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3 years ago

Can someone please please help me out ?

Chemistry

1 answer:

GarryVolchara [31]3 years ago

6 0

Answer:

4.823 x 10^-19 J

Explanation:

Energy is calculated by E = hv where h - Planck's constant in joule.s

v - frequency.

in this particular question the wave length is 4.12 x 10^-7 m. to exhaustively use this we need a relation between wave length & frequency. c=wv where C is approximately 3 x 10^8m/s

-v = c/w = 3x10^8m/s / 4.12 x 10^-7m = 7.28 x 10^14 Hz or 1/sec

now we can simply use Planck's constant in E=hv =

(6.626 x 10^-34) x (7.28 x 10^14Hz) = 4.823 x 10^-19 J.

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3 years ago

Read 2 more answers

FILL IN THE BLANKWORD BANK (can use more than once): less, increases, decreases, greaterNO2Cl(g)+NO(g)⇌NOCl(g)+NO2(g)1.Disturbin

Jlenok [28]

Explanation:

If we have the following reaction at equilibrium:

aA + bB ⇄ cC + dD

where a, b, c, and d are the stoichiometric coefficients for the reacting species A, B, C, and D. For the reaction at a particular temperature:

Kc=([C]^c *[D]^d)/([A]^a *[B]^b)

where Kc is the equilibrium constant, which holds that for a reversible reaction at equilibrium and a constant temperature, a certain ratio of reactant and product concentrations has a constant value, Kc (the equilibrium constant). Note that although the concentrations may vary, as long as the reaction in in equilibrium and temperatura don't change, the value of K remains constant.

For reactions that have not reached equilibrium, we obtain the reaction quotient (Qc), instead of the equilibrium constant by substituting the initial concentrations into the equilibrium constant expression.

Qc=([Co]^c *[Do]^d)/([Ao]^a *[Bo]^b)

To determine the direction in wich the net reaction will proceed to reach equilibrium, que compare the values of Qc and Kc.

Qc < Kc: To reach equilibrium, reactants must be converted to products (→)
Qc = Kc: The initial concentrations are equilibrium concentrations. The system in at equilibrium.
Qc > Kc: To reach equilibrium, products must be converted to reactants (←)

Solution:

We have the following reaction:

NO2Cl(g)+NO(g)⇌NOCl(g)+NO2(g)

So:

Kc=([NOCl]^1*[NO2]^1)/([NO2Cl]^1 *[NO]^1)

=([NOCl][NO2])/([NO2Cl][NO])

1. In the equation above, [NO2Cl] it's in the denominator, so if we increase it's numericall value by adding NO2Cl decreases Qc to a value less than Kc.

(From the chemical point of view, if we disturb the equilibrium adding NO2Cl (a reactant), to reach equilibrium again the system proceeds from left to right (→) consuming this reactant.)

2. To reach a new state of equilibrium (where Qc = Kc), Qc therefore increases wich means that the denominator of the expression for Qc decreases (in order to increase the denominator as mention above).

3. To accomplish this, the concentration of reageants decreases (reagents are being consumed), and the concentration of prodcuts increases (products are being formed).

4 0

3 years ago

A decorative "ice" sculpture is carved from dry ice (solid CO2) and held at its sublimation point of –78.5°C. Consider the proce

Leno4ka [110]

Answer:

The answers to the questions are;

a. The entropy of sublimation for carbon dioxide (the system) is

134.07 J/Kmol.

b. The entropy of the universe for this reversible process is 376 J/K.

Explanation:

Entropy of sublimation is the entropy change experienced following the transformation of a mole of solid to vapor at the temperature where the sublimation is taking place

a. We note that the mass of the solid CO₂ = 389 g

Molar mass of CO₂ = 44.01 g/mol

Number of moles of CO₂ in the sculpture = Mass/(Molar mass)

= (389 g)/(44.01 g/mol) = 8.84 Moles

Entropy of sublimation is given by

ΔS $_{sublimation}$ = $S_{vapor}$ - S $_{solid}$ = $\frac{\Delta H_{sublimation}}{T}$

Where:

ΔH $_{sublimation}$ = 26.1 KJ/mol

T = Temperature = –78.5°C = ‪194.65‬ K

Therefore the amount of heat required to cause the 389 g of dry ice to sublime = 26.1 KJ/mol × 8.84 Moles = 230.695 KJ

Therefore the entropy of sublimation = ΔS $_{sublimation}$ = $\frac{230.695 KJ}{194.65 K}$

= 1.185 KJ/K

= 1185 J/K = 1185/8.84 J/Kmol = 134.07 J/Kmol

b. The entropy of the universe is given by;

ΔS $_{universe}$ = $\Delta S_{system}$ + ΔS $_{surrounding}$

If the heat absorbed by the system is the same as the heat given off by the surrounding, then we have;

ΔS $_{universe}$ = $\frac{Q}{ T_{system}}$ $-\frac{Q}{T_{surrounding}}$

=1.185 KJ/K - $-\frac{230.695 KJ}{285.15K}$ = 1.185 KJ/K - 0.809 KJ/K = 0.376 KJ/K

= 376 J/K.

7 0

3 years ago

The 10x SDS gel electrophoresis buffer contains 250mM Tris HCl, 1.92M Glycine, and 1% (w/v) SDS. Buffers are always used at 1x c

olchik [2.2K]

Answer:

25 mM Tris HCl and 0.1% w/v SDS

Explanation:

A 10X solution is ten times more concentrated than a 1X solution. The stock solution is generally more concentrated (10X) and for its use, a dilution is required. Thus, to prepare a buffer 1X from a 10X buffer, you have to perform a dilution in a factor of 10 (1 volume of 10X solution is taken and mixed with 9 volumes of water). In consequence, all the concentrations of the components are diluted 10 times. To calculate the final concentration of each component in the 1X solution, we simply divide the concentration into 10:

(250 mM Tris HCl)/10 = 25 mM Tris HCl

(1.92 M glycine)/10 = 0.192 M glycine

(1% w/v SDS)/10 = 0.1% w/v SDS

Therefore the final concentrations of Tris and SDS are 25 mM and 0.1% w/v, respectively.

7 0

3 years ago