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anzhelika [568]
3 years ago
6

Amateur radio operators in the United States can transmit on several bands. One of those bands consists of radio waves with a wa

velength near 40nm.
Calculate the frequency of these radio waves.
Chemistry
1 answer:
NeTakaya3 years ago
5 0

Answer:

7.5 × 10¹⁵ Hz

Explanation:

Given data

  • Wavelength of the radio waves (λ): 40 nm = 40 × 10⁻⁹ m = 4.0 × 10⁻⁸ m
  • Frequency of the radio waves (ν): ?
  • Speed of light (c): 3.00 × 10⁸ m/s

We can determine the frequency of the radio waves using the following expression.

c = λ × ν

ν = c/λ

ν = (3.00 × 10⁸ m/s)/4.0 × 10⁻⁸ m

ν = 7.5 × 10¹⁵ s⁻¹ = 7.5 × 10¹⁵ Hz

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A 20.0 mL of 0.27 M solution of the salt NaA has a pH of 8.70.
zlopas [31]

Answer:

a) 3.969

b) 3.489

Explanation:

a) Calculate the pk, value of the acid HA

PH of salt hydrolysis

P^{H} = 1/2 ( pkw + pka + logC )

8.7 * 2 =  14 + log ( 0.27 )  + Pka

∴ Pka = 3.9686  ≈  3.969

b) Calculate the PH of a solution containing 0.3 M HA and 0.1 M NaA

PH = Pka + log ( salt / acid )

    = 3.9686 + log ( 0.1 / 0.3 )

    =  3.9686 - 0.48  = 3.489

3 0
3 years ago
An aqueous solution of ammonia is found to be basic. This observation can be explained by the net ionic equationHNO3(aq) + H2O(l
jeka57 [31]

Answer:

NH_3_{(aq)}+H_2O_{(l)}\rightarrow NH_4^+_{(aq)}+OH^-_{(aq)}

Explanation:

Bases are the species which furnish hydroxide ions in the solution or is capable of forming bonds with electron deficient species as they are electron rich species. When a base accepts a proton, it changes into a acid which is known as its conjugate acid.

Net ionic equation : In the net ionic equations, we are not include the spectator ions in the equations.  Only the species which are present in aqueous state dissociate. So, the net ionic equation of aqueous solution of ammonia is shown below as:-

NH_3_{(aq)}+H_2O_{(l)}\rightarrow NH_4^+_{(aq)}+OH^-_{(aq)}

6 0
3 years ago
When 1.04g of cyclopropane was burnt in excess oxygen in a bomb calorimeter, the temperature rose by 3.69K. The total heat capac
STatiana [176]

Answer:

\Delta _{comb}H=-2,093\frac{kJ}{mol}

Explanation:

Hello!

In this case, since these calorimetry problems are characterized by the fact that the calorimeter absorbs the heat released by the combustion of the substance, we can write:

Q_{rxn}+Q_{cal}=0

Thus, given the temperature change and the total heat capacity, we obtain the following total heat of reaction:

Q_{rxn}=-14.01kJ/K*3.69K\\\\Q_{rxn}=-51.70kJ

Now, by dividing by the moles in 1.04 g of cyclopropane (42.09 g/mol) we obtain the enthalpy of combustion of this fuel:

n=\frac{1.04g}{42.09g/mol}=0.0247mol\\\\\Delta _{comb}H=\frac{Q_{rxn}}{n}\\\\  \Delta _{comb}H=-2,093\frac{kJ}{mol}

Best regards!

4 0
2 years ago
Calculate no. Of moles in 329g of H2so4
Anestetic [448]
Mol= mass (grams) /Mr
Mr of Sulfuric Acid (H2SO4): 98
mol= 329/98
=3.36 moles
8 0
3 years ago
Omg I’ve been stressing for the past minutes can someone please help me
Alona [7]

Answer:

I think it would be the last answer

Explanation:

5 0
3 years ago
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