Is solid copper sulphate a conductor or insulator

An ionic bond is an _________ attraction between ions.

BabaBlast [244]

Answer:

An ionic bond is an attraction between ions of opposite charge in an ionic compound.

8 0

3 years ago

Identify the change in state that does not have an increase in entropy. a. Water evaporating b. Dry ice subliming c. Water boili

jarptica [38.1K]

Answer:

E. Water Freezing

Explanation:

Entropy refers to the degree of disorderliness of a system.

A. Water Evaporating: There is an increase in entropy, this is because the phase change is from liquid to gas. Gas particles are more disordered than liquid.

B. Dry Ice sublimating: Sublimating refers to a phase change from solid to gas. This is an increase in entropy, this is because the gas particles are more disordered than solid particles

C. Water Boiling: The phase change is from liquid to gaseous state. There is an increase in entropy. Gas particles are more disordered than liquid.

D. Ice melting: The phase change is from solid to liquid state. There is an increase in entropy. Liquid particles are more disordered than that of solid.

E. Water Freezing: The phase change is from liquid to solid state. There is a decrease in entropy. solid particles are less disordered than those of liquid.

4 0

3 years ago

Given that a 12.00 g milk chocolate bar contains 8.000 g of sugar, calculate the percentage of sugar present in 12.00 g of milk

neonofarm [45]

So in order for us to know the percentage of sugar present in a 12.00 g of milk chocolate, what we are going to do is that, we just have to divide 8 by 12 and multiply in by 100 and we get 66.67. Therefore, the percentage of sugar present in 12.00 g of milk chocolate bar is 66.67%. Hope this answers your question. Have a great day!

5 0

3 years ago

Read 2 more answers

A chemistry student needs of isopropenylbenzene for an experiment. He has available of a w/w solution of isopropenylbenzene in a

Lera25 [3.4K]

Question:

A chemistry student needs of 10 g isopropenylbenzene for an experiment. He has available 120 g of a 42.7% w/w solution of isopropenylbenzene in acetone. Calculate the mass of solution the student should use. If there's not enough solution, press the "No solution" button.

Answer:

The answer to the question is as follows

The mass of solution the student should use is 23.42 g.

Explanation:

To solve the question we note the following

A solution containing 42.7 % w/w of isopropenylbenzene in acetone has 42.7 g of isopropenylbenzene in 100 grams of the solution

Therefore we have 10 g of isopropenylbenzene contained in

100 g * 10 g/ 42.7 g = 23.42 g of solution

Available solution = 120 g

Therefore the quantity to used from the available solution = 23.42 g of the isopropenylbenzene in acetone solution.

8 0

4 years ago

The equilibrium constant Kp for the reaction (CH3),CCI (g) = (CH3),C=CH, (g) + HCl (g) is 3.45 at 500. K. (5.00 x 10K) Calculate

Karolina [17]

<u>Answer:</u> The value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

<u>Explanation:</u>

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = 3.45

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-1=1$

Putting values in above equation, we get:

$3.45=K_c\times (0.0821\times 500)^{1}\\\\K_c=\frac{3.45}{0.0821\times 500}=0.084$

The equation used to calculate concentration of a solution is:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

Initial moles of $(CH_3)_3CCl(g)$ = 1.00 mol

Volume of the flask = 5.00 L

So, $\text{Concentration of }(CH_3)_3CCl=\frac{1.00mol}{5.00L}=0.2M$

For the given chemical reaction:

$(CH_3)_3CCl(g)\rightarrow (CH_3)_2C=CH(g)+HCl(g)$

Initial: 0.2 - -

At Eqllm: 0.2 - x x x

The expression of $K_c$ for above reaction follows:

$K_c=\frac{[(CH_3)_2C=CH]\times [HCl]}{[(CH_3)_3CCl]}$

Putting values in above equation, we get:

$0.084=\frac{x\times x}{0.2-x}\\\\x^2+0.084x-0.0168=0\\\\x=0.094,-0.178$

Negative value of 'x' is neglected because initial concentration cannot be more than the given concentration

Calculating the concentration of reactants and products:

$[(CH_3)_2C=CH]=x=0.094M$

$[HCl]=x=0.094M$

$[(CH_3)_3CCl]=(0.2-x)=(0.2-0.094)=0.106M$

Hence, the value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

8 0

3 years ago