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3 years ago

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The energy required to split apart dichlorine according to the equation is 3.99 × 10-19 J/molecule. Cl2(g) → 2Cl(g) Calculate th

e frequency (Hz) of the photon that can dissociate dichlorine. Express answer in scientific notation.

1 answer:

Natasha2012 [34]3 years ago

5 0

Answer:

The frequency of the photon that can dissociate dichlorine is 6.02×10¹⁴ Hz

Explanation:

The energy of a photon is given by the equation:

E=h·f

E=3.99×10⁻¹⁹ J/molecule

h (Planck's constant)=6.626×10⁻³⁴ m²·kg/s

∴ f=E/h

$f=\frac{3.99*10^{-19}J}{6.626*10^{-34} m²·kg/s}$ =6.02×10¹⁴ s⁻¹= 6.02×10¹⁴ Hz

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A sample of a compound contains 60.0g C and 5.05 H it’s molar mass is 78.12g what is the compounds molecular formula

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The compound's molecular formula is C6H6

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Given the following information, what is the concentration of H2O(g) at equilibrium? [H2S](eq) = 0.671 M [O2](eq) = 0.587 M Kc =

MAVERICK [17]

<u>Answer:</u> The equilibrium concentration of water is 0.597 M

<u>Explanation:</u>

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{c}$

For a general chemical reaction:

$aA+bB\rightleftharpoons cC+dD$

The expression for $K_{eq}$ is written as:

$K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

The concentration of pure solids and pure liquids are taken as 1 in the expression.

For the given chemical reaction:

$2H_2S(g)+O_2(g)\rightleftharpoons 2S(s)+2H_2O(g)$

The expression of $K_c$ for above equation is:

$K_c=\frac{[H_2O]^2}{[H_2S]^2\times [O_2]}$

We are given:

$[H_2S]_{eq}=0.671M$

$[O_2]_{eq}=0.587M$

$K_c=1.35$

Putting values in above expression, we get:

$1.35=\frac{[H_2O]^2}{(0.671)^2\times 0.587}$

$[H_2O]=\sqrt{(1.35\times 0.671\times 0.671\times 0.587)}=0.597M$

Hence, the equilibrium concentration of water is 0.597 M

8 0

3 years ago

A sample of nitrogen gas is collected over water at temperature of 20.0˚C. What is the pressure of the nitrogen gas if atmospher

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Answer:

$P_N=0.987atm$

Explanation:

Hello there!

In this case, for these problems about collecting a gas over water, we must keep in mind that once the gas has been collected, the total pressure of the system is given by the atmospheric pressure, in this case 1.01 atm. Next, since we also have water in the mixture, we can write the following equation:

$P_T=P_w+P_N$

Thus, by solving for the pressure of nitrogen and using consistent units, we obtain:

$1.01atm=17.5torr*\frac{1atm}{760torr} +P_N\\\\P_N=1.01atm-0.023atm\\\\P_N=0.987atm$

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3 years ago

What temperature will the water reach when 10.1 g CaO is dropped into a coffee cup containing 157 g H2O at 18.0°C if the followi

Zepler [3.9K]

Answer:

Final temperature attained by water = 34.6°C

Explanation:

The reaction of CaO and H₂O is an <em>exothermic reaction</em>. The equation of reaction is given below:

CaO + H₂O ----> Ca(OH)₂

The quantity of heat given off, ΔH°rxn = 64.8KJ/mol = 64800J/mol

Number of moles of CaO = mass/molar mass, where molar mass of Ca0 = 56g/mol, mass of CaO = 10.1g

Number of moles of CaO = 10.1g/56g/mol =0.179moles

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Using the formula, <em>Quantity of heat, q = mass * specific heat capacity * temperature rise.</em>

mass of mixture = (10.1 + 157)g = 167.1g, Initial temperature = 18.0°C

Final temperature(T₂) - Initial temperature(T₁) = Temperature rise

11599.2J/mol = 167.1g * 4.18J/g·°C * ( T₂ - 18.0°C)

11599.2 = 698.478T₂ - 12572.604

11599.2 + 12572.604 = 698.478T₂

698.478T₂ = 24171.804

T₂ = 34.6°C

Therefore, final temperature attained by water = 34.6°C

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3 years ago

Many double-displacement reactions are enzyme-catalyzed via the "ping pong" mechanism, so called because the reactants appear to

zhenek [66]

Answer:

<u>D. It will decrease by a factor of 4</u>

Explanation:

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$A+B\rightarrow C+D$

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$Rate\ \alpha [A]^{a}[B]^{b}$

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According to question, <u><em>doubling the concentration of the first reactant causes the rate to increase by a factor of 2 means,</em></u>

r' = 2r if [A'] = 2[A]

Here [B] is uneffected means [B']=[B]

hence new rate =

$r'=[A']^{a}[B']^{b}$

Put the value of [A'] , [B'] and r' in the above equation:

$2r=[2A]^{a}[B]^{b}$ ...........(2)

Divide equation (1) by (2) we , get

$\frac{2r}{r}=\frac{[2A]^{2}[B]^{b}}{[A]^{a}[B]^{b}}$

$2= 2(\frac{A}{A})^{a}\times (\frac{B}{B})^{b}$

Here A and A cancel each other

B and B cancel each other

We get,

$2= 2^{a}\times 1^{b}$

1^b = 1 ( power of 1 = 1)

$2= 2^{a}$

This is possible only when a = 1

We know that : a + b = 3

1 + b = 3

b =3 -1 = 2

b = 2

Hence the rate law becomes :

$r=[A]^{a}[B]^{b}$

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Look in the question now, it is asked to calculate the concentration of [B],if cut in half

Hence

[B']=1/2[B]

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$r'=[A]^{1}[B']^{2}$

$r'=[A]^{1}(\frac{1}{2}[B])^{2}$

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But

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Compare equation (a) and (b) , we get

new rate r' =

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7 0

3 years ago