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2 years ago

Gasoline in a storage tank potential or kinetic?

Chemistry

2 answers:

cestrela7 [59]2 years ago

8 0

Potential because the gas is not yet in use or using energy

Xelga [282]2 years ago

7 0

Potential i believe.

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Chlorine gas (cl2) forms when two chlorine atoms share an electron. what type of bonding is present in chlorine gas?

Masja [62]

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3 years ago

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QUESTION 6 Consider the following reaction between the diatomic and monatomic forms of iodine: I2 (g) <-> 2I (g) When 0.09

aalyn [17]

Answer: The equilibrium constant is $3.3\times 10^{-4}$

Explanation:

Initial concentration of $I_2$ = 0.095 M

The given balanced equilibrium reaction is,

$I_2(g)\rightleftharpoons 2I(g)$

Initial conc. 0.095 M 0 M

At eqm. conc. (0.095-x) M (2x) M

Given : 2x = 0.0055

x = 0.00275

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[l]^2}{[I_2]}$

Now put all the given values in this expression, we get :

$K_c=\frac{(0.0055)^2}{(0.095-0.00275)}$

$K_c=\frac{(0.0055)^2}{0.09225}=0.00033$

Thus the equilibrium constant is $3.3\times 10^{-4}$

3 0

2 years ago

Science fair ideas? please help

egoroff_w [7]

You can Maybe do a volcano?

8 0

3 years ago

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A student dissolved 1.805g of a monoacidic weak base in 55mL of water. Calculate the equilibrium pH for the weak monoacidic base

yawa3891 [41]

Answer:

11.39

Explanation:

Given that:

$pK_{b}=4.82$

$K_{b}=10^{-4.82}=1.5136\times 10^{-5}$

Given that:

Mass = 1.805 g

Molar mass = 82.0343 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{1.805\ g}{82.0343\ g/mol}$

$Moles= 0.022\ moles$

Given Volume = 55 mL = 0.055 L ( 1 mL = 0.001 L)

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity=\frac{0.022}{0.055}$

Concentration = 0.4 M

Consider the ICE take for the dissociation of the base as:

B + H₂O ⇄ BH⁺ + OH⁻

At t=0 0.4 - -

At t =equilibrium (0.4-x) x x

The expression for dissociation constant is:

$K_{b}=\frac {\left [ BH^{+} \right ]\left [ {OH}^- \right ]}{[B]}$

$1.5136\times 10^{-5}=\frac {x^2}{0.4-x}$

x is very small, so (0.4 - x) ≅ 0.4

Solving for x, we get:

x = 2.4606×10⁻³ M

pOH = -log[OH⁻] = -log(2.4606×10⁻³) = 2.61

5 0

2 years ago

Write the correct chemical formula for potassium and sulfur

Shalnov [3]

Phosphorus + Sulfur ------> Phosphorus sulfide

2P + 3S ------> P2S3

Hope it helped!

3 0

3 years ago