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3 years ago

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]

Answer: The $\Delta G$ for the reaction is 54.425 kJ/mol

Explanation:

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:

Decomposition reaction:

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

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What does the mass of an object measure?

anyanavicka [17]

Answer:

the amount of matter it contain

5 0

3 years ago

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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

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