What is the difference between red phosphorus and white phosphorus?

Chemistry

2 answers:

RoseWind [281]2 years ago

8 0

Answer:

White phosphorusRed PhosphorusIt is insoluble in water but soluble in carbon disulphide.It is insoluble in both water and carbon disulphide.It undergoes spontaneous combustion in air.It is relatively

Explanation:

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oksano4ka [1.4K]2 years ago

3 0

Answer:

Red phosphorus and white phosphorus are two the most common and important allotropes of phosphorus. Red phosphorus is thermally more stable than white phosphorus. It is a white waxy solid. ... It turns yellow in colour in exposure to light, hence, it is also called as yellow phosphorus.

Explanation:

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HURRY I"M BEING TIMED!!!! if an unknown solution has a pH of 2. how would you classify this solution?

Sergeu [11.5K]

An acid, acids have a pH less than 7

7 0

3 years ago

You have 2.2 mol Xe and 2.0 mol F₂, but when you carry out the reaction you end up with only 0.25 mol XeF₄. What is the percent

alexira [117]

Answer:

The correct answer is 25 %

Explanation:

According to the chemical reaction:

Xe(g) + 2 F₂ (g) → XeF₄ (g)

1 mol of Xe(g) reacts with 2 mol of F₂(g), so the stoichiometric ratio os reactants is 2 mol F₂/mol Xe.

We have 2.2 mol Xe and 2.0 mol F₂, so the ratio is:

2.0 mol F₂/2.2 mol Xe = 0.909 mol F₂/mol Xe

The molar ratio of reactant we have is lower than the required, so F₂ is the limiting reactant.

By using the limiting reactant, we calculate the theoretical amount of product (XeF₄). For this, we know that 1 mol of XeF₄ is formed from 2 mol of F₂ (1 mol XeF₄/ 2 mol F₂), and we have 2.0 mol F₂:

2.0 mol F₂ x (1 mol XeF₄/ 2 mol F₂)= 1 mol XeF₄

If we only obtained 0.25 mol XeF₄, the percent yield of the experiment is:

Yield = experimental amount/theoretical amount x 100%

= 0.25 mol XeF₄/ 1 mol XeF₄ x 100% = 25 %

8 0

2 years ago

Given the equation for a system at equilibrium:

zlopas [31]

Answer: 1. $NH_3$ increases.

Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

For the given equation:

$N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)+energy$

If the concentration of $N_2$ is increased , according to the Le-Chatlier's principle, the reaction has to shift to right or forward direction. In order to do that the concentration of products has to increase.