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Flauer [41]
3 years ago
11

-

Chemistry
1 answer:
skelet666 [1.2K]3 years ago
5 0

Answer:

(ii) 1 dye

(iii) Food coloring F is insoluble in the solvent

(iv) 'E' and 'H'

(b) Food colouring G

Explanation:

Paper chromatography principle is based on the rates of migration of chemicals across a sheet of paper which are different and it consists of a stationary phase such as the water in the paper and a mobile phase such as the solvent resulting in the partitioning of the components of the mixture across the paper

The solution components are positioned to start in one place from where they migrate and separate out on the chromatography paper

(ii) The number of components into which the food colouring 'H' separates into = 1

Therefore, the number of dyes in food colouring 'H' = 1 dye

(iii) Food coloring 'F' does not move because it is insoluble in the solvent, which is the mobile phase

(iv) The food colouring that contains the dye that is likely to be most soluble in the solvent are does for which the dyes travel furthest, which are;

Food coloring 'E' and 'H'

(b) Using a similar question solution found on 'tutor my self' website, we have;

The R_f values are given as follows;

R_f = \dfrac{Distance \ moved \ by \ dye}{Distance \ moved \ by \ solvent}

The distance moved by the solvent = 5 units

The distance moved by dyes in food colouring 'E' and 'H' = 4 units

The distance moved by dye in food colouring 'G' = 3.3 units

The distance moved by the second dye in food colouring 'E' = 2.7 units

By inspection, we get;

R_f dye in food colouring 'G' = 3.3/5 = 0.66,

Therefore, the dye with R_f value closest to 0.67 is the dye in food colouring 'G'.

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Answer:

Explanation:

To determine the molecular formula of the compound, the empirical formula must be determined first. To determine the empirical formula, the percentage of each constituent is divided by its molar mass. This is shown below

Carbon = 60/12 = 5

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Hydrogen = 8/1 = 8

The next step is to divide each ratio by the smallest value. The smallest value is 2. It becomes

Carbon = 5/2 = 2.5

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Oxygen = 2/2 = 1

Hydrogen = 8/2 = 4

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C3H4O

From the given relative molecular mass of the compound, the molecular formula can be determined

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how many kilograms of a 35% m/m sodium chlorate solution is needed to react completely with 0.29 l of a 22% m/v aluminum nitrate
Stolb23 [73]

Answer:- 0.273 kg

Solution:- A double replacement reaction takes place. The balanced equation is:

3NaClO_3+Al(NO_3)_3\rightarrow 3NaNO_3+Al(ClO_3)_3

We have 0.29 L of 22% m/v aluminum nitrate solution. m/s stands for mass by volume. 22% m/v aluminium nitrate solution means 22 g of it are present in 100 mL solution. With this information, we can calculate the grams of aluminum nitrate present in 0.29 L.

0.29L(\frac{1000mL}{1L})(\frac{22g}{100mL})

= 63.8 g aluminum nitrate

From balanced equation, there is 1:3 mol ratio between aluminum nitrate and sodium chlorate. We will convert grams of aluminum nitrate to moles and then on multiplying it by mol ratio we get the moles of sodium chlorate that could further be converted to grams.

We need molar masses for the calculations, Molar mass of sodium chlorate is 106.44 gram per mole and molar mass of aluminum nitrate is 212.99 gram per mole.

63.8gAl(NO_3)_3(\frac{1mol}{212.99g})(\frac{3molNaClO_3}{1molAl(NO_3)_3})(\frac{106.44g}{1mol})

= 95.7gNaClO_3

sodium chlorate solution is 35% m/m. This means 35 g of sodium chlorate are present in 100 g solution. From here, we can calculate the mass of the solution that will contain 95.7 g of sodium chlorate  and then the grams are converted to kg.

95.7gNaClO_3(\frac{100gSolution}{35gNaClO_3})(\frac{1kg}{1000g})

= 0.273 kg

So, 0.273 kg of 35% m/m sodium chlorate solution are required.

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3 years ago
The Henry's law constant (kH) for O2 in water at 20°C is 1.28 × 10−3 mol/(L·atm). (a) How many grams of O2 will dissolve in 4.00
Burka [1]

Answer:

Solubility of O₂(g) in 4L water = 3.42 x 10⁻² grams O₂(g)

Explanation:

Graham's Law => Solubility(S) ∝ Applied Pressure(P) => S =k·P

Given P = 0.209Atm & k = 1.28 x 10⁻³mol/L·Atm

=> S = k·P = (1.28 x 10⁻³ mole/L·Atm)0.209Atm = 2.68 x 10⁻³ mol O₂/L water.

∴Solubility of O₂(g) in 4L water at 0.209Atm = (2.68 x 10⁻³mole O₂(g)/L)(4L)(32 g O₂(g)/mol O₂(g)) = <u>3.45 x 10⁻² grams O₂(g) in 4L water. </u>

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