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2 years ago

A scientific law is ___.

Chemistry

1 answer:

Daniel [21]2 years ago

3 0

Answer: a rule that describes a pattern in nature.

Explanation:

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Choose all the correct information concerning chemical reactions: Involves small amounts of energy. Some of the mass is converte

suter [353]

A chemical reactions :
Involves small amounts of energy.
 Only electrons are involved.
The mass remains the same.

8 0

3 years ago

Read 2 more answers

Materials are transported within a single celled organism by the

CaHeK987 [17]

The cytoplasm transports materials within a single celled organism. that is the answer.

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3 years ago

Consider the following reaction at a high temperature. Br2(g) ⇆ 2Br(g) When 1.35 moles of Br2 are put in a 0.780−L flask, 3.60 p

UNO [17]

Answer : The equilibrium constant $K_c$ for the reaction is, 0.1133

Explanation :

First we have to calculate the concentration of $Br_2$ .

$\text{Concentration of }Br_2=\frac{\text{Moles of }Br_2}{\text{Volume of solution}}$

$\text{Concentration of }Br_2=\frac{1.35moles}{0.780L}=1.731M$

Now we have to calculate the dissociated concentration of $Br_2$ .

The balanced equilibrium reaction is,

$Br_2(g)\rightleftharpoons 2Br(aq)$

Initial conc. 1.731 M 0

At eqm. conc. (1.731-x) (2x) M

As we are given,

The percent of dissociation of $Br_2$ = $\alpha$ = 1.2 %

So, the dissociate concentration of $Br_2$ = $C\alpha=1.731M\times \frac{1.2}{100}=0.2077M$

The value of x = 0.2077 M

Now we have to calculate the concentration of $Br_2\text{ and }Br$ at equilibrium.

Concentration of $Br_2$ = 1.731 - x = 1.731 - 0.2077 = 1.5233 M

Concentration of $Br$ = 2x = 2 × 0.2077 = 0.4154 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be :

$K_c=\frac{[Br]^2}{[Br_2]}$

Now put all the values in this expression, we get :

$K_c=\frac{(0.4154)^2}{1.5233}=0.1133$

Therefore, the equilibrium constant $K_c$ for the reaction is, 0.1133

7 0

2 years ago

A voltaic cell is constructed using the following half-reactions: Ag+(aq) + e- ---> Ag(s) EoAg+ = +0.80 V Cu2+(aq) + 2e- ---&

Anton [14]

- Standard reduction potential of Ag/Ag⁺ is 0.80 v and that of Cu⁺²(aq)/Cu⁰ is +0.34 V.
- The couple with a greater value of standard reduction potential will oxidize the reduced form of the other couple.
Ag⁺ will be reduced to Ag(s) and Cu⁰ will be oxidized to Cu²⁺
Anode reaction: Cu⁰(s) → Cu²⁺ + 2 e⁻ E⁰ = +0.34 V
Cathode reaction: Ag⁺(aq) + e → Ag(s) E⁰ = +0.80 V
Cell reaction: Cu⁰(s) + 2 Ag⁺(aq) → Cu⁺²(aq) + 2 Ag⁰(s)
E⁰ cell = E⁰ cathode + E⁰ anode
= 0.80 + (-0.34) = + 0.46 V

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2 years ago

When the temperature of a reacting mixture increases, the rate of reaction increases. Which statement explains why the rate of r

wlad13 [49]

Answer:

This is one of the factors that affects chemical reactions

Temperature:This is because when the temperature is raised energy in form of heat is supplied to the reacting particles and so the rate of reaction is increased.

3 0

3 years ago