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4 years ago

Determine the energy of a photon with a wavelength of 44

Chemistry

1 answer:

nika2105 [10]4 years ago

6 0

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Each student in a class placed a 2.00 g sample of a mixture of Cu and Al in a beaker and placed the beaker in a fume hood. The s

Aleks [24]

Answer:

Percentage mass of copper in the sample = 32%

Explanation:

Equation of the reaction producing Cu(NO₃) is given below:

Cu(s)+ 4HNO₃(aq) ---> Cu(NO₃)(aq) + 2NO₂(g) + 2H₂O(l)

From the equation of reaction, 1 mole of Cu(NO₃) is produced from 1 mole of copper. Therefore, 0.010 moles of Cu(NO₃) will be produced from 0.010 mole of copper.

Molar mass of copper = 64 g/mol

mass of copper = number of moles * molar mass

mass of copper = 0.01 mol * 64 g/mol = 0.64 g

Percentage by mass of copper in the 2.00 g sample = (0.64/2.00) * 100%

Percentage mass of copper in the sample = 32%

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3 years ago

When is the Sun directly overhead in the continental United States?

astra-53 [7]

Answer:

Explanation:

The sun is directly overhead at noon on the equator on the first day of spring, and on the first day of fall. You would have to be less than 23.5 degrees above or below the equator to have the Sun pass directly overhead. Therefore, it never occurs in the continental US.

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3 years ago

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Which element has the highest 2nd ionization energy? <br> A. Be <br> B. C<br> C. Li<br> D. B

dlinn [17]

Answer:

The answer is c.........

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3 years ago

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Potential energy is based on an objects

erma4kov [3.2K]

This is some information about the potential energy.

Explanation:

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3 0

3 years ago

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The formation of ammonia is represented by the equation N2(g) + 3H2(g) ⇌ 2NH3(g). Determine the enthalpy of formation of ammonia

Savatey [412]

Answer:

$\Delta _fH_{NH_3}=-66\frac{kJ}{mol}$

Explanation:

Hello!

In this case, since the study of the bond energy allows us to compute the enthalpies of some reactions, for this combination reaction by which ammonia is yielded, we understand the enthalpy of reaction equals the enthalpy of formation of ammonia, and, in terms of the bonds energy we can write:

$\Delta _fH_{NH_3}=Delta _rH=\Sigma \Delta H(bonds \ broken)-\Sigma \Delta H(bonds \ formed)$

Whereas the bonds enthalpy of those bonds that get broken cover the N≡N and the three H-H bonds at the reactants side and the enthalpy of those bonds that are formed cover the six N-H bonds at the products; which means we obtain:

$\Delta _fH_{NH_3}=942\frac{kJ}{mol} +3*436\frac{kJ}{mol}-6*386\frac{kJ}{mol}\\\\\Delta _fH_{NH_3}=-66\frac{kJ}{mol}$

Which differs from the theoretical value that is -46 kJ/mol.

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7 0

3 years ago

Determine the energy of a photon with a wavelength of 44​

Determine the energy of a photon with a wavelength of 44