What is number 1 for b, c, and d?

concentrated phosphoric acid is90% H3PO4 by mass and the remaining mass is water. The molarity of H3PO4 is 12.2M at temperature

blondinia [14]

Answer:

82.0 mL

Explanation:

Step 1: Given data

Concentration of concentrated acid (C₁): 12.2 M

Volume of concentrated acid (V₁): ?

Concentration of dilute acid (C₂): 1.00 M

Volume of dilute acid (V₂): 1.00 L

Step 2: Calculate the required volume of the concentrated acid

We want to prepare a dilute solution from a concentrated one. We can calculate the volume of the concentrated acid using the dilution rule.

C₁ × V₁ = C₂ × V₂

V₁ = C₂ × V₂ / C₁

V₁ = 1.00 M × 1.00 L / 12.2 M = 0.0820 L = 82.0 mL

4 0

2 years ago

What unit is a wave velocity measured in

Ivahew [28]

Answer:

radians

Explanation:

5 0

3 years ago

Read 2 more answers

Be sure to answer all parts. Styrene is produced by catalytic dehydrogenation of ethylbenzene at high temperature in the presenc

svlad2 [7]

Answer:

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

8 0

3 years ago

Identify the weak diprotic acid. identify the weak diprotic acid. h2so4 hcooh

slega [8]

Among formic acid (HCOOH ) and sulfuric acid (H₂SO₄), formic acid is the weak acid. Acidic strength of any acid is the tendency of that acid to loose proton. Among these two acids formic acid has a pKa value of 3.74 greater than that of sulfuric acid i.e. -10. Remember! Greater the pKa value of acid weaker is that acid and vice versa. Below I have drawn the Ionization of both acids to corresponding conjugate bases and protons. The structures below with charges are drawn in order to explain the reason for strength. As it is seen in charged structure of formic acid, there is one positive charge on carbon next to oxygen carrying proton. The electron density is shifted toward carbon as it is electron deficient and demands more electron hence, attracting electron density from oxygen and making the oxygen hydrogen bond more polar. While, in case of sulfuric acid it is depicted that Sulfur attached to oxygen containing proton has 2+ charge, means more electron deficient as compared to carbon of formic acid, hence, more electron demanding and strongly attracting electrons from oxygen and making the oxygen hydrogen bond very polar and highly ionizable.

7 0

3 years ago

Determine the temperature change when 150. G block of gold is supplied with 1.00 • 10 ^3 J of heat

IceJOKER [234]

Answer:

∇T = 51.68°C

Explanation:

Mass = 150g

Heat Energy (Q) = 1.0*10³J

Change in temperature ∇T = ?

Q = mc∇T

Q = heat energy

M = mass

C = specific heat capacity of the gold = 0.129j/g°C

∇T = change in temperature

Q = Mc∇T

1.0*10³ = 150 * 0.129 * ∇T

1000 = 19.35∇T

Solve for ∇T

∇T = 1000 / 19.35

∇T = 51.679°C = 51.68°C

The change in temperature of gold was 51.68°C

8 0

3 years ago