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ira [324]
3 years ago
12

A photon that has an energy of 8.63 × 10^-12 J would emit light at which wavelength (pm)

Chemistry
1 answer:
ololo11 [35]3 years ago
4 0

Answer:The photon will emit light of 2.30\times 10^{-2} pm.

Explanation:

The energy emitted by the photon,E : 8.63\times 10^{-12} J

Wavelength of the light emitted by this photon = \lambda

E=\frac{hc}{\lambda }

h = Planck's constant =6.626\times 10^{-34} Js

c = speed of the light =3\times 10^{8} m/s

8.63\times 10^{-12} J=\frac{6.626\times 10^{-34} Js\times 3\times 10^{8} m/s}{\lambda }

\lambda=\frac{6.626\times 10^{-34} Js\times 3\times 10^{8} m/s}{8.63\times 10^{-12} J}=2.30\times 10^{-14} m=2.30\times 10^{-2} pm

1 m=1\times 10^{12} pm

The photon will emit light of 2.30\times 10^{-2} pm.

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Copper has two naturally occurring isotopes, 63Cu (isotopic mass 62.9296 amu) and 65Cu (isotopic mass 64.9278 amu). If copper ha
Semenov [28]

Answer:

Percentage of first isotope = 69.152 %

Percentage of second isotope =  30.848 %

Explanation:

The formula for the calculation of the average atomic mass is:

Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})

Given that:

For first isotope:

Let % = x %

Mass = 62.9296 amu

For second isotope:

% = 100 - x  %  (Since, there are only two isotopes)

Mass = 64.9278 amu

Average mass = 63.546 amu

Thus,  

63.546=\frac {x}{100}\times {62.9296}+\frac {(100-x)}{100}\times {64.9278}

Solving,

1.9982 x = 138.18

Thus,

<u>Percentage of first isotope = x = 69.152 %</u>

<u>Percentage of second isotope = 100 - x % = 30.848 %</u>

5 0
4 years ago
What rules should you follow when adding and subtracting? Check all that apply.
GaryK [48]

Answer:

The answer to this question would be:  

2. The exponents need to be the same so the leading numbers can be added or subtracted.

3. The result has the same number of decimal places as the least precise number.

The exponent is the 10^x that used in scientific notation. .When adding 12.3* 10^5 with 34.56 * 10^4, you need to make the exponent same first. In this case, it would become 12.3* 10^5 + 3.456 * 10^5= 15.756* 10^5

The decimal place could be rounded up to the least precise number. I'm not sure with the significant figures but it seems to use the biggest one.

Explanation:

brainly.com/question/7413247

5 0
3 years ago
For the neutralization reaction involving HNO3 and Ca(OH)2, how many liters of 1.55 M HNO3 are needed to react with 45.8 mL of a
liraira [26]
2HNO3 + Ca(OH)2 ==> Ca(NO3)2 + 2H2O 
<span>mols Ca(OH)2 = M x L = ? </span>
<span>Using the coefficients in the balanced equation, convert mols Ca(OH)2 to mols HNO3. </span>
<span>Then M HNO3 = mols HNO3/LHNO3. You have mols and M, solve for L.</span>
3 0
3 years ago
Read 2 more answers
3. Light acts like (1 point)
alexandr402 [8]
Both a particle and a wave?
6 0
3 years ago
In a titration, 0.01M KOH was used to neutralize 18 milliliters of 2 M HBr. What was the volume of the base used?
NikAS [45]

Answer:

3.60 ml

Explanation:

First of all we must put down the equation of the reaction. This will serve as a guide to our solution;

KOH(aq) + HBr(aq) -----> KBr(aq) + H2O(l)

The following were given in the question;

Concentration of acid CA= 2M

Volume of acid VA= 18ml

Concentration of base CB= 0.01 M

Volume of base VB= ????

Number of moles of acid NA= 1

Number of moles of base NB= 1

From;

CAVA/CBVB = NA/NB

CAVANB= CBVBNA

Therefore;

VB= CAVANB/CBNA

Substituting values;

VB= 2 × 18 ×1 / 0.01×1

VB= 3.60 ml

Therefore; 3.60 ml of base was used.

4 0
3 years ago
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