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3 years ago

Covalent compounds are formed when

1 answer:

8 0

a nonmetal bonds with a nonmetal, eg. Carbon and Oxygen form CO2 or CO

Metals and nonmetals form ionic bonds, metals don't bond with each other, and most noble gases don't bond at all.

Hope this helps! :)

You might be interested in

How many moles of copper atoms are in 150g of copper metal

djyliett [7]

The answer to this question would be: 2.36 mol

To answer this question, you need to know the molecular weight of copper. Molecular weight determines how much the weight of 1 mol of a molecule has. Copper molecular weight about 63.5g/mol. Then, the amount of mol in 150g copper should be: 150g / (63.5g/mol)= 2.36 mol

3 0

3 years ago

08.04 MC) An unknown solution has a pH of 7.2. Which of these chemicals is likely to cause the greatest decrease in the pH of th

OLEGan [10]

Answer:

HNO3

Explanation:

i have the test

6 0

3 years ago

2. When heated above 500 ºC, potassium nitrate decomposes according to the equation below. 4KNO3 2K2O + 2N2 + 5O2A.If oxygen is

AysviL [449]

Answer:

(a) The rate of formation of K2O is 0.12 M/s.

The rate of formation of N2 is also 0.12 M/s

(b) The rate of decomposition of KNO3 is 0.24 M/s

Explanation:

(a) From the equation of reaction, the mole ratio of K2O to O2 is 2:5.

Rate of formation of O2 is 0.3 M/s

Therefore, rate of formation of K2O = (2×0.3/5) = 0.12 M/s

Also from the equation of reaction, mole ratio of N2 to O2 is 2:5.

Rate of formation of N2 = (2×0.3/5) = 0.12 M/s

(b) From the equation of reaction, mole ratio of KNO3 to O2 is 4:5.

Therefore, rate of decomposition of KNO3 = (4×0.3/5) = 0.24 M/s

3 0

3 years ago

A sample consisting of 1.00 mol of perfect gas molecules at 27 °C is expanded isothermally from an initial pressure of 3.00 atm

Evgesh-ka [11]

Answer:

a) reversibly

ΔU = 0

q = 2740.16 J

w = -2740.16 J

ΔH = 0

ΔS(total) = 0

ΔS(sys) =9.13 J/K

ΔS(surr) = -9.13 J/K

b) against a constant external pressure of 1.00 atm

ΔU = 0

w = -1.66 kJ

q = 1.66 kJ

ΔH = 0

ΔS(sys) = 9.13 J/K

ΔS(surr) = -5.543 J/K

ΔS(total) = 3.587 J/K

Explanation:

Step 1: Data given:

Number of moles = 1.00 mol

Temperature = 27.00 °C = 300 Kelvin

Initial pressure = 3.00 atm

Final pressure = 1.00 atm

The gas constant = 8.31 J/mol*K

(a) reversibly

Step 2: Calculate work done

For ideal gases ΔU depends only on temperature. So as it is an isothermal (T constant).

Since the temperature remains constant:

ΔU = 0

ΔU = q + w

q = -w

w = -nRT ln (Pi/Pf)

⇒ with n = the number of moles of perfect gas = 1.00 mol

⇒ with R = the gas constant = 8.314 J/mol*K

⇒ with T = the temperature = 300 Kelvin

⇒ with Pi = the initial pressure = 3.00 atm

⇒ with Pf = the final pressure = 1.00 atm

w =- 1*8.314 *300 * ln(3)

w = -2740.16 J

q = -w

q = 2740.16 J

Step 3: Calculate change in enthalpy

Since there is no change in energy, ΔH = 0

Step 4: Calculate ΔS

for an isothermal process

ΔS (total) = ΔS(sys) + ΔS(surr)

ΔS(sys) = -ΔS(surr)

ΔS(sys) = n*R*ln(pi/pf)

ΔS(sys) = 1.00 * 8.314 * ln(3)

ΔS(sys) = 9.13 J/K

ΔS(surr) = -9.13 J/K

ΔS (total) = ΔS(sys) + ΔS(surr) = 0

(b) against a constant external pressure of 1.00 atm

Step 1: Calculate the work done

w = -Pext*ΔV

w = -Pext*(Vf - Vi)

⇒ with Vf = the final volume

⇒ with Vi = the initial volume

We have to calculate the final and initial volume. We do this via the ideal gas law P*V=n*R*T

V = (n*R*T)/P

Initial volume = (n*R*T)/Pi

⇒ Vi = (1*0.08206 *300)/3

⇒ Vi = 8.206 L

Final volume = (n*R*T)/Pf

⇒ Vf = (1*0.08206 *300)/1

⇒ Vf = 24.618 L

The work done w = -Pext*(Vf - Vi)

w = -1.00* ( 24.618 - 8.206)

w = -16.412 atm*L

w = -16 .412 *(101325/1atm*L) *(1kJ/1000J)

w = -1662.9 J = -1.66 kJ

Step 2: Calculate the change in internal energy

ΔU = 0

q = -w

q = 1.66 kJ

ΔH = 0 because there is no change in energy

Step 3: Calculate ΔS

ΔS(sys) = n*R*ln(3)

ΔS(sys) = 1.00 * 8.314 * ln(3)

ΔS(sys) = 9.13 J/K

ΔS(surr) = -q/T

ΔS(surr) = -1662.9J/300K

ΔS(surr) = -5.543 J/K

ΔS(total) = ΔS(surr) +ΔS(sys) = -5.543 J/K + 9.13 J/K = 3.587 J/K

4 0

4 years ago

Name five solvents other than water

vovikov84 [41]

Answer:

Ethanol

Methanol

Acetone

Tetrachloroethylene

and Hexanes

Explanation:

5 0

3 years ago