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Luda [366]
3 years ago
5

What is the energy of a photon of violet light with a frequency of 7.57x10^14 s^-1?

Chemistry
1 answer:
Eddi Din [679]3 years ago
6 0

Answer:

ΔE = 5.02 x 10⁻¹⁹ j

Explanation:

ΔE (photon) = h·f = (6.63 x 10⁻³⁴ j·s)(7.57 x 10¹⁴ s⁻¹) = 5.02 x 10⁻¹⁹ j

h = Planck's Constant = 6.63 x 10⁻³⁴ j·s

f = frequency (given) = 7.57 x 10¹⁴ s⁻¹

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What is the correct notation for a calcium ion whose outershell is a full octet?
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8 0
3 years ago
A sample of a gas takes up 2.35L of space at room temperature (20.0ºC). What volume will the gas occupy at -5.00ºC? (Hint: Don’t
sergiy2304 [10]
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3 0
3 years ago
Did Chelate compounds increase the rate of reaction?
Viefleur [7K]

Answer:

Chelate, any of a class of coordination or complex compounds consisting of a central metal atom attached to a large molecule, called a ligand, in a cyclic or ring structure. An example of a chelate ring occurs in the ethylenediamine-cadmium complex:

The ethylenediamine ligand has two points of attachment to the cadmium ion, thus forming a ring; it is known as a didentate ligand. (Three ethylenediamine ligands can attach to the Cd2+ ion, each one forming a ring as depicted above.) Ligands that can attach to the same metal ion at two or more points are known as polydentate ligands. All polydentate ligands are chelating agents.

Chelates are more stable than nonchelated compounds of comparable composition, and the more extensive the chelation—that is, the larger the number of ring closures to a metal atom—the more stable the compound. This phenomenon is called the chelate effect; it is generally attributed to an increase in the thermodynamic quantity called entropy that accompanies chelation. The stability of a chelate is also related to the number of atoms in the chelate ring. In general, chelates containing five- or six-membered rings are more stable than chelates with four-, seven-, or eight-membered rings.

Explanation:

7 0
2 years ago
A compound contains only change and n combustion of 35.0mg of the compound produces 33.5mg co2 and 41.1mg h2o. What is the empir
Viefleur [7K]

Answer:

The empirical formula is CH6N2

Explanation:

A compound containing only C, H, and N yields the following data. Complete combustion of 35.0 mg of the compound produced 33.5 mg of CO2 and 41.1 mg of H2O. What is the empirical formula of the compound

Step 1: Data given

Mass of the compound = 35.0 mg = 0.035 grams

Mass of CO2 = 33.5 mg = 0.0335 grams

Mass of H2O = 41.1 mg = 0.0411 grams

Molar mass CO2 = 44.01 g/mol

Molar mass H2O = 18.02 g/mol

Molar mass C = 12.01 g/mol

Molar mass O = 16.0 g/mol

Molar mass H = 1.01 g/mol

Step 2: Calculate moles CO2

Moles CO2 = 0.0335 grams / 44.01 g/mol

Moles CO2 = 7.61 *10^-4 moles

Step 3: Calculate moles C

For 1 mol CO2 we have 1 mol C

For 7.61 *10^-4 moles CO2 we have 7.61 *10^-4 moles C

Step 4: Calculate mass C

Mass C = 7.61 *10^-4 moles * 12.01 g/mol

Mass C = 0.00914 grams = 9.14 mg

Step 5: Calculate moles H2O

Moles H2O = 0.0411 grams / 18.02 g/mol

Moles H2O = 0.00228 moles

Step 6: Calculate moles H

For 1 mol H2O we have 2 moles H

For 0.00228 moles H2O we have 2* 0.00228 = 0.00456 moles H

Step 7: Calculate mass H

Mass H = 0.00456 moles * 1.01 g/mol

Mass H = 0.00461 grams = 4.61 mg

Step 8: Calculate mass N

Mass N = 35.0 mg - 9.14 - 4.61 = 21.25 mg = 0.02125 grams

Step 9: Calculate moles N

Moles N = 0.02125 grams / 14.0 g/mol

Moles N = 0.00152 moles

Step 10: Calculate the mol ratio

We divide by the smallest amount of moes

C: 0.000761 moles / 0.000761 moles= 1

H:  0.00456 moles / 0.000761 moles = 6

N: 0.00152 moles  / 0.000761 moles = 2

For every C atom we have 6 H atoms and 2 N atoms

The empirical formula is CH6N2

5 0
3 years ago
For the reaction at 298 K, 2NO2(g) N2O4(g) the values of ΔH° and ΔS° are -58.03 kJ and -176.6 J/K, respectively. Calculate the v
Ierofanga [76]

Answer:

\Delta G^o=-5.4032 kJ

The temperature for \Delta G^o=0[/tex is [tex]T=328.6 K

Explanation:

The three thermodinamic properties (enthalpy, entropy and Gibbs's energy) are linked in the following formula:

\Delta G^o=\Delta H^o + T*\Delta S^o

Where:

\Delta G^o is Gibbs's energy in kJ

\Delta H^o is the enthalpy in kJ

\Delta S^o is the entropy in kJ/K

T is the temperature in K

Solving:

\Delta G^o=-58.03 kJ - 298K*-0.1766 kJ/K

\Delta G^o=-5.4032 kJ

For \Delta G^o=0:

0=\Delta H^o - T*\Delta S^o

\Delta H^o= T*\Delta S^o

T=\frac{\Delta H^o}{\Delta S^o}

T=\frac{-58.03 kJ}{-0.1766 kJ/K}

T=328.6 K

3 0
3 years ago
Read 2 more answers
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