PLEASE HELP!!!!!

Which of the following statements is true?

Rom4ik [11]

I think that its D.

6 0

2 years ago

How many grams of copper (I) chloride can be produced from the reaction of 73.5 g of copper (I) oxide with hydrochloric acid acc

natulia [17]

The balanced equation for the reaction is as follows

Cu₂O + 2HCl ---> 2CuCl + H₂O

Molar ratio of Cu₂O to CuCl is 1:2

mass of Cu₂O reacted - 73.5 g

Number of moles of Cu₂O reacted - 73.5 g / 143 g/mol = 0.51 mol

According to the molar ratio,

when 1 mol of Cu₂O reacts then 2 mol of CuCl is formed

therefore when 0.51 mol of Cu₂O reacts then - 2 x 0.51 mol of CuCl is formed

number of CuCl moles formed - 1.02 mol

mass of CuCl formed - 1.02 mol x 99 g/mol = 101 g

mass of CuCl formed is 101 g

6 0

1 year ago

According to the following reaction, how many moles of Fe(OH)2 can form from 175.0 mL of 0.227 M LiOH solution? Assume that ther

IgorC [24]

Answer:

0.020 moles of $Fe(OH)_{2}$ can be formed

Explanation:

1. First determine the number of moles of LiOH.

Molarity is given by the following expression:

$M=\frac{molesofsolute}{Litersofsolution}$

Solving for moles of solute:

moles of solute = M * Liters of solution

Converting 175.0mL to L:

$175.0mL*\frac{1L}{1000mL}=0.175L$

Replacing values:

moles of solute = 0.227M*0.175L

moles of solute = 0.040

Therefore there are 0.040 moles of LiOH

2. Then write the balanced chemical reaction and use the stoichiometry of the reaction to calculate the number of moles of $Fe(OH)_{2}$ produced:

$FeCl_{2}(aq)+2LiOH(aq)=Fe(OH)_{2}(s)+2LiCl(aq)$

As the problem says that there are excess of $FeCl_{2}$ , the limiting reagent is the LiOH.

$0.040molesLiOH*\frac{1molFe(OH)_{2}}{2molesLiOH}=0.020molesFe(OH)_{2}$ can be formed

3 0

2 years ago

Daniel Harris has always enjoyed eating tuna fish. Unfortunately, a study of the mercury content of canned tuna in 2010 found th

meriva

That would be 3 days 6 oz of tuna

4 0

2 years ago

The graph shows the distribution of energy in the particles of two gas samples at different temperatures, T1 and T2. A, B, and C

coldgirl [10]

Answer:

The average speed of gas particles at T₂ is lower than the average speed of gas particles at T₁.

Explanation:

Particles A and C are shown as if they are on the same vertical line, which means with the same kinetic energy. Both particle A and C are to the lett of particle B, which means that the formers have a lower kinetic energy than the latter.

Since the likelyhood of a particle to participate in the reaction increases with the kinetic energy, particle B is more likely to participate in the reaction than particles A and C. Hence, the first choice is incorrect.

The graph, although not perfectly symmetrical, does show a bell shape, hence there are many particles will low kinetic energy and many particles with high kinetic energy. You cannot assert that most of the particles of the two gases have high high speeds. Hence, second statement is incorrect, too.

At high values of kinetic energy (toward the right of the curve), the line labeled T₁ is higher than the line labeled T₂, meaning that at T₁ more particles have an elevated kinetic energy than the number of particles that have an elevated kinetic energy at T₂.

On the other hand, at low values of kinetic energy (toward the left of the curve) the line T₂ is higher than the line T₁, meaning that at T₂ more particles have a low kinetic energy than the number of particles that have low kinetic energy at T₁.

Hence, the last two paragraphs are telling that the average kinetic energy of gas particles at T₂ is is lower than the average kinetic energy of gas particles at T₁.

Since the average speed is proportional the the square root of the temperature, the same trend for the average kinetic energy is true for the average speed, and you conclude that the last statement is true: "The average speed of gas particles at T₂ is lower than the average speed of gas particles at T₁".

Since more particles at T₁ have high kinetic energy than the number of particles at T₂ that have a high kinetic energy, more particles of gas at T₁ are likely to participate in the reaction than the gas at T₂, and the third statement is incorrect.

7 0

1 year ago