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3 years ago

10

Complete combustion of 9.00 g of a hydrocarbon produced 28.9 g of co2 and 9.87 g of h2o. what is the empirical formula for the h

ydrocarbon?

1 answer:

AVprozaik [17]3 years ago

8 0

DnsigshsjeshsinVshzivfbtofns

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A student reacts 5.0 g of sodium with 10.0 g of chlorine and collect 5.24 g of sodium chloride. What is the percent yield of thi

Ede4ka [16]

Answer: The percent yield of this combination reaction is 41.3 %

Explanation : Given,

Mass of $Na$ = 5.0 g

Mass of $Cl_2$ = 10.0 g

Molar mass of $Na$ = 23 g/mol

Molar mass of $Cl_2$ = 71 g/mol

First we have to calculate the moles of $Na$ and $Cl_2$ .

$\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}$

$\text{Moles of }Na=\frac{5.0g}{23g/mol}=0.217mol$

and,

$\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}$

$\text{Moles of }Cl_2=\frac{10.0g}{71g/mol}=0.141mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation will be:

$2Na+Cl_2\rightarrow 2NaCl$

From the balanced reaction we conclude that

As, 2 mole of $Na$ react with 1 mole of $Cl_2$

So, 0.217 moles of $Na$ react with $\frac{0.217}{2}=0.108$ moles of $Cl_2$

From this we conclude that, $Cl_2$ is an excess reagent because the given moles are greater than the required moles and $Na$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaCl$

From the reaction, we conclude that

As, 2 mole of $Na$ react to give 2 mole of $NaCl$

So, 0.217 mole of $HCl$ react to give 0.217 mole of $NaCl$

Now we have to calculate the mass of $NaCl$

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

Molar mass of $NaCl$ = 58.5 g/mole

$\text{ Mass of }NaCl=(0.217moles)\times (58.5g/mole)=12.7g$

Now we have to calculate the percent yield of this reaction.

Percent yield = $\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100$

Actual yield = 5.24 g

Theoretical yield = 12.7 g

Percent yield = $\frac{5.24g}{12.7g}\times 100$

Percent yield = 41.3 %

Therefore, the percent yield of this combination reaction is 41.3 %

4 0

4 years ago

What effect does the transfer of electrons have on the nuclei of the atoms involved?

Alex777 [14]

Answer:

Chemical processes have no effect on the nucleus otherwise we would be in deep truble. GOOD LESSONS ♡

6 0

3 years ago

What is the limiting reagent when a 2.00 g sample of ammonia is mixed with 4.00 g of oxygen?

UNO [17]

Answer:

Ammonia is limiting reactant

Amount of oxygen left = 0.035 mol

Explanation:

Masa of ammonia = 2.00 g

Mass of oxygen = 4.00 g

Which is limiting reactant = ?

Balance chemical equation:

4NH₃ + 3O₂ → 2N₂ + 6H₂O

Number of moles of ammonia:

Number of moles = mass/molar mass

Number of moles = 2.00 g/ 17 g/mol

Number of moles = 0.12 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 4.00 g/ 32 g/mol

Number of moles = 0.125 mol

Now we will compare the moles of ammonia and oxygen with water and nitrogen.

NH₃ : N₂

4 : 2

0.12 : 2/4×0.12 = 0.06

NH₃ : H₂O

4 : 6

0.12 : 6/4×0.12 = 0.18

O₂ : N₂

3 : 2

0.125 : 2/3×0.125 = 0.08

O₂ : H₂O

3 : 6

0.125 : 6/3×0.125 = 0.25

The number of moles of water and nitrogen formed by ammonia are less thus ammonia will be limiting reactant.

Amount of oxygen left:

NH₃ : O₂

4 : 3

0.12 : 3/4×0.12= 0.09

Amount of oxygen react = 0.09 mol

Amount of oxygen left = 0.125 - 0.09 = 0.035 mol

3 0

3 years ago

What is the molarity if I take 10mL of 20 volumes Hydrogen Peroxide?

Kryger [21]

Answer:

1.635 M

Explanation:

Given:

10 mL of 20 volumes Hydrogen Peroxide

Here,

20 volumes of Hydrogen Peroxide means that on decomposition of 1 mL of H₂O₂ 20 mL of O₂ is obtained

also,

means 1 dm³ of H₂O₂ solution produces 20 dm³ oxygen

Now,

at 298K and 1 atm

20 dm³ oxygen = $\frac{\textup{20}}{\textup{24.47}}$ moles

or

= 0.817 moles

also,

2H₂O₂ → 2H₂O + O₂

thus,

1 dm³ of solution must contain 2 × moles of O₂ as moles of H₂O₂

thus,

Number of moles of H₂O₂ = 2 × 0.817

or

Number of moles of H₂O₂ = 1.635 moles

Hence,

For 20 volume hydrogen peroxide is 1.635 M

3 0

3 years ago

Why is the answer C for this problem?

pshichka [43]

Answer:

$\boxed{\text{(C) X}$_{3}$P$_{2}}$

Explanation:

Step 1. Identify the Group that contains X

We look at the consecutive ionization energies and hunt for a big jump between them

$\begin{array}{crc}n & IE_{n} & IE_{n} - IE_{n-1}\\1 & 730 & \\2 & 1450 & 720\\3 & 7700 & 6250\\4 & 10500 & 2800\\\end{array}$

We see a big jump between n = 2 and n = 3. This indicates that X has two valence electrons.

We can easily remove two electrons, but the third electron requires much more energy. That electron must be in the stable, filled, inner core.

So, X is in Group 2 and P is in Group 15.

Step 2. Identify the Compound

X can lose two valence electrons to reach a stable octet, and P can do the same by gaining three electrons.

We must have 3 X atoms for every 2 P atoms.

The formula of the compound is $\boxed{\text{X}$_{3}$P$_{2}}$}$ .

4 0

3 years ago