All categories

3 years ago

4. Which has the longer wavelength: light with a frequency of 7.32 X 1014 Hz or light with a frequency of 6.0

Chemistry

1 answer:

Elina [12.6K]3 years ago

5 0

Answer:

6.0 x 10^14 Hertz

Explanation:

The higher the frequency of a wave, the shorter the wavelength.

The lower the frequency of a wave, the longer the wavelength.

These waves are both light waves so no other factors will affect them since their other properties such as speed will all be the same. Due to this, the wave with the the lower frequency which is 6.0x10^14 hertz, will have the longer wavelength

You might be interested in

What is the effect of a catalyst on a reaction? It increases the entropy. It decreases the entropy. It changes the equilibrium.

Leto [7]

Answer: It increases the entropy

Explanation:

Entropy is the measure of degree of randomness or disorder.

Activation energy is the extra energy that must be supplied to reactants in order to cross the energy barrier and thus convert to products.

A catalyst is a substance which increases the rate of a reaction by taking the reaction through a different path which involves lower activation energy and thus more molecules can cross the energy barrier and more molecules convert to products. Thus leading to more randomness and more entropy.

4 0

3 years ago

5.00 g of an unknown compound was dissolved in 250.0 mL of water. The unknown compound was found to be a non-electrolyte and the

Zielflug [23.3K]

Answer:

This unknown compound has a molar mass of 33.74g/mole

Explanation:

Step 1: Given data

The formula for the osmotic pressure is:

π = MRT = 14.5 atm

⇒ with M = Concentration of the solution

⇒ with R = gas constant = 0.08206 L atm / mol K

⇒ with T = absolute temperature (in Kelvin) = 298K

Step 2: Calculate concentration

M = π/ RT

M = 14.5 / (0.08206 * 298)

M = 0.593 M this is 0.593 moles per L

Step 3: Calculate number of moles

Since the volume is only 0.250 L, the amount of moles is:

0.593 * 0.250 L = 0.1482 moles

Step 4: Calculate molar mass

The molar mass is : mass / moles

Molar mass of this unknown compund = 5g / 0.1482 moles = 33.74 g/mole

This unknown compound has a molar mass of 33.74g/ mole

4 0

3 years ago

Consider the decomposition of a metal oxide to its elements, where M represents a generic metal.

VladimirAG [237]

Answer:

For 1: The standard Gibbs free energy change of the reaction is 10.60 kJ/mol

For 2: The equilibrium constant for the given reaction at 298 K is $1.386\times 10^{-2}$

For 3: The equilibrium pressure of oxygen gas is 0.0577 atm

Explanation:

For 1:

The equation used to calculate standard Gibbs free energy change of a reaction is:

$\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]$

For the given chemical reaction:

$M_2O_3(s)\rightarrow 2M(s)+\frac{3}{2}O_2(g)$

The equation for the standard Gibbs free energy change of the above reaction is:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(M(s))})+(\frac{3}{2}\times \Delta G^o_f_{(O_2(g))})]-[(1\times \Delta G^o_f_{(M_2O_3(s))})]$

We are given:

$\Delta G^o_f_{(M_2O_3(s))}=-10.60kJ/mol\\\Delta G^o_f_{(M(s))}=0kJ/mol\\\Delta G^o_f_{(O_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (0))+(\frac{3}{2}\times (0))]-[(1\times (-10.60))]\\\\\Delta G^o_{rxn}=10.60kJ/mol$

Hence, the standard Gibbs free energy change of the reaction is 10.60 kJ/mol

For 2:

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G^o=-RT\ln K_{eq}$

where,

$\Delta G^o$ = Standard Gibbs free energy = 10.60 kJ/mol = 10600 J/mol (Conversion factor: 1 kJ = 1000 J )

R = Gas constant = 8.314 J/K mol

T = temperature = 298 K

$K_{eq}$ = equilibrium constant = ?

Putting values in above equation, we get:

$10600J/mol=-(8.314J/Kmol)\times 298K\times \ln (K_{eq})\\\\K_{eq}=1.386\times 10^{-2}$

Hence, the equilibrium constant for the given reaction at 298 K is $1.386\times 10^{-2}$

For 3:

The expression of $K_{eq}$ for above equation follows:

$K_{eq}=p_{O_2}^{3/2}$

The concentration of pure solids and pure liquids are taken as 1 in the expression. That is why, the concentration of metal and metal oxide is taken as 1 in the expression.

Putting values in above expression, we get:

$1.386\times 10^{-2}=p_{O_2}^{3/2}\\\\p_{O_2}=0.0577atm$

Hence, the equilibrium pressure of oxygen gas is 0.0577 atm

6 0

3 years ago

What liquid is sour and can dissolve metal

LUCKY_DIMON [66]

Answer: In essence, Boyle defined alkalies as substances that consume, or neutralize, acids. Acids lose their characteristic sour taste and ability to dissolve metals when they are mixed with alkalies. Alkalies even reverse the change in color that occurs when litmus comes in contact with an acid.

Explanation:

8 0

3 years ago

Water left in a a freezer is called

Masteriza [31]

Water left in a freezer is called ice

6 0

3 years ago