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aalyn [17]
3 years ago
13

How much energy does a photon of frequency 6 x 10^12 Hz have?

Chemistry
1 answer:
san4es73 [151]3 years ago
8 0

Answer:

<h2>B.3.98x10^-21 J</h2>

Explanation:

#cArryonLeArning

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Binigyan ang isang liham ng pasasalamat at gagawin ko rin siya ng kabutihan sa ibang pagkakataon​
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125 ml of nitrogen gas is collected at 70.0 degrees Celsius. The pressure
dybincka [34]

Answer: Volume of the gas at STP is 22.53 L.

Explanation:

Given : Volume = 125 mL   (as 1 mL = 0.001 L) = 0.125 L

Temperature = 70^{o}C = (70 + 273) K = 343 K

Pressure = 125 kPa = 125 kPa \times \frac{0.01 atm}{1 kPa} = 1.25 atm

According to the ideal gas equation, the volume of given nitrogen gas is calculated as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

1.25 atm \times V = 1 mol \times 0.0821 L atm/mol K \times 343 K\\V = \frac{1 mol \times 0.0821 L atm/mol K \times 343 K}{1.25 atm}\\= \frac{28.1603}{1.25} L\\= 22.53 L

Hence, volume of the gas at STP is 22.53 L.

5 0
2 years ago
Purification of copper can be achieved by electrorefining copper from an impure copper anode onto a pure copper cathode in an el
givi [52]

Answer: The amount of time needed to plate 14.0 kg of copper onto the cathode is 295 hours

Explanation:

We are given:

Moles of electron = 1 mole

According to mole concept:

1 mole of an atom contains 6.022\times 10^{23} number of particles.

We know that:

Charge on 1 electron = 1.6\times 10^{-19}C

Charge on 1 mole of electrons = 1.6\times 10^{-19}\times 6.022\times 10^{23}=96500C

Cu^{2+}+2e^-\rightarrow Cu

2\times 96500=193000C is passed to deposit = 1 mole of copper

63.5 g of copper is deposited by = 193000 C

14\times 1000g=14000g of copper is deposited by =\frac{193000}{63.5}\times 14000=42551181 C

To calculate the time required, we use the equation:

I=\frac{q}{t}

where,

I = current passed = 40.0 A

q = total charge = 42551181 C

t = time required = ?

Putting values in above equation, we get:

40.0=\frac{42551181 C}{t}\\\\t=1063779sec

Converting this into hours, we use the conversion factor:

1 hr = 3600 seconds

So, 1063779s\times \frac{1hr}{3600s}=295hr

Hence, the amount of time needed to plate 14.0 kg of copper onto the cathode is 295 hours

3 0
3 years ago
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