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Ede4ka [16]
3 years ago
13

The instrument pictured below is a manometer resistometer aerometer barometer

Chemistry
1 answer:
lidiya [134]3 years ago
3 0

that's instrument picture there is an manometer

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What is the lowest unoccupied molecular orbital in F2?
Juliette [100K]

Answer:

σ*2pₓ, also called 3\sigma_{u}\text{*}

Explanation:

I have drawn the MO diagram for fluorine below.

Each F atom contributes seven valence electrons, so we fill the MOs of fluorine with 14 electrons.

We have filled the \pi \text{*2p}_{y} and \pi \text{*2p}_{z} MOs.

They are the highest occupied molecular orbitals (HOMOs).

The next unfilled level (the LUMO) is the σ*2pₓ orbital. If you use the symmetry notation, it is called the 3\sigma_{u}\text{*} orbital.

This is the orbital that fluorine uses when it acts as an electron acceptor.

7 0
3 years ago
manganese has an atomic number of 25 and an atomic mass of 55 amu. How many particles are found in its nucleus​
saw5 [17]

The particles in the nucleus is 55

6 0
3 years ago
Which change would result in a stronger electromagnet?
lidiya [134]
Yep its B i did the test
8 0
3 years ago
Read 2 more answers
On average what is the time between collisions of a xenon atom at 300 K and (a) one torr pressure; (b) one bar pressure.
Amanda [17]

Answer:

(a). 132 × 10^-9 s = 132 nanoseconds.

(b)..176.5 pico-seconds.

Explanation:

(a). At one torr, the first thing to do is to find the speed and that can be done by using the formula below;

Speed = [ (8 × R × T)/ Mm × π]^1/2.

Where Mm = molar mass, T = temperature and R = gas constant.

Speed= [ ( 8 × 8.314 × 300)/ 131.293 × π × 10^-3)^1/2. = 220m/s.

The next thing to do now is to calculate for the degree of collision which can be calculated by using the formula below;

Degree of collision = √2 × π × speed × d^2 × pressure/ K × T.

Note that pressure = 1 torr = 133.32 N/m^2 and d = collision diameter.

Degree of collision = √2 × π × 220 × (4.9 × 10^-10)^2 × 133.32/ 1.38 × 10^-23 × 300.

Degree of collision = 7.55 × 10^6 s^-1.

Thus, 1/ 7.55 × 10^6. = 132 × 10^-9 s = 132 nanoseconds.

(b). At one bar;

1/10^5 × 10^3 × 56.65 = 1.765 × 10^-10 = 176.5 pico-seconds.

6 0
3 years ago
Most of the mass of an atom comes from the (blank) and (blank)
dsp73
Protons and neutrons make up most of an atom's mass. Electrons have almost no mass.
5 0
3 years ago
Read 2 more answers
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