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nlexa [21]
2 years ago
10

What is the value for AG at 5000 K if AH = -220 kJ/mol and S= -0.05 kJ/(mol-K)?

Chemistry
1 answer:
Serhud [2]2 years ago
4 0

Answer:

C. 30 kJ

Explanation:

Hello there!

In this case, in agreement to the thermodynamic definition of the Gibbs free energy, in terms of enthalpy of entropy:

\Delta G= \Delta H-T\Delta S

It is possible to calculate the required G by plugging in the given entropy and enthalpy as shown below:

\Delta G=-220kJ/mol-5000K*-0.05kJ/mol*K\\\\\Delta G=30kJ/mol

Therefore, the answer is C. 30 kJ .

Best regards!

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. If this same atom with 22 protons and 19 electrons were to gain 3 electrons, the net charge on the atom would be
mina [271]

Answer:

neutral

Explanation:

19+3=22

22 protons & 22 neutrons --> neutral net charge

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2 years ago
Describe briefly how to obtain the radial probability of an electron​
Kaylis [27]

Explanation:

The radial distribution function gives the probability density for an electron to be found anywhere on the surface of a sphere located a distance r from the proton. Since the area of a spherical surface is 4πr2, the radial distribution function is given by 4πr2R(r)∗R(r).

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8 0
2 years ago
Suppose the half-life of element k is 20 minutes. if you have a 200 g sample of k, how much of element k will be left after 60 m
vodka [1.7K]
T=20 min
m₀=200 g
t=60 min

the mass of element through time t is:
m=m₀*2^(-t/T)

m=200*2^(-60/20)=25 g

25 grams of element will be left after 60 minutes

5 0
3 years ago
Read 2 more answers
Answer the following for the reaction: 3AgNO3(aq)+Na3PO4(aq)→Ag3PO4(s)+3NaNO3(aq)
Brums [2.3K]

Answer:1) Volume of AgNO_3 required is 55.98 mL.

2) 0.62577 grams of Ag_3PO_4 is produced.

Explanation:

3AgNO_3(aq)+Na_3PO_4(aq)\rightarrow Ag_3PO_4(s)+3NaNO_3(aq)

1) Molarity of AgNO_3,M_1=0.225 M

Volume of AgNO_3.V_1=?

Molarity of Na_3PO_4,M_2=0.135 M

Volume of Na_3PO_4,V_2=31.1 mL=0.0311 L

Molarity=\frac{\text{number of moles}}{\text{volume of solution in liters}}

\text{number of moles }Na_3PO_4=M_2\times V_2=0.135 mol/L\times 0.0311 L=0.0041985 moles

According to reaction, 1 mole of Na_3PO_4 reacts with 3 mole of AgNO_3, then, 0.0041985 moles of Na_3PO_4 will react with:

\frac{3}{1}\times 0.0041985 moles of AgNO_3 that is 0.0125955 moles.

M_1=0.225 M=\frac{\text{number of moles of }AgNO_3}{V_1}

V_1=\frac{0.0125955 moles}{0.225 M}=0.05598 L=55.98 mL

Volume of AgNO_3 required is 55.98 mL.

2)

Molarity=0.195 M=\frac{\text{number of moles}}{\text{volume of solution in liters}}

Number of moles of AgNO_3=0.195\times 0.023 L=0.004485 moles

According to reaction, 3 moles of AgNO_3 gives 1 mole of Ag_3PO_4, then 0.004485 moles of AgNO_3 will give:\frac{1}{3}\times 0.004485 moles of Ag_3PO_4 that is 0.001495 moles.

Mass of Ag_3PO_4 =

Moles of Ag_3PO_4 × Molar Mass of Ag_3PO_4

= 0.001495 moles × 418.58 g/mol = 0.62577 g

0.62577 grams of Ag_3PO_4 is produced.

7 0
3 years ago
Oxidation Number of fe in fe3o4​
oee [108]

Oxidation number of fe in the compound...,

Let the oxidation number of (Fe) be x

The oxidation number of oxygen(o) is (-2)

; 3x + 4(-2) = 0...where zero is the total charge on the compound(fe3o4)

; 3x = 8...then divide both sides by 3

Oxidation number of fe is( 2.67 )

4 0
2 years ago
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