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3 years ago

What is the value for (delta)G at 1000 K if (delta)H = -220 kJ/mol and (delta)S = -0.05 kJ/(molK)?

Chemistry

2 answers:

lesantik [10]3 years ago

8 0

Answer:

D. -170 kJ (apex)

tankabanditka [31]3 years ago

6 0

The system is isothermal, so we use the formula:
(delta)G = (delta)H - T (delta) S

Plugging in the given values:
(delta)G = -220 kJ/ mol - (1000K) (-0.05 kJ/mol K)
(delta)G = -170 kJ/mol

If we take a basis of 1 mol, the answer is
D. -170 kJ

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A 5.000 g mixture contains strontium nitrate and potassium bromide. Excess lead(II) nitrate solution is added to precipitate out

scZoUnD [109]

Answer: The mass percent of potassium bromide in the mixture is 9.996%

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For lead (II) bromide:

Given mass of lead (II) bromide = 0.7822 g

Molar mass of lead (II) bromide = 367 g/mol

Putting values in equation 1, we get:

$\text{Moles of lead (II) bromide}=\frac{0.7822g}{367g/mol}=0.0021mol$

The chemical equation for the reaction of lead (II) nitrate and potassium bromide follows:

$2KBr+Pb(NO_3)_2\rightarrow PbBr_2+2KNO_3$

By Stoichiometry of the reaction:

1 mole of lead (II) bromide is produced from 2 moles of potassium bromide

So, 0.0021 moles of lead (II) bromide will be produced from = $\frac{2}{1}\times 0.0021=0.0042mol$ of potassium bromide

Now, calculating the mass of potassium bromide by using equation 1, we get:

Molar mass of KBr = 119 g/mol

Moles of KBr = 0.0042 moles

Putting values in equation 1, we get:

$0.0042mol=\frac{\text{Mass of KBr}}{119g/mol}\\\\\text{Mass of KBr}=0.4998g$

To calculate the percentage composition of KBr in the mixture, we use the equation:

$\%\text{ composition of KBr}=\frac{\text{Mass of KBr}}{\text{Mass of mixture}}\times 100$

Mass of mixture = 5.000 g

Mass of KBr = 0.4998 g

Putting values in above equation, we get:

$\%\text{ composition of KBr}=\frac{0.4998g}{5.000g}\times 100=9.996\%$

Hence, the percent by mass of KBr in the mixture is 9.996 %

5 0

3 years ago

29.47 mL of a solution of the acid HBr is titrated, and 72.90 mL of 0.2500-M NaOH is required to reach the equivalence point. Ca

Klio2033 [76]

The original concentration of the acid solution is 6.175 $\times$ 10^-4 mol / L.

Explanation:

Concentration is the ratio of solute in a solution to either solvent or total solution. It is expressed in terms of mass per unit volume

HBr + NaOH -----> NaBr + H2O

There is a 1:1 equivalence with acid and base.

Moles of NaOH = 72.90 $\times$ 10^-3 $\times$ 0.25

= 0.0182 mol.

[ HBr ] = moles of base / volume of a solution

= 0.0182 / 29.47

= 6.175 $\times$ 10^-4 mol / L.

4 0

2 years ago

1 i i r 1 X 1 I A mm mm nu. Inchon-Mvim u m an

Setler [38]

The answer is that the pilot was tired of life and committed suicide with hundreds of passengers.

8 0

3 years ago

13. What physical property is characteristic of all of the elements in the group located in the

zimovet [89]

Answer:

B. gas state at room temperature

Explanation:

6 0

3 years ago

By how much will the water temperature increases if 1046 J of heat energy are added. The specific heat of water is 4.184 J/g • °

Brilliant_brown [7]

Answer:

Option A

250 degrees Celcius

Explanation:

If 1046J of heat energy is added to water, the water will experience a rise in temperature, at a rate that is directly proportional to its specific heat capacity.

Mathematically, this can be seen as $Q=C\Delta T$

Where C = specific heat of water = 4.184 J/g • °C.

Q = heat energy = 1046 J

$\Delta T =1046/4.148=250 degrees Celcius$

Therefore, the increase in temperature that will be experienced, is for 250 degrees Celcius

8 0

2 years ago