the molar fraction of it.

  1. P.V=n.R.T
  2. 760 mmHg. 1 atm
  3. 17.5 mmHg (17.5 mmHg / 760 mmHg)=0.0230 atm
  4. 0.0230 m.1L=n.0.082 L.atm/mol.K.293 K(0.0230atm.1L)/(0.082mol.K/L.atm .293K)=n 9.58 × 10 ^ 4 mols = n.
  5. Molar fraction = mols )f gas/general mols.
  6. Molar fraction water vapor =9.58×10^ -four mols / 0.0413 mols
  7. Sum of molar fraction =1
  8. 1 - 9.58 × 10 ^ 4 × mols / 0.0413 ×mols = molar fraction H2
  9. 0.9767 = molar fraction H2
  10. H2 pressure / Total pressure =molar fraction H2
  11. H2 pressure / 55mmHg = =0.9767 0.9767 = h2 pressure =755 mmHg.
  12. 737,47 mmHg.
<h3>What is a mole fraction?</h3>

Mole fraction is a unit of concentration, described to be identical to the variety of moles of an issue divided through the whole variety of moles of a solution. Because it's miles a ratio, mole fraction is a unitless expression.

Thus it is clear that the partial pressure of H2 is 737,47 mmHg.

To learn  more about partial pressure refer to the link :

brainly.com/question/19813237

<h3 />

5 0
3 years ago
Two identical containers, one red and one yellow, are inflated with different gases at the same volume and pressure. Both contai
Evgen [1.6K]

Answer:

8

Explanation:

Here we're dealing with the root mean square velocity of gases. We'll provide the formula in order to calculate the root mean square velocity of a gas:

v_{rms}=\sqrt{\frac{3RT}{M}}

Here:

R = 8.314 \frac{J}{K mol} is the ideal gas law constant;

T is the absolute temperature in K;

M is the molar mass of a compound in kg/mol.

We know that the gas from the red container is 4 times faster, as it takes 4 times as long for the yellow container to leak out, this means:

\frac{v_{rms, red}}{v_{rms, yellow}} = 4

We also know that the temperature of the red container is twice as large:

\frac{T_{red}}{T_{yellow}} = 2

Write the ratio of the velocities and substitute the variables:

\frac{v_{rms, red}}{v_{rms, yellow}}=\frac{\sqrt{\frac{3RT_{red}}{M_{red}}}}{\sqrt{\frac{3RT_{yellow}}{M_{yellow}}}}=4

Then:

\frac{\sqrt{\frac{3RT_{red}}{M_{red}}}}{\sqrt{\frac{3RT_{yellow}}{M_{yellow}}}}=\sqrt{\frac{3RT_{red}}{M_{red}}\cdot \frac{M_{yellow}}{3RT_{yellow}}}=\sqrt{\frac{T_{red}}{T_{yellow}}\cdot \frac{M_{yellow}}{M_{red}}}=4

From here:

16 = \frac{T_{red}}{T_{yellow}}\cdot \frac{M_{yellow}}{M_{red}}

Then:

\frac{M_{yellow}}{M_{red}} = \frac{16}{\frac{T_{red}}{T_{yellow}}} = \frac{16}{2} = 8

5 0
3 years ago
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