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3 years ago

What is the wavelength of an X-ray that has a frequency of 7.8 x 1017 Hz

Chemistry

1 answer:

SVEN [57.7K]3 years ago

3 0

Answer:

λ = 0.38 ×10⁻⁹ m

Explanation:

Given data:

Wavelength of xray = ?

Frequency of xray = 7.8 ×10¹⁷ Hz

Solution:

Formula:

Speed of light = wavelength × frequency

speed of light = 3×10⁸ m/s

Now we will put the values in formula.

3×10⁸ m/s = λ × 7.8 ×10¹⁷ Hz

λ = 3×10⁸ m/s / 7.8 ×10¹⁷ Hz

Hz = s⁻¹

λ = 3×10⁸ m/s / 7.8 ×10¹⁷s⁻¹

λ = 0.38 ×10⁻⁹ m

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Anon25 [30]

Answer : The concentration in (M)of bromide ions in a saturated solution of mercury (II) bromide is, $2.7\times 10^{-7}M$

Explanation :

The solubility equilibrium reaction will be:

$HgBr_2\rightleftharpoons Hg^{2+}+2Br^{-}$

Let the molar solubility be 's'.

The expression for solubility constant for this reaction will be,

$K_{sp}=[Hg^{2+}][Br^{-}]^2$

$K_{sp}=(s)\times (2s)^2$

$K_{sp}=4s^3$

Given:

$K_{sp}$ = $8.0\times 10^{-20}$

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

$K_{sp}=4s^3$

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2 years ago

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DiKsa [7]

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2 years ago

For the reaction Fe3O4(s) + 4H2(g) --> 3Fe(s) + 4H2O(g)

mojhsa [17]

Answer : The value of equilibrium constant for this reaction at 328.0 K is $1.70\times 10^{15}$

Explanation :

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

$\Delta H^o$ = standard enthalpy = 151.2 kJ = 151200 J

$\Delta S^o$ = standard entropy = 169.4 J/K

T = temperature of reaction = 328.0 K

Now put all the given values in the above formula, we get:

$\Delta G^o=(151200J)-(328.0K\times 169.4J/K)$

$\Delta G^o=95636.8J=95.6kJ$

The relation between the equilibrium constant and standard Gibbs free energy is:

$\Delta G^o=-RT\times \ln k$

where,

$\Delta G^o$ = standard Gibbs free energy = 95636.8 J

R = gas constant = 8.314 J/K.mol

T = temperature = 328.0 K

K = equilibrium constant = ?

Now put all the given values in the above formula, we get:

$95636.8J=-(8.314J/K.mol)\times (328.0K)\times \ln k$

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Therefore, the value of equilibrium constant for this reaction at 328.0 K is $1.70\times 10^{15}$

3 0

3 years ago

What is the partial pressure of argon, par, in the flask? express your answer to three significant figures and include the appro

monitta

Given which are missing in your question:
the flask is filled with 1.45 g of argon at 25 C°
So according to this formula (Partial pressure):
PV= nRT
first, we need n, and we can get by substitution by:
n = 1.45/mass weight of argon
= 1.45 / 39.948 = 0.0363 mol of Ar
we have R constant = 0.0821
and T in kelvin = 25 + 273 = 298
and V = 1 L
∴ P * 1 = 0.0363* 0.0821 * 298 = 0.888 atm

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3 years ago