In the atmosphere water vapor condenses to form clouds.

Convert the following into balanced equations:

enyata [817]

Pb(NO₃)₂ + 2KI → PbI₂ + 2KNO₃

We must first convert from a word equation to a symbol equation:to a symbol equation:

Lead (II) Nitrate + Potassium Iodide → Lead (II) Iodide + Potassium Nitrate

The lead (II) ion is represented as Pb²⁺ , whilst the nitrate ion is NO⁻₃

To balance the charges, we require two nitrate ions per lead (II) ion, and so lead (II) nitrate is Pb(NO₃)₂

The potassium ion is K ⁺ and the iodide ion is I ⁻

The two charges balance in 1:1 ratio, giving a formula of KNO₃

The symbol equation is as follows:

Pb(NO₃)₂ + KI →PbI₂ + KNO ₃

The most obvious change we must make, when balancing this equation, is to increase the number of nitrate ions on the right hand side of the equation. We can to this by placing a coefficient of 2 before the potassium nitrate:

Pb(NO₃)₂ +KI →PbI₂ +2KNO₃

In doing this we have upset the balance of potassium ions on each side of the equation.

Again, we can fix this: we must simply place another coefficient of 2, this time before the potassium iodide:

Pb(NO₃)₂ +2KI → PbI₂ +2KNO ₃

Concluding :

Checking over the equations once more, you will notice that we initially had 1 iodide ion on the right hand side, but 2 on the left. However, we already dealt with this in balancing out potassium ions. Now, our equation is balanced.

And that's it! One last thing to add is that you may have noticed the irregularity in iodide ions rather than nitrate ions. In this case, you would arrived at the same answer simply by working backwards.

Learn more about balanced equation :

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3 0

2 years ago

What is the amount, in moles, of each elemental sample? a. 11.8 g Ar b. 3.55 g Zn c. 26.1 g Ta d. 0.211 g Li

spayn [35]

Answer:

A. 0.295 mole

B. 0.055 mole

C.0.144 mole

D. 0.03 mole

Explanation:

To find the amount in moles, we simply use a mathematical relation that connects mass, atomic mass and number of moles.

Number of moles = mass/atomic mass

A. Atomic mass of Argon is 40

n = 11.8/40 = 0.295 mole

B. Atomic mass of zinc is 65

n = 3.55/65 = 0.055 mole

C. Atomic mass of Tantalum is 181

n = 26.1/181 = 0.144 mole

D. Atomic mass of lithium is 7

n = 0.211/7 = 0.03 mole

8 0

3 years ago

Consider the following equilibrium:H2CO3(aq) + H2O(l) H3O+(aq) + HCO3-1(aq).What is the correct equilibrium expression?

Vanyuwa [196]

The equilibrium expression shows the ratio between products and reactants. This expression is equal to the concentration of the products raised to its coefficient divided by the concentration of the reactants raised to its coefficient. The correct equilibrium expression for the given reaction is:<span>

<span>H2CO3(aq) + H2O(l) = H3O+(aq) + HCO3-1(aq)

Kc = [HCO3-1] [H3O+] / [H2O] [H2CO3]</span></span>

4 0

3 years ago

Each of the chemically active Period 2 elements forms stable compounds in which it has bonds to fluorine.(a) What are the names

Softa [21]

The names of the compound are lithium fluoride ( LiF) , beryllium difluoride (BeF2) , Boron trifluoride (BF3) , carbon tetrafluoride (CF4) , nitrogen trifluoride (NF3) , oxygen difluoride (OF2) .

Lithium is the first element of period 2 which reacts with fluorine to form LiF ( lithium fluoride ) . it is an inorganic compound . it is also a colorless solid . it is less soluble in water . it is chemically stable because of its comparable molecular mass .

Beryllium is the second element of period 2 which reacts with fluorine to give beryllium difluoride (BeF2) . it is inorganic compound . it is highly soluble in water. it is also a stable compound . it have low melting point .

Boron is the third element of period 2 which reacts with fluorine to form

BF3 (Boron trifluoride ) . it is a inorganic compound . it is colorless and toxic gas forms . it is stable in dry atmosphere but its octet is not satisfied .

Carbon is the 4th element of the period 2 which reacts with fluorine to form carbon tetrafluoride (CF4) . it is not soluble in water . it is a greenhouse gas . it dissolves in oil. it is very stable compound .it forms covalent bond .

Nitrogen is the 5th element of period 2 which reacts with fluorine to form nitrogen trifluoride (NF3) . it is also a inorganic compound . it is colorless and non-flammable . it is a stable gas at room temperature .

Oxygen is the 6th element of period 2 which reacts with fluorine to form oxygen difluoride (OF2) . it is colorless poisonous gas . it is partially stable or relatively stable .

Neon is a noble gas and also a stable element . it is odorless and colorless . so it is nonreactive . so it doesn't form bond with fluorine .

<h3>Learn more about Fluorine here :</h3>

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8 0

2 years ago

Consider the reaction: A <=> B. Under standard conditions at equiliubrium, the concentrations of the compounds are [A] = 1

Katen [24]

Answer:

$Keq'>1\\\Delta G'$

Explanation:

Hello,

In this case, for the given reaction, the equilibrium constant turns out:

$Keq=\frac{[B]}{[A]}=\frac{0.5M}{1.5M} =1/3$

Nonetheless, we are asked for the reverse equilibrium constant that is:

$Keq'=\frac{1}{Keq}=3$

Which is greater than one.

In such a way, the Gibbs free energy turns out:

$\Delta G'=-RTln(Keq')\\$

Now, since the reverse equilibrium constant is greater than zero its natural logarithm is positive, therefore with the initial minus, the Gibbs free energy is less than zero, that is, negative.

7 0

3 years ago

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