True or False: Electrons in each lower level are filled before electrons fill

The ideal gas constant, R has several different values that could be used. Which quantity causes these differences?

givi [52]

Answer:

$\boxed{\text{Pressure and volume}}$

Explanation:

The measure of R is always the same, but the numbers may differ depending on the units you use.

For example, in SI units, R = 8.314 Pa·m³K⁻¹mol

If your measurement uses different units, you must either convert your units to SI or use a value of R consistent with your units.

If you use bars and litres, R = 0.083 14 bar·L·K⁻¹mol⁻¹.

If you use kilopascals and litres, R = 8.314 kPa·L·K⁻¹mol

If you use atmospheres and litres, R = 0.082 06 L·atm·K⁻¹mol⁻¹.

If you use Torr and cubic centimetres, R = 62 368 Torr·cm³ K⁻¹mol⁻¹.

The only units that don't change are "K⁻¹mol⁻¹".

$\text{The quantities that affect the value of R are }\boxed{\textbf{pressure and volume}}$

4 0

4 years ago

I2(g) + Cl2(g)2ICl(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.62 mol

galina1969 [7]

Answer:

The change in entropy of the surrounding is -146.11 J/K.

Explanation:

Enthalpy of formation of iodine gas = $\Delta H_f_{(I_2)}=62.438 kJ/mol$

Enthalpy of formation of chlorine gas = $\Delta H_f_{(Cl_2)}=0 kJ/mol$

Enthalpy of formation of ICl gas = $\Delta H_f_{(ICl)}=17.78 kJ/mol$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

For the given chemical reaction:

$I_2(g)+Cl_2(g)\rightarrow 2ICl(g),\Delta H_{rxn}=?$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(2\times \Delta H_f_{(ICl)})]-[(1\times \Delta H_f_{(I_2)})+(1\times \Delta H_f_{(Cl_2)})]$

$=[2\times 17.78 kJ/mol]-[1\times 0 kJ/mol+1\times 62.436 kJ/mol]=-26.878 kJ/mol$

Enthaply change when 1.62 moles of iodine gas recast:

$\Delta H= \Delta H_{rxn}\times 1.62 mol=(-26.878 kJ/mol)\times 1.62 mol=-43.542 kJ$

Entropy of the surrounding = $\Delta S^o_{surr}=\frac{\Delta H}{T}$

$=\frac{-43.542 kJ}{298 K}=\frac{-43,542 J}{298 K}=-146.11 J/K$

1 kJ = 1000 J

The change in entropy of the surrounding is -146.11 J/K.

4 0

3 years ago

1. If a person's mass is 56 kg, what is this weight in pounds?<br> 56kg

Natali [406]

Answer:

123lb

Explanation:

lb= 2.2046 kg

56×2.2046= 123lb

3 0

3 years ago

What is formed when oxygen and hydrogen unite chemically

gavmur [86]

A. An element is the answer

6 0

4 years ago

I have a 270 g cup of coffee at 94.1 degrees Celsius. Assume that the specific heat of coffee is the same as pure water, 4.18 J/

natulia [17]

Answer:

90.64 degrees

Explanation:

Lets list out the parameters given in the question.

Mass of Cup = 270g

Temp. of cup = 94.1C

Specific heat capacity of C = 4.18J/gC

Temp. of Silver spoon = 26.8C

heat capacity of the silver spoon = 61.2 J/C

The fact that thermal equilibrium was mentioned simply means that;

Heat Lost by coffee= Heat gained by spoon

The relationship between the parameters mentioned is given as;

Q = m* c * dT

Where;

m is the mass, c is the specific heat capacity and dT is the "change" in temperature.

Assume the final temperature of the whole system is "x". They would both have the same final temperature since they are in thermal equilibrium.

Heat lost by coffee = 270 * 4.18 * (94.1 - x)

(94.1 -x) because coffee will loose heat, hence temperature of coffee will fall.

Heat gained by spoon = c *dT = 61.2 * (x - 26.8).

dT = Final Temp. - Initial Temp.

dT = x - 26.8

Heat lost = Heat gained :

270 * 4.18 * (94.1 - x) = 61.2 * (x - 26.8),

1128.6 (94.1 - x) = 61.2 (x- 26.8)

18.44 ( 94.1 - x) = x - 26.8

1735.2 - 18.44 x = x - 26.8

1735.2 + 26.8 = x + 18.44 x

1762 = 19.44 x

x = 90.64 degrees

8 0

3 years ago