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marishachu [46]
3 years ago
7

What is the value for AG at 100 Kif AH = 27 kJ/mol and AS = 0.09 kJ/(mol-K)?

Chemistry
1 answer:
REY [17]3 years ago
8 0

Answer:

ΔG =   18KJ/mol

Explanation:

Given data:

ΔG = ?

ΔS = 0.09 Kj/mol.K

ΔH = 27 KJ/mol

Temperature  = 100 K

Solution:

Formula:

ΔG = ΔH - TΔS

ΔS = entropy

ΔH = enthalpy

by putting values,

ΔG =  27 KJ/mol - 100K(0.09 Kj/mol.K)

ΔG =  27 KJ/mol - 9 KJ/mol

ΔG =   18KJ/mol

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What system is responsible for transporting food, waste, and gases throughout the body?
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3. Do distilled water conduct electricity? What will happen if we add sugar to it and
nikitadnepr [17]

Answer:

It cannot conduct electricity, however adding salt or sugar will make the water have impurities/other substance making it easier to conduct electricity

Explanation:

Distilled water by itself does not contain impurities, thus, it cannot <em>conduct </em>electricity.

When you put salt in water, the water molecules pull the sodium and chlorine ions apart so they are floating freely, increasing the conductivity.

For more information, please refer to the internet :D

Have fun studying, and goodluck!

If you are satisfied with this answer, please rate it or give <u><em>brainliest.</em></u>

8 0
2 years ago
Read 2 more answers
If a car can go from 0 to 60 mi/hr in 8.0 seconds, what would be its final speed after 5.0 seconds if its starting speed were 50
tangare [24]

Answer:

87.5 mi/hr

Explanation:

Because a = Δv / Δt (a = vf - vi/ Δt), we need to find the acceleration first to know the change in velocity so we can determine the final velocity.

vf = 60 mi/hr

vi = 0 mi/hr

Δt = 8 secs

a = vf - vi/ Δt

= 60 mi/hr - 0 mi/hr/ 8 secs

= 60 mi/hr / 8 secs

= 7.5 mi/hr^2

Now that we know the acceleration of the car is 7. 5 mi/hr^2, we can substitute it in the acceleration formula to find the final velocity when the initial velocity is 50 mi/hr after 5 secs.

vi = 50 mi/ hr

Δt = 5 secs

a = 7.5 mi/ hr^2

a = vf - vi/ Δt

7.5 = vf - 50 mi/hr / 5 secs

37.5 = vf - 50

87.5 mi/ hr = vf

7 0
3 years ago
Determine the mass of CaCO3 required to produce 40.0 mL CO2 at STP. Hint use molar volume of an ideal gas (22.4 L)
cupoosta [38]

Answer:

m_{CaCO_3}=0.179gCaCO_3

Explanation:

Hello,

In this case, since the undergoing chemical reaction is:

CaCO_3(s)\rightarrow CaO(s)+CO_2(g)

The corresponding moles of carbon dioxide occupying 40.0 mL (0.0400 L) are computed by using the ideal gas equation at 273.15 K and 1.00 atm (STP) as follows:

PV=nRT\\\\n=\frac{PV}{RT}=\frac{1.00 atm*0.0400L}{0.082\frac{atm*L}{mol*K}*273.15 K})=1.79x10^{-3} mol CO_2

Then, since the mole ratio between carbon dioxide and calcium carbonate is 1:1 and the molar mass of the reactant is 100 g/mol, the mass that yields such volume turns out:

m_{CaCO_3}=1.79x10^{-3}molCO_2*\frac{1molCaCO_3}{1molCO_2} *\frac{100g CaCO_3}{1molCaCO_3}\\ \\m_{CaCO_3}=0.179gCaCO_3

Regards.

3 0
3 years ago
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