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katovenus [111]
3 years ago
5

A chemist is using radiation with a frequency of 6.0 x 10^13 Hz

Chemistry
1 answer:
kirza4 [7]3 years ago
4 0

Answer:

λ = 0.5×10⁻⁵ m

E = 39.78×10⁻²¹ J

The given radiation is gamma ray.

Explanation:

Given data:

Frequency of radiation = 6.0 ×10¹³ Hz

Wavelength of radiation = ?

Type of radiation = ?

Energy of radiation = ?

Solution:

Formula:

speed of light = wavelength × frequency

3 ×10⁸ m/s = λ × 6.0 ×10¹³ Hz

Hz = s⁻¹

λ = 3 ×10⁸ m/s /  6.0 ×10¹³  s⁻¹

λ = 0.5×10⁻⁵ m

Energy of radiation:

E = hf

h = planck's constant

f = frequency

E = 6.63×10⁻³⁴ j.s × 6.0 ×10¹³  s⁻¹

E = 39.78×10⁻²¹ J

The given radiation is gamma ray.

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Calculate the molecular mass of the element
Elanso [62]

Answer:

Cu3 (PO4)2

3×64 + 2×(31+ 4×16)

192 + 2×(31+64)

192 + 2×(95)

192+190

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4 0
2 years ago
Without doing any calculations, match the following thermodynamic properties with their appropriate numerical sign for the follo
Mrac [35]

<u>Answer:</u> \Delta H of the reaction will be negative, \Delta S of the reaction will be positive and \Delta G of the reaction will be negative.

<u>Explanation:</u>

Thermodynamic properties are enthalpy change (\Delta H), entropy change (\Delta S) and Gibbs free energy(\Delta G)

Exothermic reactions are defined as the reactions in which energy is released in the form of heat. The enthalpy change (\Delta H) of the reaction comes out to be negative for this kind of reaction.

Entropy change is defined as the change in the measure of randomness in the reaction. It is represented as (\Delta S). Randomness of gaseous particles is more than that of liquid which is further more than that of solids.

For the given exothermic reaction:

2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)

As, number of gaseous particles on the product side is more than the number of gaseous particles on the reactant side. So, the entropy change is positive. Hence, \Delta S is positive.

The above reaction is spontaneous. Thus, the Gibbs free energy will be negative.

For the given reaction:

  • \Delta H = -ve
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6 0
3 years ago
When H2SO4 is added to PbI2, a precipitate of PbSO4 forms. The PbSO4 is then filtered from the solution, dried, and weighed. If
Nookie1986 [14]

Answer:

n_{I^-}=3.11x10^{-4}molI^-

Explanation:

Hello,

In this case, the undergoing chemical reaction is shown below:

H_2SO_4(aq)+PbI_2(aq)\rightarrow PbSO_4(s)+2HI(aq)

Thus, we obtain 0.0471 g of lead (II) sulfate, the iodide ions  in the original solution are computed below by stoichiometry, taking into account that into 1 mole of lead (II) iodide there are 2 moles of iodide ions:

n_{I^-}=0.0471gPbSO_4*\frac{1molPbSO_4}{303.26gPbSO_4}*\frac{1molPbI_2}{1molPbSO_4}*\frac{2molI^-}{1molPbI_2} \\n_{I^-}=3.11x10^{-4}molI^-

Best regards.

3 0
3 years ago
SKJIEDHISJLKDSKSJKA HELPP
Vinil7 [7]

Answer:

agree

disagree

agree

hope it help u

6 0
2 years ago
Which are the intermolecular forces that can act between non-polar molecules?
Licemer1 [7]
The answer is london dispersion forces. It is one of two intermolecular forces that is present in all molecules! 


8 0
2 years ago
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