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3 years ago

How to separate nuts and petrol

Chemistry

2 answers:

IrinaVladis [17]3 years ago

8 0

I just need to comment on some stuff sorry

Law Incorporation [45]3 years ago

3 0

Explanation:

The boiling point of petrol is 483K and that of water in 373K. Since the difference between the boiling point temperatures is high, we can use the method of distillation as explained above to obtain petrol and water from the liquid mixture. Therefore, the correct answer is option (A).

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Which solution will form a precipitate when mixed with a solution of aqueous na2co3? which solution will form a precipitate when

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Answer:

$CuCl_{2}(aq.)$ will form a precipitate of insoluble $CuCO_{3}$ when aqueous $Na_{2}CO_{3}$ is added.

Explanation:

According to solubility rule-

all carbonates are insoluble except group IA compounds and $NH_{4}^{+}$

all salts of sodium are soluble

When $Na_{2}CO_{3}$ is added to given solutions, a double displacement reaction takes place in each solution to form a sodium salt and a carbonate salt.

So, in accordance with solubility rule, addition of $Na_{2}CO_{3}$ into $CuCl_{2}(aq.)$ will result precipitation of insoluble $CuCO_{3}$

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3 years ago

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Answer:

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3 years ago

Copper has two naturally occurring isotopes, 63Cu Gsotopic mass 629396 arnu) and 65Cu Osotopic mass 64.9278 amu). If copper has

madam [21]

Answer: The percentage abundance of $_{29}^{63}\textrm{Cu}$ and $_{29}^{65}\textrm{Cu}$ isotopes are 75.77% and 24.23% respectively

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

Let the fractional abundance of $_{29}^{63}\textrm{Cu}$ isotope be 'x'. So, fractional abundance of $_{29}^{63}\textrm{Cu}$ isotope will be '1 - x'

For $_{29}^{63}\textrm{Cu}$ isotope:

Mass of $_{29}^{63}\textrm{Cu}$ isotope = 62.9396 amu

Fractional abundance of $_{29}^{63}\textrm{Cu}$ isotope = x

For $_{29}^{65}\textrm{Cu}$ isotope:

Mass of $_{29}^{65}\textrm{Cu}$ isotope = 64.9278 amu

Fractional abundance of $_{29}^{65}\textrm{Cu}$ isotope = 1 - x

Average atomic mass of copper = 63.546 amu

Putting values in equation 1, we get:

$63.546=[(62.9396\times x)+(64.9278\times (1-x))]\\\\x=0.6950$

Percentage abundance of $_{29}^{63}\textrm{Cu}$ isotope = $0.6950\times 100=69.50\%$

Percentage abundance of $_{29}^{65}\textrm{Cu}$ isotope = $(1-0.6950)=0.305\times 100=30.50\%$

Hence, the percentage abundance of $_{29}^{63}\textrm{Cu}$ and $_{29}^{65}\textrm{Cu}$ isotopes are 69.50% and 30.50% respectively.

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3 years ago

How to separate nuts and petrol​

How to separate nuts and petrol