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4 years ago

What is the most stable resonance structure if oxygen is the central atom in the CON– ion?

Chemistry

1 answer:

alexandr1967 [171]4 years ago

8 0

Answer:

The most stable resonance structure shown in fig (II).

Explanation:

Given that,

If oxygen is the central atom in the CON– ion

We know that,

Resonance :

The movement of electron is called resonance.

If the electronegativity is more of element then the element is more stable.

We need to find the most stable resonance structure

According to figure,

Electronegativity of N is more than C

So, N is the more stable in fig (II).

Hence, The most stable resonance structure shown in fig (II).

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Ou have a 5 ml sample of a protein in 0.5 m nacl. you place the protein/salt sample inside dialysis tubing (see fig. 2-14) and p

Oduvanchick [21]

Answer:

Procedure (2)

Explanation:

Assume the dialyses come to equilibrium in the allotted times.

Procedure (1)

If you are dialyzing 5 mL of sample against 4 L of water, the concentration of NaCl will be decreased by a factor of

$\dfrac{5}{4000} = \dfrac{1}{800}$

Procedure (2)

For the first dialysis, the factor is

$\dfrac{5}{1000} = \dfrac{1}{200}$

After a second dialysis, the original concentration of NaCl will be reduced by a factor of

$\dfrac{1}{200} \times \dfrac{1}{200} = \dfrac{1}{40000}$

Procedure (2) is more efficient by a factor of

$\dfrac{40000}{800} = \mathbf{50}$

4 0

3 years ago

1826.5g of methanol (CH3OH), molar mass = 32.0 g/mol is added to 735 g of water, what is the molality of the methane 0.0348 m 1.

Snowcat [4.5K]

Answer:

Molality = 1.13 m

Explanation:

Molality is defined as the moles of the solute present in 1 kilogram of the solvent.

Given that:

Mass of $CH_3OH$ = 26.5 g

Molar mass of $CH_3OH$ = 32.04 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{26.5\ g}{32.04\ g/mol}$

$Moles\ of\ CH_3OH= 0.8271\ moles$

Mass of water = 735 g = 0.735 kg ( 1 g = 0.001 kg )

So, molality is:

$m=\frac {0.8271\ moles}{0.735\ kg}$

<u>Molality = 1.13 m</u>

4 0

3 years ago

A material has an ASTM grain size number of 7. Determine the magnification, if the number of grains per square inch observed is:

Luden [163]

Answer:

A) M = 100X

B) M = 36X

C) M = 178.88X

Explanation:

Given data:

ASTM grain size number 7

a) total grain per inch^2 - 64 grain/inch^2

we know that number of grain per square inch is given as

$Nm = 2^{n-1} (\frac{100}{M})^2$

where M is magnification, n is grain size

therefore we have

$64 = 2^{7-1}(\frac{100}{M})^2$

solving for M we get

M = 100 X

B) total grain per inch^2 = 500 grain/inch^2

we know that number of grain per square inch is given as

$Nm = 2^{n-1} (\frac{100}{M})^2$

where M is magnification, n is grain size

therefore we have $500 = 2^{7-1}(\frac{100}{M})^2$

solving for M we get

M = 36 X

C) Total grain per inch^2 = 20 grain/inch^2

we know that number of grain per square inch is given as

$Nm = 2^{n-1} (\frac{100}{M})^2$

where M is magnification, n is grain size

therefore we have $20 = 2^{7-1}(\frac{100}{M})^2$

solving for M we get

M = 178.88 X

8 0

3 years ago

A military jet cruising at an altitude of 12.0km and speed of 1300./kmh burns fuel at the rate of 46.2/Lmin. How would you calcu

ira [324]

Answer:

2,452.12 L of fuel the jet consumes on a 1150 km mission.

Explanation:

Speed of the jet = 1300 km/h

1 hour = 60 minutes

$1300 km/h=\frac{1300 km}{60 min}=21.667 km/min$

Rate at which jet consumes fuel = 46.2 L/min

Distance covered by jet by consuming 1 liter fuel or mileage = R

R = $\frac{21.667 km/min}{46.2 L/min}=0.4690 km/L$

The amount of fuel the jet consumes on a 1150 km mission will = V

Amount of fuel = $\frac{Distance}{Mileage}$

$V=\frac{1150 km}{0.4690 km/L}=2,452.12 L$

2,452.12 L of fuel the jet consumes on a 1150 km mission.

3 0

3 years ago

DNA in a cell replicates.

Shtirlitz [24]

Answer:

I'm assuming it would be B, since it makes sense. If not, I'm really sorry.

Explanation:

5 0

3 years ago

Read 2 more answers