All categories

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

You might be interested in

I need help in chemistry 10th grade (sophomore)

Goryan [66]

Answer:

1) 2

2)4

3)5

4)2

5)2

6)1

7)3

8)5

9)4

10)4

11)3

12)5

13)6

14)2

15)2

16)1

17)2

3 0

2 years ago

Read 2 more answers

Set 2 - Ec

qwelly [4]

Venus has the least amount

4 0

3 years ago

Read 2 more answers

Some velomobile seats have been designed to hold ice packs inside their cushions. If you started a ride with ice packs that held

Cerrena [4.2K]

Answer : The heat energy absorbed will be, $1.23\times 10^5cal$

Solution :

The process involved in this problem are :

$(1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(32.0^oC)$

The expression used will be:

$\Delta H=m\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]$

where,

m = mass of ice = 1100 g

$c_{p,l}$ = specific heat of liquid water = $4.18J/g^oC$

$\Delta H_{fusion}$ = enthalpy change for fusion = $6.01kJ/mole=6010J/mole=\frac{6010J/mole}{18g/mole}J/g=333.89J/g$

Molar mass of water = 18 g/mole

Now put all the given values in the above expression, we get:

$\Delta H=1100g\times 333.89J/g+[1100g\times 4.18J/g^oC\times (32.0-0)^oC]$

$\Delta H=514415J=122948.1358cal=1.23\times 10^5cal$

Conversion used : (1 cal = 4.184 J)

Therefore, the heat energy absorbed will be, $1.23\times 10^5cal$

7 0

3 years ago

C3H8 + 5O2 → 3CO2+ 4H2O, if 5.75L of oxygen are consumed in the above reaction, how many L of carbon dioxide are produced?

anzhelika [568]

Answer: 3.45 L carbon dioxide are produced

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given volume}}{\text {Molar volume}}=\frac{5.75L}{22.4L}=0.257moles$

$C_3H_8+5O_2(g)\rightarrow 3CO_2+4H_2O$

According to stoichiometry :

5 moles of $O_2$ produce = 3 moles of $CO_2$

Thus 0.257 moles of $O_2$ will produce= $\frac{3}{5}\times 0.257=0.154moles$ of $CO_2$

Volume of $CO_2=moles\times {\text {Molar volume}}=0.154moles\times 22.4L/mol=3.45$

Thus 3.45 L carbon dioxide are produced

6 0

2 years ago

Which of the following statements is true about cellular respiration?

amid [387]

It turns sugar into H2O and CO2. So I guess you could say it breaks down food to release energy.

6 0

3 years ago

Read 2 more answers