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2 years ago

The relationship between kinetic energy and speed is

Chemistry

1 answer:

Aleksandr-060686 [28]2 years ago

6 0

Answer:

The energy transferred is known as kinetic energy, and it depends on the mass and speed achieved.

This equation reveals that the kinetic energy of an object is directly proportional to the square of its speed.

KE = 0.5 • m • v2

where m = mass of object

v = speed of object

That means that for a twofold increase in speed, the kinetic energy will increase by a factor of four. For a threefold increase in speed, the kinetic energy will increase by a factor of nine.

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Sulfur and fluorine react in a combination reaction to produce sulfur hexafluoride:

mixer [17]

In a particular experiment, the per cent yield is 79.0%. This means that in this experiment, a 7.90-g sample of fluorine yields is 7g of SF6.

<h3>How is Sulphur hexafluoride formed?</h3>

Sulfur Hexafluoride is a disparity agent formed of an inorganic fluorinated inert gas comprised of six fluoride atoms bound to one sulfur atom, with possible diagnostic activity upon imaging.

Thus, a sample of fluorine yields 7g of SF6.

To learn more about Sulfur Hexafluoride click here;

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3 0

1 year ago

The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures

QveST [7]

Answer:

For A: The $K_p$ for the given reaction is $4.0\times 10^1$

For B: The $K_c$ for the given reaction is 1642.

Explanation:

The given chemical reaction follows:

$2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

For A:

The expression of $K_p$ for the above reaction follows:

$K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}$

We are given:

$p_{NOCl}=0.24 atm\\p_{NO}=9.10\times 10^{-2}atm=0.0910atm\\p_{Cl_2}=0.174atm$

Putting values in above equation, we get:

$K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1$

Hence, the $K_p$ for the given reaction is $4.0\times 10^1$

For B:

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = $4.0\times 10^1$

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642$

Hence, the $K_c$ for the given reaction is 1642.

7 0

2 years ago

I need to know what the answer to number 4 letter a.

QveST [7]

I believe it is C. I hope

4 0

2 years ago

Read 2 more answers

The total pressure of gases a, b, and c in a closed container is 4.1 atm. if the mixture is 36% a, 42% b, and 22% c by volume, w

Natali [406]

The partial pressure of gas C is 0.902 atm

calculation

partial pressure of gas c =[( percent by volume of gas C / total percent) x total pressure]

percent by volume of gas C= 22%

Total percent = 36% +42% + 22% = 100 %

Total pressure = 4.1 atm

partial pressure of gas C is therefore = 22/100 x 4.1 atm = 0.902 atm

3 0

3 years ago

Read 2 more answers

Calculate the entropy change for the reaction: Fe2O3(s) +3C(s) -> 2Fe(s) + 3CO(g)Entropy data:Fe2O3(s): 90 J/K molC(s): 5.7 J

Annette [7]

Explanation:

We are given: entropy of Fe2O3 = 90J/K.mol

: entropy of C = 5.7J/K.mol

: entropy of Fe = 27.2J/K.mol

: entropy of CO = 198J/K.mol

$\begin{gathered} \Delta S\text{ = S}_{products}-S_{reactants} \\ \\ \text{ = \lparen3}\times198+2\times27.2)-(3\times5.7+90) \\ \\ \text{ = 541.3J/K.mol} \end{gathered}$

Answer:

The correct answer is C.

5 0

7 months ago