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3 years ago

The wavelength of light that has a frenquency of 1.20x10^13s^-1

Chemistry

1 answer:

fredd [130]3 years ago

8 0

speed of light=wavelength*frequency

3*10^8m/s=wavelength*1.20*10^13

wavelength=3*10^8/1.2*10^13

=10^-5/0.4=2.5*10^-5m

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When 1.2383 g of an organic iron compound containing Fe, C, H, and O was burned in O2, 2.3162 g of CO2 and 0.66285 g of H2O were

uysha [10]

Answer:

$\boxed{\text{C$_{15}$H$_{21}$FeO$_{6}$}}$

Explanation:

Let's call the unknown compound X.

1. Calculate the mass of each element in 1.23383 g of X.

(a) Mass of C

$\text{Mass of C} = \text{2.3162 g } \text{CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{0.632 07 g C}$

(b) Mass of H

$\text{Mass of H} = \text{0.66285 g }\text{H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g } \text{{H$_{2}$O}}} = \text{0.074 157 g H}$

(c)Mass of Fe

(i)In 0.4131g of X

$\text{Mass of Fe} = \text{0.093 33 g Fe$_{2}$O$_{3}$}\times \dfrac{\text{111.69 g Fe}}{\text{159.69 g g Fe$_{2}$O$_{3}$}} = \text{0.065 277 g Fe}$

(ii) In 1.2383 g of X

$\text{Mass of Fe} = \text{0.065277 g Fe}\times \dfrac{1.2383}{0.4131} = \text{0.195 67 g Fe}$

(d)Mass of O

Mass of O = 1.2383 - 0.632 07 - 0.074 157 - 0.195 67 = 0.336 40 g

2. Calculate the moles of each element

$\text{Moles of C = 0.63207 g C}\times\dfrac{\text{1 mol C}}{\text{12.01 g C }} = \text{0.052 629 mol C}\\\\\text{Moles of H = 0.074157 g H} \times \dfrac{\text{1 mol H}}{\text{1.008 g H}} = \text{0.073 568 mol H}\\\\\text{Moles of Fe = 0.195 67 g Fe} \times \dfrac{\text{1 mol Fe}}{\text{55.845 g Fe}} = \text{0.003 5038 mol Fe}\\\\\text{Moles of O = 0.336 40} \times \dfrac{\text{1 mol O}}{\text{16.00 g O}} = \text{0.021 025 mol O}$

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

$\text{C: } \dfrac{0.052629}{0.003 5038}= 15.021\\\\\text{H: } \dfrac{0.073568}{0.003 5038} = 20.997\\\\\text{Fe: } \dfrac{0.003 5038}{0.003 5038} = 1\\\\\text{O: } \dfrac{0.021025}{0.003 5038} = 6.0006$

4. Round the ratios to the nearest integer

C:H:O:Fe = 15:21:1:6

5. Write the empirical formula

$\text{The empirical formula is } \boxed{\textbf{C$_{15}$H$_{21}$FeO$_{6}$}}$

5 0

3 years ago

A balloon filled with helium gas occupies 2.50 L at 25°C and 1.00 atm. When released, it rises to an altitude where the temperat

stira [4]

Answer:

8,19 L

Explanation:

The combined gas law is a equation that can be used when the initial and final conditions -pressure,volume,amount of moles,temperature-, of a gas change during a process. It can be written:

$\frac{P1V1}{n1T1}=\frac{P2V2}{n2T2}$

If the amount of the gas remains constant, then n1=n2 and we have:

$\frac{P1V1}{T1}=\frac{P2V2}{T2}$

For the problem we have:

Note: When working with gases is important to use absolute temperature values (°K, °K=°C+273,15):

P1=1 atm, V1=2,5L, T1=25+273,15=298,15°K

P2=0,3 atm, V2=?, T2=20+273,15=293,15°K

$V2=\frac{P1V1T2}{T1P2}=\frac{1atm*2,5L*293,15K}{298,15K*0,3atm}=8,19L$

The new volume of the balloon is 8,19 L.

5 0

2 years ago

If element A has an ionic charge of +2 and element B has an ionic charge of -3, what will be the formula for the compound create

Softa [21]

A3B2 because the oxidation numbers are the same as ionic charge just switch symbol and number. Then use the cross cross method and you get A3B2.

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3 years ago

Amino acids are the building blocks of proteins. The simplest amino acid is glycine (H2NCH2COOH). Draw a Lewis structure for gly

VashaNatasha [74]

Answer:

Explanation:

The lewis structure (indicating all the atoms and patterns provided as hint in the question) of glycine can be seen in the attachment below. While the chemical structure of glycine can be seen below

H₂N - C - C =O

| \

H OH

The structure (of glycine) above provides a "fair idea" of how the lewis structure will be.

3 0

2 years ago

The picture below is a great example of __________ biodiversity.

ki77a [65]

Answer:

number three is the answer moderate

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2 years ago

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