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3 years ago

HEL PHELP HE LP HELPPP ILL MARK U THE BRAINLIEST :)

Chemistry

2 answers:

Serga [27]3 years ago

8 0

Answer:

C3H6

Explanation:

3 Carbon atoms

6 Hydrogen atoms

Kitty [74]3 years ago

8 0

Answer:

C3H6

Explanation:

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Why do scientists measure and record the mass of objects rather than their weight?

Zepler [3.9K]

Because the mass is how much of something is in an object, gold and silver can have the same mass but have very different weights.

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3 years ago

By changing the number of _______ you get the same element that has a different atomic mass. An example of this would be carbon-

iren2701 [21]

Answer:

neutrons, isotope

Explanation:

Isotopes are atoms of the same element with different atomic mass. The difference in atomic mass is as a result of difference in the number of neutrons in the nuclei of the atoms.

Hence, changing the number of neutrons of an atom with result in having different atomic mass but the number of protons and electrons remain the same.

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What kind of mixture is a solution? A suspension? A colloid?

hjlf

* The mixture is a type of homogeneous material, whose uniformity is found in any proportion of its constituent substances.

* Colloid medium is a type of heterogeneous material, whose diversity is observed only by means of instruments for high resolution.

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3 years ago

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What makes metals, in particular, good conductors of electricity?

hichkok12 [17]

Answer:

the answer is A(the ability of electrons to flow through out the metal)

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3 years ago

Gaseous methane (CH₄) reacts with gaseous oxygen gas (O₂) to produce gaseous carbon dioxide (CO₂) and gaseous water (H₂O) If 0.3

AveGali [126]

Answer : The percent yield of $CO_2$ is, 68.4 %

Solution : Given,

Mass of $CH_4$ = 0.16 g

Mass of $O_2$ = 0.84 g

Molar mass of $CH_4$ = 16 g/mole

Molar mass of $O_2$ = 32 g/mole

Molar mass of $CO_2$ = 44 g/mole

First we have to calculate the moles of $CH_4$ and $O_2$ .

$\text{ Moles of }CH_4=\frac{\text{ Mass of }CH_4}{\text{ Molar mass of }CH_4}=\frac{0.16g}{16g/mole}=0.01moles$

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{0.84g}{32g/mole}=0.026moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CH_4+2O_2\rightarrow CO_2+2H_2O$

From the balanced reaction we conclude that

As, 2 mole of $O_2$ react with 1 mole of $CH_4$

So, 0.026 moles of $O_2$ react with $\frac{0.026}{2}=0.013$ moles of $CH_4$

From this we conclude that, $CH_4$ is an excess reagent because the given moles are greater than the required moles and $O_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $CO_2$

From the reaction, we conclude that

As, 2 mole of $O_2$ react to give 1 mole of $CO_2$

So, 0.026 moles of $O_2$ react to give $\frac{0.026}{2}=0.013$ moles of $CO_2$

Now we have to calculate the mass of $CO_2$

$\text{ Mass of }CO_2=\text{ Moles of }CO_2\times \text{ Molar mass of }CO_2$