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3 years ago

What is the difference between a compound and a mixture?

Chemistry

2 answers:

Artyom0805 [142]3 years ago

3 0

Answer:

Compound are substances which can be formed by chemically combining two or more elements. Mixtures are substances that are formed by physically mixing two or more substances. Compounds can be of three types, which are: covalent compounds, metallic compounds and ionic compounds.

sveta [45]3 years ago

3 0

Answer:

The answer is a mixture can be physically separated into its components; a compound cannot be physically separated into its components.

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A reaction was performed in which 3.6 g 3.6 g of benzoic acid was reacted with excess methanol to make 1.3 g 1.3 g of methyl ben

Anit [1.1K]

Answer: The percent yield of the reaction is 32.34 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For benzoic acid:

Given mass of benzoic acid = 3.6 g

Molar mass of benzoic acid = 122.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of benzoic acid}=\frac{3.6g}{122.12g/mol}=0.0295mol$

The chemical equation for the reaction of benzoic acid and methanol is:

$\text{Benzoic acid + methanol}\rightarrow \text{methyl benzoate}$

By Stoichiometry of the reaction

1 mole of benzoic acid produces 1 mole of methyl benzoate

So, 0.0295 moles of benzoic acid will produce = $\frac{1}{1}\times 0.0295=0.108$ moles of methyl benzoate

Now, calculating the mass of methyl benzoate from equation 1, we get:

Molar mass of methyl benzoate = 136.15 g/mol

Moles of methyl benzoate = 0.0295 moles

Putting values in equation 1, we get:

$0.0295mol=\frac{\text{Mass of methyl benzoate}}{136.15g/mol}\\\\\text{Mass of methyl benzoate}=(0.0295mol\times 136.15g/mol)=4.02g$

To calculate the percentage yield of methyl benzoate, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of methyl benzoate = 1.3 g

Theoretical yield of methyl benzoate = 4.02 g

Putting values in above equation, we get:

$\%\text{ yield of methyl benzoate}=\frac{1.3g}{4.02g}\times 100\\\\\% \text{yield of methyl benzoate}=32.34\%$

Hence, the percent yield of the reaction is 32.34 %

6 0

3 years ago

How can you demonstrate waves?

Lady bird [3.3K]

Shaking a phone cord, strumming a guitar string, playing a trumpet

7 0

3 years ago

determine the volume of your 0.25 m stock solution that is needed to make 200 ml of 0.010 m nacl. show all calculations

rusak2 [61]

The volume of your 0.25 m stock solution that is needed to make 200 ml of 0.010 m NaCl is 0.008 L

Concentration is the abundance of a constituent divided by way of the overall volume of an aggregate. several sorts of mathematical descriptions may be outstanding: mass concentration, molar concentration, variety concentration, and extent awareness.

Calculation:-

C₁ = 0.25 M

V₁ = ?

C₂ = 0.010 M

V₂ = 200 ml = 0.2 L

V₁ = C₂V₂/C₁

= 0.010 × 0.2 / 0.25

= 0.008 L

The concentration of a substance is the quantity of solute found in a given amount of solution. Concentrations are normally expressed in terms of molarity, defined because of the variety of moles of solute in 1 L of answer.

The Concentration of an answer is a measure of the quantity of solute that has been dissolved in a given amount of solvent or answer. A concentrated answer is one that has a rather huge quantity of dissolved solute.

Learn more about concentration here:-brainly.com/question/26255204

#SPJ4

5 0

1 year ago

Calculate the volume of 2.65 moles of N2O gas at STP.

Reptile [31]

Answer:

63.6 dm³

Explanation:

Volume of gas= no. of moles ×24dm³

Thus, volume of N₂O

= 2.65 × 24

= 63.6 dm³

3 0

3 years ago

Hemoglobin molecules in blood bind oxygen and carry it to cells, where it takes part in metabolism. The binding of oxygen hemogl

Alex73 [517]

Without wasting much of our time, Here is the correct question.

Hemoglobin molecules in blood bind oxygen and carry it to cells, where it takes part in metabolism. The binding of oxygen hemoglobin(aq) + O2(aq) -------> hemoglobin O2(aq) is first order in hemoglobin and first order in dissolved oxygen, with a rate constant of 4 × 10⁷ L mol⁻¹ s⁻¹. Calculate the initial rate at which oxygen will be bound to hemoglobin if the concentration of hemoglobin is 2 × 10⁻⁹ M and that of oxygen is 5 × 10⁻⁵M.

Answer:

4 × 10⁻⁶ M s⁻¹

Explanation:

The equation for the reaction between Hemoglobin molecules in blood that binds with oxygen molecule can be represent by:

$hemoglobin_{(aq)$ + $O_{2(aq)$ ---------> $hemoglobin.O_{2(aq)$

Now, we are also being told to calculate only!, the initial rate at which oxygen will be bound to hemoglobin.

So, If it is first order in hemoglobin and also first order in Oxygen molecule at the initial rate of the the reaction, therefore, the rate for the reaction can be expressed as :

rate = k $[hemoglobin_{(aq)}][O_{2(aq)}]$

Let's not forget that we are so given some parameters;

where

k (rate constant) = 4 × 10⁷ L mol⁻¹ s⁻¹

[ $hemoglobin_{(aq)$ ] = 2 × 10⁻⁹ M

[ $O_{2(aq)$ ] = 5 × 10⁻⁵ M

Substituting our data given into the above rate formula, we have:

rate = (4 × 10⁷ L mol⁻¹ s⁻¹) × (2 × 10⁻⁹ M) × (5 × 10⁻⁵ M)

rate = 4 × 10⁻⁶ M s⁻¹ ( given that 1 M = 1 mol L⁻¹ )

∴ the initial rate at which oxygen will be bound to hemoglobin = 4 × 10⁻⁶ M s⁻¹

7 0

3 years ago