HELP ME PLEASE!!!! I NEEDZ HELP!!!

So no one go help me wit my chemistry huh

kramer

Answer:

depends...

on how difficult it is

5 0

3 years ago

Read 2 more answers

Classify each of the following reactions as either exothermic or endothermic: 2H,O(l) + heat → 2H₂(g) + O₂(g)

Murljashka [212]

Answer:

endothermic

Explanation:

It is endothermic since it absorbs heat. If the reaction RELEASED heat (exothermic) the "+ heat" would be on the right side of the equation.

6 0

1 year ago

According to which theory or law is a chemical reaction most likely to occur when two particles with the proper energy and orien

Nat2105 [25]

<h2>Collision Theory </h2>

Explanation:

<h3>The given statement is related to the collision theory - </h3>

Collision theory was given by William Lewis in 1916.

This theory explains in a qualitative manner that in what way any chemical reaction occurs and the reason for the different reaction rates for different reactions.

<h3>According to the collision theory - </h3>

Molecules must collide in order to react
Sufficient amount of energy is needed for collisions (kinetic energy) so that the chemical bonds should break
This energy used is known as the activation energy
On the increase in the temperature, the kinetic energy of the molecule increases and the molecules move faster and collide with a proper orientation at an increased speed
This increases the rate of a collision that in turn increases the breaking of the bond

6 0

2 years ago

3 mol of nitrogen gas reacts with 6 mol of hydrogen gas. how many mol of gaseous product is produced

jenyasd209 [6]

Your answer would 6 Moles

6 0

3 years ago

A mixture of 15.0 g of the anesthetic halothane (C2HBrClF3 197.4 g/mol) and 22.6 g of oxygen gas has a total pressure of 862 tor

AlexFokin [52]

Answer : The partial pressure of $C_2HBrClF_3$ and $O_2$ are, 84 torr and 778 torr respectively.

Explanation : Given,

Mass of $C_2HBrClF_3$ = 15.0 g

Mass of $O_2$ = 22.6 g

Molar mass of $C_2HBrClF_3$ = 197.4 g/mole

Molar mass of $O_2$ = 32 g/mole

First we have to calculate the moles of $C_2HBrClF_3$ and $O_2$ .

$\text{Moles of }C_2HBrClF_3=\frac{\text{Mass of }C_2HBrClF_3}{\text{Molar mass of }C_2HBrClF_3}=\frac{15.0g}{197.4g/mole}=0.0759mole$

and,

$\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{22.6g}{32g/mole}=0.706mole$

Now we have to calculate the mole fraction of $C_2HBrClF_3$ and $O_2$ .

$\text{Mole fraction of }C_2HBrClF_3=\frac{\text{Moles of }C_2HBrClF_3}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.0759}{0.0759+0.706}=0.0971$