# Solve for x . 875 = 5 x 3 Express the answer to the hundredths place (i.e., two digits after the decimal point).

Answer : The value of 'x' for this expression is, 5.59

Explanation :

The given expression is:

Now we have to determine the value of 'x' by solving the above expression.

Thus, the value of 'x' for this expression is, 5.59

## Related Questions

300.0 mL of a 0.335 M solution of NaI is diluted to 700.0 mL. What is the new concentration of the solution?

Answer: The new concentration of the solution is 0.143 M.

Explanation:

Given: = 300.0 mL,     = 0.335 M

= 700.0 mL,         = ?

Formula used is as follows.

Substitute values into the above formula as follows.

Thus, we can conclude that the new concentration of the solution is 0.143 M.

To find the new concentration of the solution, you can use the formula C1V1 = C2V2. Plugging in the given values, the new concentration of the solution is 0.144 M.

### Explanation:

To find the new concentration of the solution, we can use the formula:

C1V1 = C2V2

Where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.

Plugging in the given values, we get:

(0.335 M)(300.0 mL) = C2(700.0 mL)

Solving for C2, we find the new concentration of the solution to be 0.144 M.

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Be sure to answer all parts. Styrene is produced by catalytic dehydrogenation of ethylbenzene at high temperature in the presence of superheated steam. (a) Given these data, find ΔH o rxn , ΔG o rxn , and ΔS o rxn at 298 K: ΔH o f (kJ/mol) ΔG o f (kJ/mol) S o (J/mol·K) Ethylbenzene, C6H5−CH2CH3 −12.5 119.7 255 Styrene, C6H5−CH=CH2 103.8 202.5 238 ΔH o rxn ΔG o rxn ΔS o rxn kJ kJ J/K (b) At what temperature is the reaction spontaneous? °C (c) What are ΔG o rxn and K at 600.°C? ΔG o rxn K kJ/mol × 10 Enter your answer in scientific notation.

a) ΔHºrxn = 116.3 kJ, ΔGºrxn = 82.8 kJ,  ΔSºrxn =  0.113 kJ/K

b) At 753.55 ºC or higher

c )ΔG =  1.8 x 10⁴ J

K = 8.2 x 10⁻²

Explanation:

a)                                 C6H5−CH2CH3  ⇒  C6H5−CH=CH2  + H₂

ΔHf kJ/mol                    -12.5                           103.8                      0

ΔGºf kJ/K                        119.7                         202.5                      0

Sº J/mol                          255                          238                      130.6*

Note: This value was not given in our question, but is necessary and can be found in standard handbooks.

Using Hess law to calculate  ΔHºrxn we have

ΔHºrxn  = ΔHfº C6H5−CH=CH2 +  ΔHfº H₂ - ΔHºfC6H5−CH2CH3

ΔHºrxn =     103.8 kJ + 0 kJ  - (-12.5 kJ)

ΔHºrxn = 116.3 kJ

Similarly,

ΔGrxn = ΔGºf C6H5−CH=CH2 +  ΔGºfH₂ - ΔGºfC6H5CH2CH3

ΔGºrxn=   202.5 kJ + 0 kJ - 119.7 kJ  = 82.8 kJ

ΔSºrxn = 238 J/mol + 130.6 J/mol -255 J/K = 113.6 J/K = 0.113 kJ/K

b) The temperature at which the reaction is spontaneous or feasible occurs when ΔG becomes negative and using

ΔGrxn =  ΔHrxn -TΔS

we see that will happen when the term  TΔS  becomes greater than ΔHrxn since ΔS  is positive  , and so to sollve for T we will make ΔGrxn equal to zero and solve for T. Notice here we will make the assumption that  ΔºHrxn and ΔSºrxn remain constant at the higher temperature  and will equal the values previously calculated for them. Although this assumption is not entirely correct, it can be used.

0 = 116 kJ -T (0.113 kJ/K)

T = 1026.5 K  =  (1026.55 - 273 ) ºC = 753.55 ºC

c) Again we will use

ΔGrxn =  ΔHrxn -TΔS

to calculate ΔGrxn   with the assumption that ΔHº and ΔSºremain constant.

ΔG =  116.3 kJ - (600+273 K) x 0.113 kJ/K =  116.3 kJ - 873 K x 0.113 kJ/K

ΔG =  116.3 kJ - 98.6 kJ =  17.65 kJ = 1.8 x 10⁴ J ( Note the kJ are converted to J to necessary for the next part of the problem )

Now for solving for K, the equation to use is

ΔG = -RTlnK and solve for K

- ΔG / RT = lnK  ∴ K = exp (- ΔG / RT)

K = exp ( - 1.8 x 10⁴ J /( 8.314 J/K  x 873 K)) = 8.2 x 10⁻²

The change in enthalpy, entropy, and free energy were calculated for the dehydrogenation reaction of ethylbenzene into styrene. The reaction was found to be endothermic and results in a decrease in overall disorder. Under the given conditions, the reaction will never be spontaneous.

