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3 years ago

What is the empirical formula for c12h24o6? what is the empirical formula for c12h24o6? ch2o cho c2h5o c2h4o cho2?

1 answer:

4 0

Empirical formula: The formula consist of proportions of the elements which is present in the compound or the simplest whole number ratios of atoms.

Now, molecular formula is equal to the product of n (ratio) and empirical formula.

Molecular formula = $n\times empirical formula$ (1)

molecular formula = $C_{12}H_{24}O_{6}$ (given)

Since, 6 is the smallest subscript in above molecular formula to get the simpler whole number of atoms. Therefore, divide all the subscripts i.e. number of carbon atoms (12), number of hydrogen atoms (24) and number of oxygen atoms (6) by 6.

empirical formula becomes $C_{2}H_{4}O$

Thus, according to the formula (1)

$C_{12}H_{24}O_{6} = 6\times C_{2}H_{4}O$

Hence, empirical formula of given molecular formula is $C_{2}H_{4}O$

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anyanavicka [17]

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8 0

3 years ago

ANSWER ASAP PLEASE

user100 [1]

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2 years ago

When a compound containing cesium ions is heated in a Bunsen burner flame, photons with an energy of 4.30 x 10-19 J are emitted.

inessss [21]

<u>Answer:</u> The wavelength of the flame is 462 nm and color of cesium flame is blue.

<u>Explanation:</u>

To calculate the wavelength, we use Planck's equation, which is:

$E=\frac{hc}{\lambda}$

where,

E = Energy of 1 photon = $4.30\times 10^{-19}J$

h = Planck's constant = $6.625\times 10^{-34}J.s$

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength = ?

Putting values in above equation, we get:

$4.30\times 10^{-19}=\frac{6.625\times 10^{-34}J.s\times 3\times 10^8m/s}{\lambda }\\\\\lambda=\frac{6.625\times 10^{-34}J.s\times 3\times 10^8m/s}{4.30\times 10^{-19}}=4.62\times 10^{-7}m=462nm$

The range of wavelength of blue light lies in range of 500 nm - 435 nm

The calculated wavelength lies in the above range. So, the color of the cesium flame is 462 nm

Hence, the wavelength of the flame is 462 nm and color of cesium flame is blue.

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3 years ago

How much will a student earns in 19 weeks if she works for 5 hours each week at a rate of $7 / hour?

koban [17]

Answer:

665

Explanatio

8 0

3 years ago

Read 2 more answers

A mixture of pure agcl and pure agbr is found to contain 60.94% ag by mass. what are the mass percents of cl and br in the mixtu

Annette [7]

First step is to calculate the mass of Ag in each compound separately:
From the periodic table:
molar mass of Ag is 107.87 gm
molar mass of Cl is 35.45 gm
molar mass of Br is 79.9 gm
For AgCl, mass % of Ag = [107.87/143.32] x 100 = 75.26%
For AgBr, mass % of Ag = [107.87/187.77] x 100 = 57.45 %

Second step is to calculate the mass % of each compound in the mixture:
Assume mass % of AgCl is y and that of AgBr is (1-y) as the total percentage is 100% or 1
0.6094 = 0.7526 y + 0.5745 (1-y)
y = 0.8716
This means that the mixture is almost 87% AgCl and 13% AgBr
The mass % of chlorine and bromine together is (100%-60.94%) which is 39.06%
mass % of chlorine = (1-0.6094)(0.8716) x 100 = 34.044%
mass % of bromine = 39.04 - 34.044 = 5.056%

7 0

3 years ago