What is the entropy of a solid based on

A helium balloon with an internal pressure of 1atm and a volume of 4.20 L at 18.0°C is released. What volume will the balloon oc

kari74 [83]

Answer:

7.59 L

Explanation:

- Use combined gas law formula and rearrange.

- Change C to K

- Hope that helped! Please let me know if you need further explanation.

3 0

2 years ago

During the combustion of 2.00 g of coal, the temperature of 500 g of water inside the calorimeter increased from 25.0°c to 43.7°

pantera1 [17]

Answer is: 39,083kJ.
m(coal) = 2,00g.
m(water) = 500g.
ΔT = 43,7°C - 25°C = 18,7°C, <span>difference at temperatures.</span>
c(water) = 4,18 J/g·°C, <span>specific heat of water
</span>Q = m(water)·ΔT·c(water), heat of reaction.
Q = 500g·18,7°C·4,18J/g·°C.
Q = 39083J = 39,083kJ.

6 0

3 years ago

what is the molecular formula for a compound with the empirical formula: K2SO4 and a molecular mass of 696g

LekaFEV [45]

Answer:

K8S4O16 or K8(SO4)4 depending on if the SO4 is supposed to represent sulfate or not

Explanation:

Find the molar mass of K2SO4 first:

2K + S + 4O ≈ 174 g/mol

Divide the goal molar mass of 696 by the molar mass of the empirical formula:

696 / 174 = 4

This means you need to multiply everything in the empirical formula by 4:

K2SO4 --> K8S4O16 or K8(SO4)4 depending on if the SO4 is for sulfate or not

4 0

2 years ago

You weigh out 0.1183 g of a complex salt to analyze for the percentage of cyanide ion in your complex salt. After dissolving the

Stels [109]

Answer:

25.35%

Explanation:

Again let me restate the the equation of the reaction;

H2O (ℓ) + 2 MnO4 - (aq) + 3 CN- (aq) → 2 MnO2 (s) + 3 CNO- (aq) + 2 OH- (aq)

Amount of potassium permanganate reacted = 10.2/1000 * 0.08035 = 8.1957 * 10^-4 moles

If 2 moles of MnO4 - reacts with 3 moles of CN-

8.1957 * 10^-4 moles of MnO4 - reacts with 8.1957 * 10^-4 * 3/2

= 1.229 * 10^-3 moles of CN-

Mass of CN- reacted = 1.229 * 10^-3 moles of CN- * 26.02 g/mol

= 0.03 g

Hence, percentage of the cyanide = 0.03 g/0.1183 g * 100

= 25.35%

8 0

2 years ago

A 10.21 mol sample of argon gas is maintained in a 0.7564 L container at 296.9 K. What is the pressure in atm calculated using t

soldi70 [24.7K]

Answer:

The pressure in atm calculated using the van der Waals' equation, is 337.2atm

Explanation:

This is the Van der Waals equation for real gases:

(P + a/v² ) ( v-b) = R .T

where P is pressure

v is Volume/mol

R is the gas constant and T, T° in K

a y b are constant for each gas, so those values are data, from the statement.

[P + 1.345 L²atm/mol² / (0.7564L/10.21mol)² ] (0.7564L/10.21mol - 3.219×10-2 L/mol ) = 0.082 L.atm/mol.K . 296.9K

[P + 1.345 L²atm/mol² / 5.48X10⁻³ L²/mol²] (0.074 L/mol - 3.219×10-2 L/mol ) = 0.082 L.atm/mol.K . 296.9K

(P + 245.05 atm) (0.04181L/mol) = 0.082 L.atm/mol.K . 296.9K

(P + 245.05 atm) (0.04181L/mol) = 24.34 L.atm/mol

0.04181L/mol .P + 10.24 L.atm/mol = 24.34 L.atm/mol

0.04181L/mol .P = 24.34 L.atm/mol - 10.24 L.atm/mol

0.04181L/mol. P = 14.1 L.atm/mol

P = 14.1 L.atm/mol / 0.04181 mol/L

P = 337.2 atm

4 0

2 years ago