CHEMISTRY HIGH SCHOOL

Pick the answer that most likely represents the substances’ relative solubilities in water.(A) CH₃CH₂CH₂CH₃ < CH₃CH₂CH₂OH < HOCH₂CH₂OH
(B) CH₃CH₂CH₂OH < CH₃CH₂CH₂CH₃ < HOCH₂CH₂OH
(C) CH₃CH₂CH₂CH₃ < HOCH₂CH₂OH < CH₃CH₂CH₂OH
(D) HOCH₂CH₂OH < CH₃CH₂CH₂OH < CH₃CH₂CH₂CH₃
(E) HOCH₂CH₂OH < CH₃CH₂CH₂OH < CH₃CH₂CH₂CH₃

Answers

Answer 1

Answer: the naswer should be A

Answer 2

Answer:

Answer:

Explanation:

eighty-three

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What is the mass/volume percent of a solution with 75g of Na2CO3 dissolved in 20.0 g of water?11%

9.3 %

30%

7.5%

Answers

Answer:

percent by mass = 79%

Explanation:

Data Given:

mass of solute (Na₂CO₃) = 75 g

mass of solvent ( water ) = 20 g

mass percent of solution = ?

Solution:

Formula used to find percent by mass

mass of solute / mass of solution x 100 . . . . . . (1)

first we have to find mass of solution

mass of solution = mass of solute + mass of Solvent . . . . (2)

put values in equation 2

mass of solution = 75 g + 20 g

mass of solution = 95 g

Now,

put values in equation 1

Percent by mass = mass of solute / mass of solution x 100

Percent by mass = 75 g / 95 g x 100

Percent by mass = 79 %

so,

The percent of the solution by mass = 79%

***Note:

The same work can be used for the other values too i.e. if you have 20 g of solute and 75 g solvent.

Answer:

The answer is 78.9%

Explanation:

The percentage weight in volume indicates the mass of solute per volume. We will use the% weight/volume formula to calculate what the exercise asks for, as follows: $%weight/volume=(solute weight)/(solution volume)x100%= (75 g)/(75+20)= 78.9%$

$percent Weight/volume=(solute weight)/(solution volume)x100=(75 g)/(75+20)x100=78.9$ %

Consider a generic redox reaction?Consider a generic redox reaction X(s) + Y^+(aq) <----> X^+(aq) + Y(s) then suppose that the coefficients in the redox reaction are doubled. How will the following quantities be affected for a voltaic cell under nonstandard conditions? 1.Doubled 2. Halved 3. Squared 4. No Change The options are InQ, E^o, E, n, Q

Answers

The Nernst equation allows us to predict the cell potential for voltaic cells under conditions other than the standard conditions of 1M, 1 atm, 25°C. The effects of different temperatures and concentrations may be tracked in terms of the Gibbs energy change ΔG. This free energy change depends upon the temperature & concentrations according to ΔG = ΔG° + RTInQ where ΔG° is the free energy change under conditions and Q is the thermodynamic reaction quotient. The free energy change is related to the cell potential Ecell by ΔG= nFEcell

so for non-standard conditions
-nFEcell = -nFE°cell + RT InQ

or

Ecell = E°cell - RT/nF (InQ)

which is called Nernst equation.

Answer:

n and lnQ - doubled

Q - Squared

E, Edegree - No change

How many electron cloud energy levels are present in aluminum?

Answers

i think the answer is three

A sample of salt (sodium chloride) is placed on the tip of a platinum wire. When the sample is heated on blue flame the flame turned yellow. Explain how this can be done to electrons in sodium atoms

Answers

So platinum is a transition metal. In general transition metals are reducers, which means they can give the electrons they have, to the sodium atoms. Also in chemistry we look at sub orbitals rather that shells(2,8,8). So due to the energy from heat, the d orbital split as electrons move to a higher energy level. Some of the electrons are given to the sodium ions and therefore the flame changes colour to yellow.
The excitation of the electrons is caused by them getting energy and so moving up an energy level. This energy is released and the electron returns to it's original state. The energy released, however, does not release in the same direction, but in different/various directions. Therefore the colour of the light changes as some energy is released in the surrounding.

The molar mass of calcium is 40.08, oxygen is 16.0, and hydrogen is 1.01. What is the mass of one mole of calcium hydroxide (Ca(OH)2)?

Answers

$Ca(OH)_2$

$MM=Molar~mass$

$MM_(Ca(OH)_2)=MM_(Ca)+2*(MM_O+MM_H)$

$MM_(Ca(OH)_2)=40.08+2*(16.0+1.01)$

$MM_(Ca(OH)_2)=40.08+2*(17.01)$

$MM_(Ca(OH)_2)=40.08+34.02$

$\boxed{\boxed{MM_(Ca(OH)_2)=74.1~g/mol}}$

What is the total mass of solute in 1000. grams of a solution having a concentration of 5 parts per million?(1) 0.005 g (3) 0.5 g
(2) 0.05 g (4) 5 g