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3 years ago

How do substances and mixtures diifer

Chemistry

1 answer:

vlabodo [156]3 years ago

5 0

Answer:

hello hope this helps :)

a pure substance consists only of one element or one compound. a mixture consists of two or more different substances, not chemically joined together.

Explanation:

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3 years ago

Liquid hexane

maks197457 [2]

Answer: The mass of $H_2O$ produced is 2.52 g

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

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

For hexane:

Given mass of hexane = 1.72 g

Molar mass of hexane = 86.18 g/mol

Putting values in equation 1, we get:

$\text{Moles of hexane}=\frac{1.72g}{86.18g/mol}=0.020mol$

For oxygen gas:

Given mass of oxygen gas = 8.0 g

Molar mass of oxygen gas= 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of water}=\frac{8.0g}{32g/mol}=0.25mol$

The chemical equation for the combustion of hexane follows:

$2C_6H_{14}+19O_2\rightarrow 12CO_2+14H_2O$

By stoichiometry of the reaction:

If 2 moles of hexane reacts with 19 moles of oxygen gas

So, 0.020 moles of hexane will react with = $\frac{19}{2}\times 0.020=0.19mol$ of oxygen gas

As the given amount of oxygen gas is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Thus, hexane is considered a limiting reagent because it limits the formation of the product.

By the stoichiometry of the reaction:

If 2 moles of hexane produces 14 moles of $H_2O$

So, 0.020 moles of hexane will produce = $\frac{14}{2}\times 0.020=0.14mol$ of $H_2O$

We know, molar mass of $H_2O$ = 18 g/mol

Putting values in above equation, we get:

$\text{Mass of }H_2O=(0.14mol\times 18g/mol)=2.52g$

Hence, the mass of $H_2O$ produced is 2.52 g

4 0

3 years ago

An exponent of "2" means that if we double the concentration of the reactant the rate doubles as well Exponents in rate laws are

Karo-lina-s [1.5K]

Answer:

- False.

- True.

Explanation:

Hello, for each statement we state:

- An exponent of "2" means that if we double the concentration of the reactant the rate doubles as well.

FALSE because considering a rate law like:

$-r=kC^2$

The exponent of "2" powers the concentration to the second power, not doubles the rate law, thus, if C is 3, for k=1, r will be -9. On the other hand if the rate is like:

$-r=kC$

The rate will be -3, that is why the rate is not doubled when the "2" in concentration is present.

- Exponents in rate laws are based on the coefficients from the balanced equation.

FALSE because for nonelemental chemical reactions, the exponents do not match with each species' stoichiometric coefficients in the rate law.

- The rate constant, k, takes into account the effect of activation energy and temperature on the reaction.

TRUE, since the Arrhenius equation allows us to prove the effect of the activation energy and the temperature:

$k=Aexp(-\frac{Ea}{RT})$

- Differential rate laws allow us to compare concentration and time.

TRUE as they are given like:

$\frac{1}{\nu _A} \frac{dC_A}{dt} =\frac{1}{\nu _B} \frac{dC_B}{dt} =...$

Best regards.

5 0

2 years ago