### Explanation:

The processes involved in the production of styrene from ethylbenzene are fairly complex and require knowledge of thermodynamics. We'll begin with ΔH°rxn, which is found by subtracting the enthalpy (ΔH) of the reactants from that of the products: ΔH°rxn = [ΔH°f(styrene)] - [ΔH°f(ethylbenzene)] = 103.8 kJ/mol - (-12.5 kJ/mol) = 116.3 kJ/mol. This means the reaction is endothermic, as heat is absorbed.

The change in entropy ΔS°rxn, obtained likewise, is [S°(styrene) - S°(ethylbenzene)] = (238 J/mol·K - 255 J/mol·K) = -17 J/mol·K. This indicates a decrease in disorder in the system.

With these, we can calculate the change in free energy ΔG°rxn at a given temperature (T) using the equation ΔG°rxn = ΔH°rxn - TΔS°rxn. Substituting the known values at 298 K, ΔG°rxn = 116.3 kJ/mol - (298 K)(-17 J/mol·K) = 121.2 kJ/mol, indicating a non-spontaneous reaction.

For the reaction to be spontaneous, ΔG°rxn must be less than zero. Solving for T in the above equation with ΔG°rxn = 0, yields T = ΔH°rxn / ΔS°rxn = 116.3 kJ/mol / -17 J/mol·K ≈ -6840 K. This value is negative, implying the reaction is never spontaneous under the given conditions.

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How many moles of Al are necessary to form 23.6 g of AlBr₃ from this reaction: 2 Al(s) + 3 Br₂(l) → 2 AlBr₃(s) ?

0.085 moles of Al are required to form 23.6 g of AlBr₃.

Let's consider the following balanced equation for the synthesis reaction of AlBr₃.

2 Al(s) + 3 Br₂(l) → 2 AlBr₃(s)

First, we will convert 23.6 g to moles using the molar mass of AlBr₃ (266.69 g/mol).

The molar ratio of Al to AlBr₃ is 2:2. The moles of Al required to form 0.0885  moles of AlBr₃ are:

0.085 moles of Al are required to form 23.6 g of AlBr₃.

0.088 mole of Al.

Explanation:

First, we shall determine the number of mole in 23.6 g of AlBr₃.

This is illustrated below:

Mass of AlBr₃ = 23.6 g

Molar Mass of AlBr₃ = 27 + 3(80) = 267 g/mol

Mole of AlBr₃ =.?

Mole = mass/Molar mass

Mole of AlBr₃ = 23.6 / 267

Mole of AlBr₃ = 0.088 mol

Next, we shall writing the balanced equation for the reaction.

This is given below:

2Al(s) + 3Br₂(l) → 2AlBr₃(s)

From the balanced equation above,

2 moles of Al reacted with 3 mole of Br₂ to 2 moles AlBr₃.

Finally, we shall determine the number of mole of Al needed for the reaction as follow:

From the balanced equation above,

2 moles of Al reacted to 2 moles AlBr₃.

Therefore, 0.088 mole of Al will also react to produce 0.088 mole of AlBr₃.

Which of the following is in intensive property a. mass b. magnetism c shape D. volume

Intensive properties are physical properties that do not depend on the amount or size of the material being measured. In other words, they remain constant regardless of the quantity of the substance.

The correct answer is b. magnetism.

Out of the options provided:

a. mass is an extensive property because it depends on the amount of the substance. If you have more of a substance, you will have a greater mass.

b. magnetism is an intensive property because it remains the same regardless of the size or amount of the material with the magnetic property.

c. shape is not a standard property used to classify intensive or extensive properties. It is more of a description of the object's form.

d. volume is an extensive property because it depends on the size and amount of the substance. If you have more of a substance, you will have a larger volume.

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b. Magnetism (sorry im very late)

Explanation:

Intensive properties do not depend on size, no matter what it doesn't. For example, magnetism, density, melting and boiling points, and color. All of those support intensive property.

What energy transfer happens when wood is burning?

Mechanical to Heat

explanation:

The wood itself can make mechanical energy but when it's on fire it makes heat energy

Answer: Chemical to heat and light

Explanation: The energy transforms from chemical energy to heat and light energy. Because when the candle burns a chemical reaction occurs and produces heat and light.

The radius of an atom of gold (Au) is about 1.35 Å.How many gold atoms would have to be lined up to span 5.5 mm ?

The number of gold atoms that would be needed to span this distance is 20,370.4 atoms.

### How many gold atoms would have to be lined up?

To calculate how many gold atoms would need to be lined up to span a given distance, we will us the following method.

The number of gold atoms that would be needed to span this distance:

Distance  = Diameter of a gold atom

Distance = 2 x 1.35 Å

Number of gold atoms = Total distance / Distance spanned by a single atom

Number of gold atoms = (5.5 x 10⁻⁴ cm) / (2 x 1.35 Å)

1 Å = 10⁻⁸ cm.

Number of gold atoms = (5.5 x 10⁻⁴ cm) / (2 x 1.35 x 10⁻⁸ cm)

Number of gold atoms = 20,370.4 atoms