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2 years ago

In any atom, the size of the nucleus with respect to the size of

Chemistry

1 answer:

iragen [17]2 years ago

6 0

Answer:

Size of the nucleus of an atom is very small as compared to the size of the atom.

According to Rutherford gold foil experiment, nucleus is very small in size as compared to the size of the atom as a whole. Nucleus is very hard, dense and positively charged which consists of protons and neutrons.

Explanation:

You might be interested in

Which of the following is accurate in describing the placement and classification of iodine?

fenix001 [56]

You can answer this question by only searching the element in the periodic table.

The atomic number of iodine, I, is 53. It is placed in the column 17 (this is the Group) and row 5 (this is the Period).

The conclusion is that the iodine is located in Period 5, Group 17, and is classified as a nonmetal.

4 0

2 years ago

Suppose Gabor, a scuba diver, is at a depth of 15m15m. Assume that: The air pressure in his air tract is the same as the net wat

alexandr1967 [171]

Answer:

The ration of molar concentration is "2.5".

Explanation:

The given values are:

Average density of salt water,

= $1.03 \ g/cm^3$

Net pressure,

= $2.00 \ atm$

Increase in pressure,

= $1.00 \ atm$

Now,

The under water pressure will be:

= $\frac{15 \ m}{10}\times 1 \ atm +1 \ atm$

= $1.5\times 1+1$

= $1.5+1$

= $2.5 \ atm$

hence,

The ratio will be:

= $\frac{(\frac{n}{V})_{15m} }{(\frac{n}{V})_{surface} }$

or,

= $\frac{P}{P_s}$

= $\frac{2.5}{1}$

= $2.5$

7 0

3 years ago

What is the volume at stp of 720.0 ml of a gas collected at 20.0 °c and 3.00 atm pressure?

MArishka [77]

Combined gas law is
PV/T = K (constant)

P = Pressure
V = Volume
T = Temperature in Kelvin

For two situations, the combined gas law can be applied as,
P₁V₁ / T₁ = P₂V₂ / T₂

P₁ = 3.00 atm P₂ = standard pressure = 1 atm
V₁ = 720.0 mL T₂ = standard temperature = 273 K
T₁ = (273 + 20) K = 293 K

By substituting,
3.00 atm x 720.0 mL / 293 K = 1 atm x V₂ / 273 K
V₂ = 2012.6 mL

hence the volume of gas at stp is 2012.6 mL

5 0

3 years ago

A flask contains a mixture of hydrogen gas and water vapor at STP. If the pressure of the water vapor is 19.5 mmHg, then what is

horrorfan [7]

Answer:

I would say A but not sure

Explanation:

7 0

2 years ago

Sulfuric acid is produced in larger amounts by weight than any other chemical. It is used in manufacturing fertilizers, oil refi

Fed [463]

Answer:

A. -166.6 kJ/mol

B. -127.7 kJ/mol

C. -133.9 kJ/mol

Explanation:

Let's consider the oxidation of sulfur dioxide.

2 SO₂(g) + O₂(g) → 2 SO₃(g) ΔG° = -141.8 kJ

The Gibbs free energy (ΔG) can be calculated using the following expression:

ΔG = ΔG° + R.T.lnQ

where,

ΔG° is the standard Gibbs free energy

R is the ideal gas constant

T is the absolute temperature (25 + 273.15 = 298.15 K)

Q is the reaction quotient

The molar concentration of each gas ([]) can be calculated from its pressure (P) using the following expression:

$[]=\frac{P}{R.T}$

Calculate ΔG at 25°C given the following sets of partial pressures.

Part A 130atm SO₂, 130atm O₂, 2.0atm SO₃. Express your answer using four significant figures.

$[SO_{2}]=[O_{2}]=\frac{130atm}{(0.08206atm.L/mol.K).298K} =5.32M$

$[SO_{3}]=\frac{2.0atm}{(0.08206atm.L/mol.K).298K} =0.0818M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0818^{2} }{5.32^{3} } =4.44 \times 10^{-5}$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln (4.44 × 10⁻⁵) = -166.6 kJ/mol

Part B 5.0atm SO₂, 3.0atm O₂, 30atm SO₃ Express your answer using four significant figures.

$[SO_{2}]=\frac{5.0atm}{(0.08206atm.L/mol.K).298K}=0.204M$

$[O_{2}]=\frac{3.0atm}{(0.08206atm.L/mol.K).298K}=0.123M$

$[SO_{3}]=\frac{30atm}{(0.08206atm.L/mol.K).298K}=1.23M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{1.23^{2} }{0.204^{2}.0.123 } =296$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 296 = -127.7 kJ/mol

Part C Each reactant and product at a partial pressure of 1.0 atm. Express your answer using four significant figures.

$[SO_{2}]=[O_{2}]=[SO_{3}]=\frac{1.0atm}{(0.08206atm.L/mol.K).298K}=0.0409M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0409^{2} }{0.0409^{3}} =24.4$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 24.4 = -133.9 kJ/mol

7 0

2 years ago