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2 years ago

Which of the following is not a nanoscale object?

Chemistry

2 answers:

Mekhanik [1.2K]2 years ago

7 0

Answer:

The correct answer will be option-A muscle cell.

Explanation:

Nanoscales are the structures less than 100 nm in length which is considered a billionth of a meter scale. The nanoscales lead to the formation of a new field of study called nanotechnology which finds its use in all perspectives of science.

In the given question, the nanoscale objects are enzymes (3-7) nm) length, the water molecule (0.275 nm), and xenon atom. Since the muscle cell length is usually measured in mm(1-40 mm)

Thus, option-A is the correct answer.

KiRa [710]2 years ago

6 0

If I am correct I would believe that it would be a muscle cell.

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In the best Lewis structure for the fulminate ion, CNO–, what is the formal charge on the central nitrogen atom?

Andrews [41]

Answer : The formal charge on the central nitrogen atom is, (+1)

Explanation :

Resonance structure : Resonance structure is an alternating method or way of drawing a Lewis-dot structure for a compound.

Resonance structure is defined as any of two or more possible structures of the compound. These structures have the identical geometry but have different arrangements of the paired electrons. Thus, we can say that the resonating structure are just the way of representing the same molecule.

First we have to determine the Lewis-dot structure of $CNO^-$ .

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

In the Lewis-dot structure the valance electrons are shown by 'dot'.

The given molecule is, $CNO^-$

As we know that carbon has '4' valence electrons, nitrogen has '5' valence electrons and oxygen has '6' valence electrons.

Therefore, the total number of valence electrons in $CNO^-$ = 4 + 5 + 6 + 1= 16

According to Lewis-dot structure, there are 8 number of bonding electrons and 8 number of non-bonding electrons.

Now we have to determine the formal charge for each atom.

Formula for formal charge :

$\text{Formal charge}=\text{Valence electrons}-\text{Non-bonding electrons}-\frac{\text{Bonding electrons}}{2}$

For structure 1 :

$\text{Formal charge on O}=6-6-\frac{2}{2}=-1$

$\text{Formal charge on C}=4-2-\frac{6}{2}=-1$

$\text{Formal charge on N}=5-0-\frac{8}{2}=+1$

For structure 2 :

$\text{Formal charge on O}=6-4-\frac{4}{2}=0$

$\text{Formal charge on C}=4-4-\frac{4}{2}=-2$

$\text{Formal charge on N}=5-0-\frac{8}{2}=+1$

For structure 3 :

$\text{Formal charge on O}=6-2-\frac{6}{2}=+1$

$\text{Formal charge on C}=4-6-\frac{2}{2}=-3$

$\text{Formal charge on N}=5-0-\frac{8}{2}=+1$

The best Lewis-dot structure is, structure 1.

Thus, the formal charge on the central nitrogen atom is, (+1)

3 0

2 years ago

) Starting with 5.00 g barium chloride n hydrate yields 4.26 g of anhydrous barium chloride after heating. Determine the integer

RSB [31]

Answer:

Explanation:

Mass of water molecule = mass of hydrated salt - mass of anhydrous salt

Mass of water molecule = 5.00 - 4.26 = 0.74g of water molecule.

Number of moles = mass / molarmass

Molar mass of water = 18.015g/mol

No. of moles of water = 0.74 / 18.015 = 0.0411 moles.

Mass of BaCl2 present =?

1 mole of BaCl2 = 208.23 g

X mole of BaCl2 = 4.26 g

X = (4.26 * 1) / 208.23

X = 0.020

0.020 moles is present in 4.26g of BaCl2

Mole ratio between water and BaCl2 =

0.0411 / 0.020 = 2

Therefore 2 molecules of water is present the hydrated salt.

4 0

2 years ago

Calculate the specific heat capacity for a 22.7-g sample of lead that absorbs 237 J when its temperature increases from 29.8 °C

soldier1979 [14.2K]

Answer:

$\boxed {\boxed {\sf c\approx 0.159 \ J/ g \textdegree C}}$

Explanation:

We are asked to find the specific heat capacity of a sample of lead. The formula for calculating the specific heat capacity is:

$c= \frac{Q}{m \times \Delta T}$

The heat absorbed (Q) is 237 Joules. The mass of the lead sample (m) is 22.7 grams. The change in temperature (ΔT) is the difference between the final temperature and the initial temperature. The temperature increases from 29.8 °C to 95.6 °C.

ΔT = final temperature -inital temperature
ΔT= 95.6 °C - 29.8 °C = 65.8 °C

Now we know all three variables and can substitute them into the formula.

Q= 237 J
m= 22.7 g
ΔT = 65.8 °C

$c= \frac {237 \ J}{22.7 \ g \ \times \ 65.8 \textdegree C}$

Solve the denominator.

22.7 g * 65.8 °C = 1493.66 g °C

$c= \frac {237 \ J}{1493.66 \ g \textdegree C}$

Divide.

$c= 0.1586706479 J /g \textdegree C$

The original values of heat, temperature, and mass all have 3 significant figures, so our answer must have the same. For the number we found that is the thousandth place. The 6 in the ten-thousandth place tells us to round the 8 up to a 9.

$c \approx 0.159 \ J/g \textdegree C$