Ask question

Ask question

All categories

Romashka-Z-Leto [24]

3 years ago

5

Liquid sodium is being considered as an engine coolant. How many grams of liquid sodium (minimum) are needed to absorb 1.30 MJ o

f energy in the form of heat if the temperature of the sodium is not to increase by more than 10.0 °C? Use C P = 30.8 J / ( K ⋅ mol ) for Na ( l ) at 500 K

2 answers:

Sunny_sXe [5.5K]3 years ago

7 0

Answer:

97 000 g Na

Explanation:

The absortion (or liberation) of energy in form of heat is expressed by:

q=m*Cp*ΔT

The information we have:

q=1.30MJ= 1.30*10^6 J

ΔT = 10.0°C = 10.0 K (ΔT is the same in °C than in K)

Cp=30.8 J/(K mol Na)

If you notice, the Cp in the question is in relation with mol of Na. Before using the q equation, we can find the Cp in relation to the grams of Na.

To do so, we use the molar mass of Na= 22.99g/mol

$Cp= \frac{30.8J}{K*mol Na}*\frac{1 mol Na}{22.99 g Na}=1.34\frac{J}{K*g Na}$

Now, we are able to solve for m:

$m=\frac{1.30*10^6 J}{1.34\frac{J}{K*g Na} *10.0K}=\frac{1.30*10^6J}{13.4\frac{J}{g Na} } = 9.70*10^4 g Na=97 Kg Na$ =97 000 g Na

V125BC [204]3 years ago

4 0

Explanation:

Below is an attachment containing the solution.

You might be interested in

Choose all that apply. Nitrogen has an atomic number of 7. Its atomic mass is 14. Nitrogen has

Goshia [24]

The atomic number of calculated as number of protons + number of neutrons. The number of protons are the same as atomic number. This implies that Nitrogen has seven protons. To get the the number of neutron we minus atomic mass from number of protons
that is 14-7=7 neutrons

6 0

3 years ago

Write the symbol for every chemical element that has atomic number less than 75 and atomic mass greater than 173.5 u.

bearhunter [10]

Following are the chemical elements that have atomic mass greater than 173.5 u and atomic number less than 75.

1. Lutetium
Atomic Number = 71
Atomic Mass = 174.97 u

2. Hafnium
Atomic Number = 72
Atomic Mass = 178.49 u

3. Tantalum
Atomic Number = 73
Atomic Mass = 180.95 u

4. Tungsten
Atomic Number = 74
Atomic Mass = 183.85 u

8 0

3 years ago

Structural formulas show the arrangement of blank in molecules.

zlopas [31]

The blank is atoms. Structural formulas show the atomic arrangement in a molecule.

7 0

3 years ago

2. How will the equilibrium shift if the following changes are made? State if the reaction will shift

Kamila [148]

Please give me brainleist. :)

Answer:

2a. If the temperature is increased, the reaction will shift to the right in an attempt to release some of the heat. As the forward reaction loses heat while the reverse would create more heat.

2b. If the pressure is increased, it would shift to the left to counteract the increase in pressure as the left side will have fewer molecules.

2c. If Cl2 is added the reaction will shift to the left in order to remove the stress of the extra Cl2 and favor the production of more reactant.

2d. If PCl3 is removed, the reaction will shift to the right. When part of the equation is removed the reaction learns to adapt to the loss by trying to make more Pcl3 and counteract the effects of losing the PCl3.

3a. The reaction will shift to the right to produce more heat and counter the negative effects of losing the heat.

3b. It will shift to the left to get rid of the excess HCl being produced and form more reactant from the breakdown of the HCl.

3c. It would shift to the right in order to get rid of the excess form products from it.

3d. If pressure is decreased there will be no effect on the shift of the reaction because there is an even amount of moles of gas on each side.

4a. K=[N2O4(g0] / [NO2(g)]2

4b. (Below)

K=[N2O4(g)] / [NO2(g)]2

0.4 / 0.5(2)

0.4/0.25 = 1.6

Keq= 1.6

7 0

3 years ago

Substance ΔG°f(kJ/mol) M3O4(s) −8.80 M(s) 0 O2(g) 0 Consider the decomposition of a metal oxide to its elements, where M represe

baherus [9]

Answer:

The equilibrium constant is $K =0.02867$

The equilibrium pressure for oxygen gas is $P_{O_2} = 0.09367\ atm$

Explanation:

From the question we are told that

The equation of the chemical reaction is

$M_2 O_3 _{(s)} ----> 2M_{(s)} + \frac{3}{2} O_2_{(g)}$

The Gibbs free energy for $M_3 O_4_{(s)}$ is $\Delta G^o_{1} = -8.80 \ kJ/mol$

The Gibbs free energy for $M{(s)}$ is $\Delta G^o_{2} = 0 \ kJ/mol$

The Gibbs free energy for $O_2{(s)}$ is $\Delta G^o_{3} = 0 \ kJ/mol$

The Gibbs free energy of the reaction is mathematically represented as

$\Delta G^o_{re} = \sum \Delta G^o _p - \sum G^o _r$

$\Delta G^o_{re} = \sum \Delta G^o _1 - \sum( G^o _2 +G^o _3)$

Substituting values

From the balanced equation

$\Delta G^o_{re} =[ (2 * 0) + (\frac{3}{2} * 0 )] - [1 * - 8.80]$

$\Delta G^o_{re} = 8.80 kJ/mol =8800J/mol$

The Gibbs free energy of the reaction can also be represented mathematically as

$\Delta G^o_{re} = -RTln K$

Where R is the gas constant with a value of $R = 8.314 J/mol \cdot K$

T is the temperature with a given value of $T = 298 K$

K is the equilibrium constant

Now equilibrium constant for a reaction that contain gas is usually expressed in term of the partial pressure of the reactant and products that a gaseous in state

The equilibrium constant for this chemical reaction is mathematically represented as

$K_p =[ P_{O_2}]^{\frac{3}{2} }$

Where $[ P_{O_2}]$ is the equilibrium pressure of oxygen

The p subscript shows that we are obtaining the equilibrium constant using the partial pressure of gas in the reaction

Now equilibrium constant the subject on the second equation of the Gibbs free energy of the reaction

$K = e^{- \frac{\Delta G^o_{re}}{RT} }$

Substituting values

$K= e^{\frac{8800}{8.314 * 298} }$

$K =0.02867$

Now substituting this into the equation above to obtain the equilibrium of oxygen

$0.02867 = [P_{O_2}]^{[\frac{3}{2} ]}$

multiplying through by $1 ^{\frac{2}{3} }$

$P_{O_2} = [0.02867]^{\frac{2}{3} }$

$P_{O_2} = 0.09367\ atm$

3 0

3 years ago