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3 years ago

In which of these would a nonpolar covalent bond be present? HBr H2O HCI Br2

Chemistry

1 answer:

cluponka [151]3 years ago

5 0

Br2 is the correct answer

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A sample of carbon dioxide gas that has a volume of 14.6 L at STP contains how many

Nana76 [90]

Answer:

n = 0.651 mol

Explanation:

Given data:

Volume of CO₂ = 14.6 L

Temperature = standard = 273.15 K

Pressure = standard = 1 atm

Number of moles of CO₂ = ?

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K

T = temperature in kelvin

by putting values,

1 atm × 14.6 L = n × 0.0821 atm.L/ mol.K × 273.15 K

14.6 L. atm = n × 22.43atm.L/ mol

n = 0.651 mol

7 0

2 years ago

A compound x is an important ingredient of an antacid. It is also used in fine extinguishers. Identify x.

Maslowich

NaHCO3 is the required substance often used in antacid and is used in Soda Acid Fire Extinguisher.

3 0

3 years ago

Read 2 more answers

If she suddenly dropped the books, this would increase their energy. If she instead decided to hold the books over her head, thi

Anna35 [415]

Answer:

the answer is is is true

7 0

3 years ago

When 1.045 g of CaO is added to 50.0 mL of water at 25.0 °C in a calorimeter, the temperature of the water increases to 32.3 °C.

Rus_ich [418]

Answer:

1.71 kJ/mol

Explanation:

The expression for the calculation of the enthalpy change of a process is shown below as:-

$\Delta H=m\times C\times \Delta T$

Where,

$\Delta H$ is the enthalpy change

m is the mass

C is the specific heat capacity

$\Delta T$ is the temperature change

Thus, given that:-

Mass of CaO = 1.045 g

Specific heat = 4.18 J/g°C

$\Delta T=32.3-25.0\ ^0C=7.3\ ^0C$

So,

$\Delta H=1.045\times 4.18\times 7.3\ J=31.88713\ J$

Also, 1 J = 0.001 kJ

So,

$\Delta H=0.03189\ kJ$

Also, Molar mass of CaO = 56.0774 g/mol

$Moles=\frac{Mass}{Molar\ mass}=\frac{1.045}{56.0774}\ mol=0.01863\ mol$

Thus, Enthalpy change in kJ/mol is:-

$\Delta H=\frac{0.03189}{0.01863}\ kJ/mol=1.71\ kJ/mol$

8 0

2 years ago

Upper n subscript 2 (g) plus 3 upper H subscript 2 (g) double-headed arrow 2 upper N upper H subscript 3 (g). At equilibrium, th

artcher [175]

Answer:

The <u>equilibrium constant</u> is:

$k_c=0.0030M^{-2}$

Explanation:

The correct equation is:

N₂(g) + 3H₂(g) ⇄ 2NH₃(g)

Thus, with the equilibrium concentrations you can calculate the equilibrium constant, Kc.

The equation for the equilibrium constant is:

$k_c=\dfrac{[NH_3]^2}{[N_2]\cdot [H_2]^3}$

Substituting:

$k_c=\dfrac{(0.105M)^2}{(1.1M)\cdot (1.50M)^3}$

$k_c=0.0030M^{-2}$

6 0

3 years ago