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3 years ago

Which formula represents lead (ii) phosphate? 1. pbpo4 2. pb4po4 3. pb3(po4)2 4. pb2(po4)3?

1 answer:

6 0

3.) Charge on one molecule of PO4 is -3. So on two molecule it'll be 2(-3) = -6. Let oxidation number of Pb be x. Therefore, 3x+(-6) = 0. 3x=6. x=2.

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It is 762 miles from here to Chicago. An obese physics teacher jogs at a rate of 5.0 miles every 20.0 minutes. How long would it

Lina20 [59]

Answer:

3,048 minutes

Explanation:

762 divided by 5

that number times 20

5 0

3 years ago

Covalent bonds involve the sharing of?

4vir4ik [10]

Answer:

A covalent bond is a chemical bond that involves the sharing of electron pairs between atoms. These electron pairs are known as shared pairs or bonding pairs, and the stable balance of attractive and repulsive forces between atoms, when they share electrons, is known as covalent bonding.

Explanation:

I hope this helps

7 0

3 years ago

The vapor pressure of benzene at 298 K is 94.4 mm of Hg. The standard molar Gibbs free energy of formation of liquid benzene at

horsena [70]

Answer:

ΔfG°(C₆H₆(g)) = 129.7kJ/mol

Explanation:

Bringing out the parameters mentioned in the question;

Vapor pressure = 94.4 mm of Hg

The vaporization reaction is given as;

C₆H₆(l) ⇄ C₆H₆(g)

Equilibrium in terms of activities is given by:

K = a(C₆H₆(g)) / a(C₆H₆(l))

Activity of pure substances is one:

a(C₆H₆(l)) = 1

Assuming ideal gas phase activity equals partial pressure divided by total pressure. At standard conditions

K = p(C₆H₆(g)) / p°

Where p° = 1atm = 760mmHg standard pressure

We now have;

K = 94mmHg / 760mmHg = 0.12421

Gibbs free energy is given as;

ΔG = - R·T·ln(K)

where R = gas constant = 8.314472J/molK

So ΔG° of vaporization of benzene is:

ΔvG° = - 8.314472 · 298.15 · ln(0.12421)

ΔvG° = 5171J/mol = 5.2kJ/mol

Gibbs free energy change of reaction = Gibbs free energy of formation of products - Gibbs free energy of formation of reactants:

ΔvG° = ΔfG°(C₆H₆(g)) - ΔfG°(C₆H₆(l))

Hence:

ΔfG°(C₆H₆(g)) = ΔvG°+ ΔfG°(C₆H₆(l))

ΔfG°(C₆H₆(g)) = 5.2kJ/mol + 124.5kJ/mol

ΔfG°(C₆H₆(g)) = 129.7kJ/mol

6 0

4 years ago

NEED HELP!!!

Mekhanik [1.2K]

Answer:

b. The shorter the half-life, the more dangerous the radioisotope.

8 0

3 years ago

Calculate the average reaction rate expressed in moles h2 consumed per liter per second

djyliett [7]

Answer:

Average rate of reaction expressed in moles H₂ consumed per liter per second = 0.0025 M/s = 0.0025 mol/L.s

Explanation:

The complete, correct Question, is presented in the attached image to this answer.

The average rate of reaction in terms of the reactant is defined as the total amount of reactant consumed over a period of time divided by total period of time.

Mathematically,

Average rate of reaction = (change in concentration of H₂ over a period of time) ÷ (total period of time)

Average rate of reaction = (-ΔC)/(Δt)

The minus sign is there because the concentration of reactants reduce with time.

change in concentration of H₂ = -ΔC

= -(0.02 - 0.03) = 0.01 M

Time = Δt = 4 - 0 = 4.0 s

Average rate of reaction = (0.01/4) = 0.0025 M/s

We could solve for the average rate of reaction expressed in moles Cl₂ consumed per liter per second

change in concentration of Cl₂ = -ΔC

= -(0.04 - 0.05) = 0.01 M

Time = Δt = 4 - 0 = 4.0 s

Average rate of reaction = (0.01/4) = 0.0025 M/s = 0.0025 mol/L.s

Hope this Helps!!!

8 0

4 years ago