All categories

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 $C_{2}H_{4}O$

You might be interested in

A dilute solution is prepared by transferring 40.00 ml of a 0.3433 m stock solution to a 750.0 ml volumetric flask and diluting

alina1380 [7]

We are given with the initial volume of the substance and the molarity. The first thing that needs to be done is to multiply the equation in order to obtain the number of moles such as shown below.

number of moles = (40 mL) x (1 L / 1000 mL) x (0.3433 moles / L)
number of moles = 0.013732 moles

To get the value of the molarity of the diluted solution, we divide the number of moles by the total volume.
molarity = (0.013732 moles) / (750 mL / 1000 mL/L) = 0.0183 M

Similarly, we can solve for the molarity by using the equation,
M₁V₁ = M₂V₂
Substituting the known values in the equation,
(0.3433 M)(40 mL) = M₂(750 mL)
M₂ = 0.0183 M

5 0

3 years ago

A mixture of 1.374g of H2 and 70.31g of Br2 is heated in a 2.00 L vessel at 700 K. These substances react as follows: H2(g) + Br

gregori [183]

Answer:

a. 0.139 M → [H₂] ; 0.217 M → [Br₂] ; 0.01 M → [HBr]

b. Kc = 3.31x10⁻³

Explanation:

H2(g) + Br2(g) ⇄ 2HBr(g)

Initial 1.374 g 70.31 g -

reacts X X 2x

eq. (1.374 - x) (70.31-x) 2x

In equilibrium I see, the grams I initially had minus some mass which has reacted. In products I have the double of that mass, because the stoichiometry.

So I have the mass in equilibrium, of H2 and of course I can know the mass which has reacted.

1.374g - x = 0.556 g

1.374g - 0.556 g = x = 0.808 g (This is the mass which has reacted)

70.31g - 0.808 g = 69.502 g (Mass in equilibrium of Br2)

2 . 0.808 g = 1.616 g (Mass in equibrium of HBr)

By molar mass, we can kwow the moles.

Molar mass H2: 2 g/m

Moles = mass / molar mass → 0.556 g / 2 g/m = 0.278 moles

Molar mass Br2: 159.80 g/m

Moles = mass / molar mass → 69.502 g / 159.80 g/m = 0.434 moles

Molar mass HBr: 80.9 g/m

Moles = mass / molar mass → 1.616 g / 80.9 g/m = 0.02 moles

The moles are not molarity. In equilibrium, to calculate Kc we need molarity (moles/L). The moles we have calculated are in 2 L of mixture so:

moles / L = molarity

0.278 moles / 2L = 0.139 M → [H₂]

0.434 moles / 2L = 0.217 M → [Br₂]

0.02 moles / 2L = 0.01 M → [HBr]

Kc = [HBr]² / ([H₂] . [Br₂])

Kc = 0.01² / (0.139 . 0.217) = 3.31x10⁻³

6 0

3 years ago

When a person breathes the lungs take in oxygen and release carbon dioxide. What system is this?

Svetllana [295]

It's called a respiratory system. :)

3 0

3 years ago

Read 2 more answers

Hypothesis: This is your testable statement. Usually an "If, then"

In-s [12.5K]

Chloroplast or cell wall... hope this helps

5 0

3 years ago

Identify the missing daughter nucleus in the β– emission decay of 106ru below.

kotykmax [81]

Answer is: the missing daughter nucleus is rhodium (Rh).

Nuclear reaction: ¹⁰⁶Ru → ¹⁰⁶Rh + e⁻(electron) + ve(electron antineutrino).

Beta decay is radioactive decay in which a beta ray and a neutrino are emitted from an atomic nucleus.There are two types of beta decay: beta minus and beta plus.

In beta minus decay, neutron is converted to a proton and an electron and an electron antineutrino.
In beta plus decay, a proton is converted to a neutron and positron and an electron neutrino, so mass number does not change.

6 0

3 years ago