Calculate the mole fraction of methane in the inital mixture assuming the temperature and volume remain constant

What is the energy of a photon that emits a light of frequency 6.42 x 1014 Hz?<br><br>

svetlana [45]

Answer:

Option B. 4.25×10¯¹⁹ J

Explanation:

From the question given above, the following data were obtained:

Frequency (f) = 6.42×10¹⁴ Hz

Energy (E) =?

Energy and frequency are related by the following equation:

Energy (E) = Planck's constant (h) × frequency (f)

E = hf

With the above formula, we can obtain the energy of the photon as follow:

Frequency (f) = 6.42×10¹⁴ Hz

Planck's constant (h) = 6.63×10¯³⁴ Js

Energy (E) =?

E = hf

E = 6.63×10¯³⁴ × 6.42×10¹⁴

E = 4.25×10¯¹⁹ J

Thus, the energy of the photon is 4.25×10¯¹⁹ J

3 0

3 years ago

Read 2 more answers

An atom has 5 electrons in its outer shell. how many unpaired electrons does it have?

lions [1.4K]

The answer is 3.

That is if an atom has 5 electrons in its outer shell, then it has 3 unpaired electrons.

As the outer shell have 1 s orbital too and that is fully filled and not available for bonding so it must have 3 unpaired electrons.

So the number of unpaired electrons in an atom that is having 5 electrons in its outer shell is 3.

4 0

3 years ago

PLEASE HELP!! I REALLY NEED HELP

Allushta [10]

Answer:

Explanation:

1. find the molar mass (amu) of each element and add them to get the whole molar mass.

2. divide the 1 element molar mass with the whole molar mass

3. multiple by 100 and that gives you the % composition.

1. Na(22.99amu) + Cl (35.453amu)=58.443

2(Na): $\frac{22.99}{58.443}$ = .393

2(Cl): $\frac{35.453}{58.443}$ = .607

3(Na): .393 * 100=39.3%

3(Cl): .607 * 100= 60.7%

1. K: (39.098)(2)=78.196

_ C: (12.011)(1)= 12.011

_O: (15.99)(3) = 47.997

78.196+12.011+47.997= 138.204

2:K: $\frac{78.196}{138.204}$ = .566 <u>Step </u>3: (.566)(100)= 56.6%

2: C: $\frac{12.011}{138.204}$ = .087 <u>Step 3</u>: (.087)(100)= 8.7%

2: O: $\frac{47.997}{138.204}$ = .347 <u>Step 3</u>: (.347)(100) = 34.7%

1. Fe (55.845)(3)= 167.535

_ O (15.999)(4) = 63.996

167.535+63.996=231.531

2: Fe: $\frac{167.535}{231.531}$ = .724 Step 3: (.724)(100)= 72.4%

2: O : $\frac{63.996}{231.531}$ = .276 Step 3: (.276)(100) = 27.6%

1.

C(12.011*3)=36.033

H(1.008*5)=5.04 + (1.008*3)=3.024 so its 8.064

O(15.999*3)=47.997

add them: 92.094

2: C: $\frac{36.033}{92.094}$ = .391 Step 3: (.391)(100) = 39.1%

2: H: $\frac{8.064}{92.094}$ = .088 step 3: (.088)(100) = 8.8%

2: O: $\frac{47.997}{92.094}$ = .521 step 3: (.521)(100) = 52.1%

3 0

3 years ago

The stability of atomic nuclei is related to the

Gekata [30.6K]

Ratio of neutrons & protons .

6 0

3 years ago

A solution of permanganate is standardized by titration with oxalic acid, . To react completely with mol of oxalic acid required

Zielflug [23.3K]

Answer:

$M=0.120M$

Explanation:

Hello,

In this case, the undergone chemical reaction is:

$MnO_4^-(aq)+H_2C_2O_4(aq)\rightarrow Mn^{+2}+CO_2$

In such a way, the acidic redox balance turns out:

$(Mn^{+7}O_4)^-+5e^-+8H^+\rightarrow Mn^{+2}+4H_2O\\H_2C_2O_4\rightarrow2CO_2+2H^++2e^-$

Which leads to the total balanced equation as follows:

$2(MnO_4)^-(aq)+6H^+(aq)+5H_2C_2O_4(aq)\rightarrow2Mn^{+2}(aq)+8H_2O(l)+10CO_2(g)$

Thus, as the mass of oxalic acid is not given, one could suppose a value of 1 g (which you can modify based on the actual statement) in order to compute the oxalic acid moles as shwon below:

$1gH_2C_2O_4*\frac{1molH_2C_2O_4}{90.04gH_2C_2O_4} *\frac{2mol(MnO_4)^-}{5molH_2C_2O_4} =0.00444mol(MnO_4)^-$

Whereby the molality results:

$M=\frac{0.00444mol(MnO_4)^-}{0.03702L} =0.120M$

Remember you can modify the oxalic acid mass as you desire.

Best regards.

5 0

3 years ago