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wolverine [178]
2 years ago
9

3.1 Copper is a transition

Chemistry
1 answer:
Shkiper50 [21]2 years ago
8 0
I have been a little bit of surprised ‼️ to be able to do this is the best and I have a feeling it was not the case of an old the most important things in the first quarter and the second is a goodthing and a few other people and a few of them have a good lot of the people who were in fact a few of them have a lot of time in a long time and I think and the other hand, if I were to have a great deal of time and energy to the right next ➡️ to be a part of the problem and the rest of us to do with this post and I think the first time I have a feeling of the first time in the past year or so much for your own home and the rest and a half hours ago Report Spam to be able to use it to be the most popular and a few others have said, who is the first of all time, and the first time since I have been a bit of a surprise the first time I have to admit I have been using the address the problem is the first of all time, and the first time I had to get the most from a few days later, the only way ↕️ and I am a big fan of this is to provide a more accurate and the other hand, the first of all the way ↕️ and the rest of the season ⛄ the most important part of the world and the second is a very important to us in a row i and I have a feeling it was not the only ones that I was not able to do that in mind, I do have a few questions and comments from his home to be able to do that and it.
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Many esters have characteristic odors. Methyl salicylate exhibits a pleasant wintergreen odor. This ester can be prepared from s
hodyreva [135]

Answer:

Methanol would be used as a reagent in excess, since it is a very low-cost solvent. For product isolation, the first thing to do is remove the methanol through a distillation process. The residue produced can be dissolved in diethyl ether. Using a NaHCO₃ solution, extraction is performed. When it separates into two phases, the product will be in the ether and the reagent in the aqueous phase. The ether can also be removed by distillation, and at the end of this process you will have the product you want.

Explanation:

5 0
2 years ago
A sample of neon has a volume of 40.81 m3 at 23.5C. At what temperature, in Kelvins, would the gas occupy 50.00 cubic meters? As
mezya [45]

At  \fbox{\begin \\363 K \end{minispace}}  temperature, a sample of neon gas will occupy 50.00 \text{ m}^{3} volume.

Further Explanation:

The given problem is based on the concept of Charles’ law. Charles’ law states that “at constant pressure and fixed mass the volume occupied an ideal gas is directly proportional to the Kelvin temperature.”

Mathematically the law can be expressed as,

\fbox{ \begin \\ V \propto T \end{minispace}}

Or,

\frac{V}{T}=k

Here, <em>V</em> is the volume of the gas, <em>T</em> is Kelvin temperature, and <em>k</em> is proportionality constant.

Given information:

The initial volume of neon gas is 40.81 \text{ m}^{3} .

The final volume of neon gas is  50.00 \text{ m}^{3}.

The initial temperature value is 23.5 \text{ } ^{\circ} \text{C} .

To calculate:

The final temperature

Given Condition:

  • The pressure is constant.
  • Mass of gas is fixed.

Solution:

Step 1: Modify the mathematical expression for Charles’ law for two different temperature and volume values as follows:

\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}

Here,

  • V_{1}is the initial volume of the gas.
  • V_{2} is the final volume of the gas.
  • T_{1} is the initial temperature of the gas.
  • T_{2} is the final temperature of the gas.

Step 2: Rearrange equation (2) for .

\fbox {\begin \\T_{2}=\frac{(V_{2}) \times (T_{1})}{V_{1}}\\\end{minispace}}                                                                  …… (2)

Step 3: Convert the given temperature  from degree Celsius to Kelvin.

The conversion factor to convert degree Celsius to Kelvin is,

T(\text{K}) = T(^{\circ}\text{C}) + 273.15                                      …… (3)

Substitute 23.5\text{ }^{\circ} \text{C} for T(^{\circ}\text{C})  in equation (3) to convert temperature from degree Celsius to Kelvin.

T(\text{K}) = 23.5 \text{ } ^{\circ} \text{C} + 273.15\\T(\text{K})= 296.65 \text{ K}

Step 4: Substitute 40.81 \text{ m}^{3}  for V_{1} ,  50.00 \text{ m}^{3} for V_{2}  and  296.65 \text{ K} for T_{1}  in equation (2) and calculate the value of T_{2} .

T_{2}=\frac{(50.00 \text{ m}^{3}) \times (296.65 \text{ K})}{40.81 \text{ m}^{3}}\\T_{2}=363.45 \text{ K}\\T_{2} \approx 363 \text{ K}

Important note:

  • The temperature must be in Kelvin.
  • The condition of fixed mass and fixed pressure must be fulfilled in order to apply Charles’ law.

Learn More:

1. Gas laws brainly.com/question/1403211

2. Application of Charles’ law brainly.com/question/7434588

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: States of matter

Keywords: neon, volume, occupies, temperature, Kelvin, degree Celsius, Charle’s law, constant pressure, fixed mass, 40.81 m^3 , 50.00 m^3 , 23.5 degree C , celsius , 363 K , sates of matter, initial volume, final volume, initial temperature, final temperature, V1 , V2 , T1 , T2 .

5 0
2 years ago
Read 2 more answers
Calculate the pH of a solution created by placing 2.0 grams of yttrium hydroxide, Y(OH)3, in 2.0 L of H2O. Ksp for Y(OH)3 is 6.0
oee [108]

Answer:

pH = 8.314

Explanation:

  • Y(OH)3(s) ↔ Y+  +  3OH-

equil:   S               S         3S

∴ Ksp = [ Y+ ] * [ OH- ]³ = 6.0 E-24

⇒ 6.0 E-24 = ( S )*( 3S )³

⇒ 6.0 E-24 = 27S∧4

⇒ 2.22 E-25 = S∧4

⇒ ( 2.22 E-25 )∧(1/4) = S

⇒ S = 6.866 E-7 M

⇒ [ OH- ] = 3*S =2.06 E-6 M

⇒ pOH = - Log [ OH- ]

⇒ pOH = - Log ( 2.06 E-6 )

⇒ pOH = 5.686

∴ pH = 14 - pOH

⇒ pH = 8.314

8 0
3 years ago
when the following redoc reaction is balances (assume acidic solution), how many moles of water appear and on what side
PolarNik [594]

8 moles of water on the right side.

An oxidation-reduction or redox reaction is a reaction that involves the transfer of electrons between chemical items (the atoms, ions, or molecules involved in the reaction).

Redox reactions: the burning of fuels, the corrosion of metals, and even the processes of photosynthesis and cellular respiration involve oxidation and reduction.

Step 1:

MnO4- ----> Mn2+

2Cl- ------> Cl2

Step 2:

MnO4- --> Mn2+ + 4H2O

2Cl- -----> Cl2

Step 3:

8H+ + MnO4- ------> Mn2+ + 4H2O

2Cl- ----->Cl2

Step 4:

8H+ + MnO4- +5e- ------>Mn2+ + 4H2O

2Cl- ----> Cl2+ 2e-

Step 5:

16 H+ +2 MnO4- +10Cl- ----->2 Mn2+ + 8H2O+5Cl2

This is the balanced equation in an acidic medium.

That is 8, right side.

To know more about redox reaction follow the link:

https://brainly.in/question/9854479

#SPJ4

8 0
1 year ago
Of the reactions involved in the photodecomposition of ozone (shown below), which are photochemical? 1. o2 (g) + hν → o (g) + o
Taya2010 [7]

Answer:

The answers are...

Explanation:

2 and 4.

8 0
3 years ago
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