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Luba_88 [7]
3 years ago
6

a student has a green dye mixture on which she performs a TLC experiment with a silica gel plate and acetone as the solvent. she

obtains a yellow spot with a Rf of 0.75 and a teal spot with an Rf of 0.45 explain her significance of her data.​
Chemistry
1 answer:
Stels [109]3 years ago
4 0
If he put green die into the 0.75 then it would not have make ants encounter into the internet.so ur answer is 54:
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A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required 10
masya89 [10]

<u>Answer:</u> The molar mass of unknown gas is 367.12 g/mol

<u>Explanation:</u>

Rate of a gas is defined as the amount of gas displaced in a given amount of time.

\text{Rate}=\frac{V}{t}

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation:

\text{Rate of effusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}

So,

\left(\frac{\frac{V_{X}}{t_{X}}}{\frac{V_{O_2}}{t_{O_2}}}\right)=\sqrt{\frac{M_{O_2}}{M_{X}}}

We are given:

Volume of unknown gas (X) = 1.0 L

Volume of oxygen gas = 1.0 L

Time taken by unknown gas (X) = 105 seconds

Time taken by oxygen gas = 31 seconds

Molar mass of oxygen gas = 32 g/mol

Molar mass of unknown gas (X) = ? g/mol

Putting values in above equation, we get:

\left(\frac{\frac{1.0}{105}}{\frac{1.0}{31}}\right)=\sqrt{\frac{32}{M_X}}\\\\M_X=367.12g/mol

Hence, the molar mass of unknown gas is 367.12 g/mol

3 0
4 years ago
Lisa places a substance into a jar. The substance does not change shape to match the shape of the jar.What is the state of matte
mart [117]
The matter is solid, a liquid or a gas would change shape to match the jar.
7 0
4 years ago
In which of the following situations is the greatest amount of work accomplished?
Dahasolnce [82]

<span>work = force x distances </span>

<span>A - moving 2 newton's up 0.6 meter = 1.2 joules
B - moving 4 newton's up 0.6 meter = 2.4 joules
C - moving 6 newton's up 0.3 meter = 1.8 joules
D - moving 9 newton's up 0.3 meter = 2.7 joules
The greatest amount of work is in example D. 
</span>

<span>
 D is your answer. </span>

7 0
4 years ago
Read 2 more answers
Which pair of compounds will form a buffer in aqueous solution?
nikdorinn [45]
Buffer solutions consist of an aqueous solution of a weak acid and its conjugate base or a weak base and its conjugate acid. Buffer solutions are used to have better control of the pH of solutions.

An example of a weak acid and its conjugate base is acetic acid (CH3COOH) and sodium acetate (CH3COONa).
7 0
3 years ago
Calculate the molar mass of each of the following:
Allushta [10]

Explanation:

Molar mass

The mass present in one mole of a specific species .

The molar mass of a compound , can easily be calculated as the sum of the all the individual atom multiplied by the number of total atoms .

(a) S₈

Molar mass of of the atoms are -

sulfur, S = 32 g/mol.

Molar mass of  S₈ = 8 * 32 g/mol.  = 256 g/mol.

(b) C₂H₁₂

Molar mass of of the atoms are -

Hydrogen , H = 1 g/mol

Carbon , C = 12 g/mol

Molar mass of C₂H₁₂ = ( 2 * 12 ) + (12 * 1 ) = 36 g /mol

(c) Sc₂(SO₄)₃

Molar mass of of the atoms are -

sulfur, S = 32 g/mol.

oxygen , O = 16 g/mol.

scandium , Sc = 45 g/mol.

Molar mass of Sc₂(SO₄)₃ = (2 * 45 ) + ( 3 *32 ) + ( 12 * 16 ) = 378 g /mol  

(d) CH₃COCH₃ (acetone)

Molar mass of of the atoms are -

Carbon , C = 12 g/mol

oxygen , O = 16 g/mol.

Hydrogen , H = 1 g/mol

Molar mass of CH₃COCH₃ (acetone) = (3 * 12 ) + ( 1 * 16 ) + ( 6 * 1 ) = 58g/mol

(e) C₆H₁₂O₆ (glucose)

Molar mass of of the atoms are -

Carbon , C = 12 g/mol

oxygen , O = 16 g/mol.

Hydrogen , H = 1 g/mol

Molar mass of C₆H₁₂O₆ (glucose) = ( 6 * 12 ) + ( 12 * 1 ) + ( 6 * 16 ) = 108g/mol.

6 0
3 years ago
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