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3 years ago

What is shown in the diagram below?

Chemistry

2 answers:

Sholpan [36]3 years ago

6 0

Answer:

The Correct Answer is B

Sea breeze

Explanation:

Sea breeze

The cold air from the sea interacts with the warm air from the land and it produces a slight cold front.
The Sun warmth begins to heat the earth faster than the water.
following that day the temperature of the land Improves while the temperature of the water remains at the same temperature.

Colt1911 [192]3 years ago

3 0

B I think is right. Hope this helps!

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Extended networks of pi bonds can cause a delocalization of electrons. Explain how this occurs.

algol [13]

The answer is: Electrons are shared in each pi bond. If the pi bonds flip back and forth between the adjacent p-orbitals on the two sides of an atom, the shared electrons in the p-orbitals can become delocalized.

Hope this help

7 0

3 years ago

a sample of 3.00 g of so2 (g)originally in a 5.00 L vesselat 21 degee Celsius is transferred to a 10.0 L vessel at 26 degree Cel

eimsori [14]

Answer:

1) The partial pressure of SO₂ gas in the larger container = 0.115 atm.

2) The partial pressure of N₂ gas in the larger container = 0.206 atm.

3) The total pressure in the vessel = 0.321 atm.

Explanation:

To calculate the partial pressure of each gas, we can use the general law of ideal gas: PV = nRT.

where, P is the partial pressure of the gas in atm,

V is the volume of the vessel in L,

n is the no. of moles of the gas,

R is the general gas constant (R = 0.082 L.atm/mol.K),

T is the temperature of the gas in K.

1) What is the partial pressure of SO₂ gas in the larger container?

∵ P = nRT/V.

n = mass/molar mass = (3.0 g)/(64.066 g/mol) = 0.047 mol.

R = 0.082 L.atm/mol.K.

T = 26 °C + 273.15 = 299.15 K.

V = 10.0 L. (The volume of the new container)

∴ P = nRT/V = (0.047 mol)(0.082 L.atm/mol.K)(299.15 K)/(10.0 L) = 0.115 atm.

2) What is the partial pressure of N₂ gas in the larger container?

∵ P = nRT/V.

n = mass/molar mass = (2.35 g)/(28.0 g/mol) = 0.084 mol.

R = 0.082 L.atm/mol.K.

T = 26 °C + 273.15 = 299.15 K.

V = 10.0 L. (The volume of the new container)

∴ P = nRT/V = (0.084 mol)(0.082 L.atm/mol.K)(299.15 K)/(10.0 L) = 0.206 atm.

3) What is the total pressure in the vessel?

According to Dalton's law the total pressure exerted is equal to the sum of the partial pressures of the individual gases.

∵ The total pressure in the vessel = the partial pressure of SO₂ + the partial pressure of N₂.

∴ The total pressure in the vessel = 0.115 + 0.206 = 0.321 atm.

5 0

3 years ago

_H2SO4 + __ B(OH)3 = _ B2(SO4)3 + __ H2O

Nana76 [90]

Answer:

3H2SO4 + 2B(OH)3 -> B2(SO4)3 + 6H2O

Explanation:

3 0

3 years ago

The density of nitric oxide (NO) gas at 0.866 atm and 46.2 degrees celcius is ____ g/L

Artyom0805 [142]

Answer:

d = 0.992 g/L

Explanation:

Data Given:

Pressure of nitric oxide (NO) = 0.866 atm

Temperature of a gas = 46.2° C

Convert the temperature to kelvin = 46.2° C + 273

temperature in kelvin = 319.2 K

density of nitric oxide (NO) = ?

Solution:

Density of a gas can be calculated by

d = PM /RT

Where

d = density

P = Pressure

M = molar mass of gas

R = ideal gas constant = 0.0821 L atm mol⁻¹ K⁻¹

T = temperature

So,

Molar mass of NO = 30 g/mol

Put values in the formula:

d = PM /RT

d = 0.866 atm × 30 g/mol / 0.0821 L atm mol⁻¹ K⁻¹ × 319.2 K

d = 25.98 atm. g/mol / 26.2 L atm mol⁻¹

d = 0.992 g/L

8 0

3 years ago

A 2kg metal cylinder is supplied with 1600J of energy to heat it from 5*C to 13*C. What is the SHC of the metal?

gtnhenbr [62]

Answer:

100Jkg/°C

Explanation:

Given parameters:

Mass of metal = 2kg

Amount of heat energy = 1600J

Initial temperature = 5°C

Final temperature = 13°C

Unknown:

Specific heat capacity of the metal = ?

Solution:

Specific heat capacity of a body is the amount of heat needed to raise the temperature of unit mass of a body by 1°C.

H = m x C x (T₂ - T₁ )

H is the amount of heat

m is the mass

C is the unknown specific heat capacity

T is the temperature

Insert the parameters and solve;

1600 = 2 x C x (13 - 5)

1600 = 16C

C = 100Jkg/°C

3 0

3 years ago