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3 years ago

What is meant by a “ chain reaction “ ?

Chemistry

2 answers:

lilavasa [31]3 years ago

5 0

Answer:

a chemical reaction or other process in which the products themselves promote or spread the reaction, which under certain conditions may accelerate dramatically.

Explanation:

the self-sustaining fission reaction spread by neutrons which occurs in nuclear reactors and bombs.

a series of events, each caused by the previous one.

Lelechka [254]3 years ago

5 0

Answer:

When the energy released in a reaction causes new reactions to occur, it is called a chain reaction. In a chain reaction, the reaction keeps on occurring in a sequence, producing huge amounts of energy.

Explanation:

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What is heat?<br> O Fire<br> O Hot<br> O Energy<br> O Potential

Law Incorporation [45]

Answer:

Energy

Explanation:

Heat is a form of energy.

5 0

3 years ago

1. The solubility of lead(II) chloride at some high temperature is 3.1 x 10-2 M. Find the Ksp of PbCl2 at this temperature.

solniwko [45]

Answer:

1) The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

2) The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

3)The given statement is false.

Explanation:

Solubility of lead chloride = $S=3.1\times 10^-2M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

S 2S

The solubility product of the lead(II) chloride = $K_{sp}$

$K_{sp}=[Pb^{2+}][Cl^-]^2$

$K_{sp}=S\times (2S)^2=4S^3=4\times (3.1\times 10^{-2})^3=1.2\times 10^{-4}$

The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

Concentration of aluminium nitrate = 0.000010 M

Concentration of aluminum ion = $1\timed 0.000010 M=0.000010 M$

Solubility of aluminium hydroxide in aluminum nitrate solution = $S$

$Al(OH)_3(aq)\rightleftharpoons Al^{3+}(aq)+3OH^-(aq)$

S 3S

The solubility product of the aluminium nitrate = $K_{sp}=1.0\times 10^{-33}$

$K_{sp}=[Al^{3+}][OH^-]^3$

$1.0\times 10^{-33}=(0.000010+S)\times (3S)^3$

$S=1.6\times 10^{-10} M$

The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

$Molarity=\frac{Moles}{Volume (L)}$

Mass of NaCl= 3.5 mg = 0.0035 g

1 mg = 0.001 g

Moles of NaCl = $\frac{0.0035 g}{58.5 g/mol}=6.0\times 10^{-5} mol$

Volume of the solution = 0.250 L

$[NaCl]=\frac{6.0\times 10^{-5} mol}{0.250 L}=0.00024 M$

1 mole of NaCl gives 1 mole of sodium ion and 1 mole of chloride ions.

$[Cl^-]=[NaCl]=0.00024 M$

Moles of lead (II) nitrate = n

Volume of the solution = 0.250 L

Molarity lead(II) nitrate = 0.12 M

$n=0.12 M]\times 0.250 L=0.030 mol$

1 mole of lead nitrate gives 1 mole of lead (II) ion and 2 moles of nitrate ions.

$[Pb^{2+}]=[Pb(NO_2)_3]=0.030 M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

Solubility of lead(II) chloride = $K_{sp}=1.2\times 10^{-4}$

Ionic product of the lead chloride in solution :

$Q_i=[Pb^{2+}][Cl^-]^2=0.030 M\times (0.00024 M)^2=1.7\times 10^{-9}$

$Q_i$ ( no precipitation)

The given statement is false.

3 0

3 years ago

C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)

stiv31 [10]

¹/3 C3H8(g) + ⁵/3 O2(g)

Explanation:

The coefficient before every molecule is representative of the number of moles. We can represent it in ration form so as to calculate the question;

C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(l) means;

For every 1 mole of C₃H₈(g) and 5 moles of O₂(g) produces 3 moles of CO₂(g) and 4 moles of H₂O(l).

Therefore to produce 1.00 mole of CO₂(g);

We represent it in ratio;

C₃H₈(g) : CO₂(g)

1 : 3

What about ;

? (x) : 1

We cross multiply;

3x = 1 * 1

X = 1/3

We evaluate the same for O₂;

O₂(g) : CO₂(g)

5 : 3

What about

? (x) : 1

3x = 5 * 1

x = 5/3

Learn More:

For more on evaluating moles in chemical reactions check out;

brainly.com/question/13967925

brainly.com/question/13969737

#LearnWithBrainly

7 0

3 years ago

What are prevailing winds?

Tems11 [23]

Answer:

D) winds that blow in the same direction at a consistent speed

Explanation:

i took the quiz got it right so i know the answer please trust me i know this is right i promise with all my heart

7 0

3 years ago

A gas occupying a volume of 656.0 mL at a pressure of 0.884 atm is allowed to expand at constant temperature until its pressure

7nadin3 [17]

Answer:

1.14 × 10³ mL

Explanation:

Step 1: Given data

Initial volume of the gas (V₁): 656.0 mL

Initial pressure of the gas (P₁): 0.884 atm

Final volume of the gas (V₂): ?

Final pressure of the gas (P₂): 0.510 atm

Step 2: Calculate the final volume of the gas

If we assume ideal behavior, we can calculate the final volume of the gas using Boyle's law.

P₁ × V₁ = P₂ × V₂

V₂ = P₁ × V₁/P₂

V₂ = 0.884 atm × 656.0 mL/0.510 atm = 1.14 × 10³ mL

5 0

3 years ago