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3 years ago

Which process is characterized by the movement of particles?.

Chemistry

2 answers:

mart [117]3 years ago

8 0

Answer:

Diffusion is the movement of particles from a high to low particle concentration, while osmosis is the movement of water from a high to a low water concentration

Explanation:

Diffusion is one principle method of movement of substances within cells, as well as the method for essential small molecules to cross the cell membrane. Gas exchange in gills and lungs operates by this process.

Serjik [45]3 years ago

6 0

Answer: Diffusion is the movement of particles from a high to low particle concentration, while osmosis is the movement of water from a high to a low water concentration.

Explanation:

You might be interested in

Calculate Kc for the reaction: 2 HI(g) ⇄ H2(g) + I2(g) given that the concentrations of each species at equilibrium are as follo

Paul [167]

Answer: The value of $K_c$ for the given reaction is 0.224

Explanation:

For the given chemical equation:

$2HI(g)\rightleftharpoons H_2(g)+I_2(g)$

The expression of $K_c$ for given equation follows:

$K_c=\frac{[H_2][I_2]}{[HI]^2}$

We are given:

$[HI]_{eq}=0.85M$

$[H_2]_{eq}=0.27M$

$[I_2]_{eq}=0.60M$

Putting values in above expression, we get:

$K_c=\frac{0.27\times 0.60}{(0.85)^2}=0.224$

Hence, the value of $K_c$ for the given reaction is 0.224

4 0

4 years ago

Suppose you mix 75 g of water at 15 ℃ with 25 g of water at 75 ℃. Predict the final temperature from the choices below. Explain

k0ka [10]

It’s probs a because 75-25 is 30 so it’s probs 30°C

7 0

3 years ago

An element with the valence electron configuration 4s1 would form a monatomic ion with a charge of_______ . In order to form thi

qwelly [4]

Answer:

+1, lose, 1, 4s, 4s and 3d

Explanation:

An element with the valence electron configuration 4s¹ would form a monatomic ion with a charge of +1. In order to form this ion, the element will lose 1 electron from the 4s subshell.

The corresponding oxidation reaction is:

K ⇒ K¹⁺ + 1 e⁻

[Ar] 4s¹ ⇒ [Ar]

If an element with the valence configuration 4s² 3d⁶ loses 3 electrons, these electrons would be removed from the 4s and 3d subshell(s).

The corresponding oxidation reaction is:

Fe ⇒ Fe³⁺ + 3 e⁻

[Ar] 4s² 3d⁶ ⇒ [Ar] 4s⁰ 3d⁵

3 0

3 years ago

Which of the following is an oxidation-reduction reaction? a. HCl(aq) + LiOH(aq) → LiCl(aq) + H2O(l) b. Pb(C2H3O2)2(aq) + 2 NaCl

saw5 [17]

Answer: Option (d) is the correct answer.

Explanation:

A reaction in which there occurs change in oxidation state of reacting species is known as an oxidation-reduction reaction.

(a) $HCl(aq) + LiOH(aq) \rightarrow LiCl(aq) + H_{2}O(l)$

Will be written as:

$H^{+} + Cl^{-} + Li^{+} + OH^{-} \rightarrow Li^{+} + Cl^{-} + H^{+} + OH^{-}$

In this reaction, there occurs no change in oxidation state of reacting species. Hence, it is not an oxidation-reduction reaction.

(b) $Pb(C_{2}H_{3}O_{2})_{2}(aq) + 2NaCl(aq) \rightarrow PbCl_{2}(s) + 2 NaC_{2}H_{3}O_{2}(aq)$

Will be written as:

$Pb^{2+} + 2C_{2}H_{3}OO^{-} + 2Na^{+} + 2Cl^{-} \rightarrow Pb^{2+} + 2Cl^{-} + 2Na^{+} + 2C2H3OO^{-}$

Similarly here, there occurs no change in oxidation state of reacting species. Hence, it is not an oxidation-reduction reaction.

Will be written as:

$Na^{+} + I^{-} + Ag^{2+} + NO^{2-}_{3} \rightarrow AgI(s) + Na^{+} + NO^{2-}_{3}(aq)$

Here, also there occurs no change in oxidation state of reacting species. Hence, it is not an oxidation-reduction reaction.

(d) $Mg(s) + 2 HCl(aq) \rightarrow MgCl_{2}(aq) + H_{2}(g)$

So, here there occurs change in oxidation state of Mg from 0 to +2 and oxidation state of H changes from +1 to 0. Hence, it is an oxidation-reduction reaction.

Thus, we can conclude that $Mg(s) + 2 HCl(aq) \rightarrow MgCl_{2}(aq) + H_{2}(g)$ is an oxidation-reduction reaction.

7 0

3 years ago

What is the percent yield if 0.3 mol ba(no3)2 and 0.25 mol na3po4 react to produce 0.095 mol ba3(po4)2?

Lyrx [107]

The chemical reaction is expressed as:

3Ba(NO3)2 + 2Na3PO4 = Ba3(PO4)2 + 6NaNO3

To determine the percent yield, we need to determine the theoretical yield of the reaction from the given amounts of the reactants. We do as follows:

0.3 mol 3Ba(NO3)2 ( 2 mol Na3PO4 / 3 mol Ba(NO3)2) = 0.2 mol Na3PO4

Therefore, the limiting reactant would be Ba(NO3)2 since it is consumed completely in the reaction.

Theoretical yield = 0.3 mol 3Ba(NO3)2 ( 1 mol Ba3(PO4)2 / 3 mol Ba(NO3)2) = 0.1 mol Ba3(PO4)2

Percent yield = actual yield / theoretical yield = 0.095 mol Ba3(PO4)2 / 0.1 mol Ba3(PO4)2 x 100 = 95%

4 0

3 years ago