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Sav [38]
3 years ago
10

What is the distance if traveling 6.5 m/s for 25 seconds

Chemistry
1 answer:
Nesterboy [21]3 years ago
4 0

Explanation:

In this problem, we are given a conversion factor.

6.5 m=1s

So, since we know this, we can simply multiply 25 by 6.5 to see what the distance will be.

25 * 6.5 = 162.5

So, the distance traveled for 25 seconds is 162.5 miles.

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CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7
padilas [110]

<u>Answer:</u> The \Delta G for the reaction is 54.425 kJ/mol

<u>Explanation:</u>

For the given balanced chemical equation:

CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)

We are given:

\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol

To calculate \Delta G^o_{rxn} for the reaction, we use the equation:

\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]

For the given equation:

\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]

Putting values in above equation, we get:

\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J

Conversion factor used = 1 kJ = 1000 J

The expression of K_p for the given reaction:

K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}

We are given:

p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm

Putting values in above equation, we get:

K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85

To calculate the gibbs free energy of the reaction, we use the equation:

\Delta G=\Delta G^o+RT\ln K_p

where,

\Delta G = Gibbs' free energy of the reaction = ?

\Delta G^o = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = 8.314J/K mol

T = Temperature = 25^oC=[25+273]K=298K

K_p = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol

Hence, the \Delta G for the reaction is 54.425 kJ/mol

7 0
3 years ago
Represent the decomposition of aluminum oxide using the same number of atoms of molecules as are
pantera1 [17]

Answer:

Decomposition of aluminium oxide forms  aluminium atoms and  oxygen atoms.

Explanation:

<u>Decomposition reaction:</u>

When a single compound break down into two or more simpler products.

For example "AB" reactant undergoes decomposition to form "A" and "B" products.

The chemical reaction is as follows.

AB\rightarrow A+B

The given compound is aluminium oxide.

The decomposition reaction of aluminium oxide is a follows.

Al_{2}O_{3}\rightarrow Al+O_{2}

The balanced equation is as follows.

2Al_{2}O_{3}\rightarrow 4Al+3O_{2}

Therefore, Decomposition of aluminium oxide forms aluminium atoms and  oxygen atoms.

4 0
3 years ago
What is the average kinetic energy of any substance at 0 K?
Aloiza [94]
The answer is zero!

let me know if you need help with anything else! :)
5 0
3 years ago
Read 2 more answers
What does the mass of an object measure?
anyanavicka [17]

Answer:

the amount of matter it contain

5 0
3 years ago
Read 2 more answers
How many moles of NaCl are in 75.0 g of NaCl ?
LenaWriter [7]
Number of moles is found by formula n=mass/molar mass, or m/M. the molar mass is found by adding together the atomic masses of Na and Cl (22.99 + 35.45) to give 58.44 g/mol. Since the mass of NaCl is 75.0g, we find the number of moles as follows:
n = 75.0 / 58.44 = 1.28 mol
6 0
2 years ago
Read 2 more answers
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