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2 years ago

(C)

Chemistry

1 answer:

kari74 [83]2 years ago

6 0

Answer:

177.3kg C₂₁H₄₄

Explanation:

Based on the chemical reaction:

C₂₁H₄₄ → 3C₂H₄ + C₁₅H₃₂

<em>Where 1 mole of C₂₁H₄₄ produce 3 moles of ethene, C₂H₄.</em>

<em />

To solve this question we need to determine the moles of ethene in 50.4kg. 1/3 these moles are the moles of C₂₁H₄₄ that must be added:

<em>Moles Ethene -Molar mass: 28.05g/mol-</em>

50.4kg = 50400g * (1mol / 28.05g) = 1796.8 moles of ethene

<em>Moles C₂₁H₄₄:</em>

1796.8 moles of ethene * (1 mol C₂₁H₄₄ / 3 mol C₂H₄) = 589.93 moles C₂₁H₄₄

589.93 moles C₂₁H₄₄ * (296g / mol) = 177283g =

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When 1.57 mol o2 reacts with h2 to form h2o, how many moles of h2 are consumed in the process?

natita [175]

Here we have to get the moles of hydrogen (H₂) consumed to form water (H₂O) from 1.57 moles of oxygen (O₂)

In this process 3.14 moles of H₂ will be consumed.

The balanced reaction between oxygen (O₂) and hydrogen (H₂); both of which are in gaseous state to form water, which is liquid in nature can be written as-

2H₂ (g) + O₂ (g) = 2H₂O (l).

Thus form the equation we can see that 1 mole of oxygen reacts with 2 moles of hydrogen to form 2 moles of water.

So, 1.57 moles of oxygen will consume (1.57×2) = 3.14 moles of hydrogen to form water.

3 0

2 years ago

Read 2 more answers

The factor 0.01 corresponds to which prefix? A) milli B) deci C) deka D) centi

BARSIC [14]

The answer is: D) centi

5 0

2 years ago

What is the empirical formula of a compound containing 5.03 grams carbon, 0.42 grams hydrogen, and 44.5 grams chlorine?

Thepotemich [5.8K]

Answer:

CHCl₃

Explanation:

Given parameters:

Carbon = 5.03g

Hydrogen = 0.42g

Chlorine = 44.5g

The empirical formula shows the simplest formula of a compound.

To deduce the empirical, we need two pieces of information:

> Mass of the elements or the percentage composition of the compound

>The relative atomic masses of the elements

In order to derive the empirical formula from these parameters,

>>> find the number of moles of elements by dividing the mass given by the relative atomic mass of the respective atom

>>> Divide through by the smallest mole

>>> Approximate or multiply by a factor that would make it possible for whole numbers to be obtained

From the question, we have been given the mass of each element.

Now using the period table, we can obtain the relative atomic masses of each atom:

Carbon = 12gmol⁻¹

Hydrogen = 1gmol⁻¹

Chlorine = 37.5gmol⁻¹

C H Cl

Mass(in g) 5.03 0.42 44.5

Moles 5.03/12 0.42/1 44.5/37.5

0.42 0.42 1.19

Dividing

smallest 0.42/0.42 0.42/0.42 1.19/0.42

Mole ratio 1 1 2.83

Approximate 1 1 3

The empirical formula is CHCl₃

8 0

2 years ago

Be sure to answer all parts. The sulfate ion can be represented with four S―O bonds or with two S―O and two S═O bonds. (a) Which

VMariaS [17]

Answer:

a) Representation - (in attachment)

b) Tetrahedral geometry and

c) $sp^{3}$ hybrid orbitals invovle sigma bonding.

$\pi$ orbitals are formed by the overlapping of d-orbitals of sulfur with p-orbitals of oxygen.

Explanation:

Representation in attachment.

The overall charge in the both structures are -2. The structure (b) is favored by the resonance structures since the formal charge with in the species are remained.

Therefore, structure (b) is the better representation of sulfate ion.

In sulfate ion, sulfur atom is attached with four different oxygen atoms. According to the VSEPR theory , the sufate ion has tetrahedral geometry.

There is four sigma bonds and zero lone pairs present on the central metal atom.

Hence, the hybridization of the sulfur atom is $sp^{3}$

The s and p orbitals of sulfur invovles hybridization and form sigma bonds.The orbitals of sulfur involves $\pi$ bonding.

Therefore, $\pi$ bonds are formed by the overlapping of d-orbitals of sulfur with p-orbitals of oxygen.

5 0

2 years ago

A compound is made up of 28 g N, 24 g C, 48 g O, and 8 g H .What is the empirical formula?

vovikov84 [41]

Answer:

$\rm C_2H_8N_2O_3$ .

Explanation:

<h3>Step One: calculate the coefficients. </h3>

Look up the relative atomic mass of these four elements on a modern periodic table:

$\rm C$ : approximately $12$ .
$\rm H$ : approximately $1$ .
$\rm N$ : approximately $14$ .
$\rm O$ : approximately $16$ .

The relative atomic mass of an element is numerically equal to the mass (in grams, $\rm g$ ,) of one mole of atoms of this element.

For example, the relative atomic mass of $\rm C$ is approximately $12$ . Therefore, each mole of $\rm C\!$ atoms would have a mass of $12\; \rm g$ .

This sample contains $24\; \rm g$ of carbon. That would correspond to approximately $\displaystyle \left(\frac{24}{12}\right)\; \rm mol = 2\; \rm mol$ of $\rm C$ atoms.

Similarly, for the other three elements:

$\displaystyle n(\mathrm{H}) \approx \frac{8\; \rm g}{1\; \rm g \cdot mol^{-1}} = 8\; \rm mol$ .

$\displaystyle n(\mathrm{N}) \approx \frac{28\; \rm g}{14\; \rm g \cdot mol^{-1}} = 2\; \rm mol$ .

$\displaystyle n(\mathrm{O}) \approx \frac{48\; \rm g}{16\; \rm g \cdot mol^{-1}} = 3\; \rm mol$ .

Hence, the ratio between these elements in this compound would be:

$n(\mathrm{C}): n(\mathrm{H}): n(\mathrm{N}):n(\mathrm{O}) = 2: 8 : 2 : 3$ .

In the empirical formula of a compound, the coefficients should represent the smallest possible integer ratio between the number of atoms of these elements.

$n(\mathrm{C}): n(\mathrm{H}): n(\mathrm{N}):n(\mathrm{O}) = 2: 8 : 2 : 3$ is indeed the smallest possible integer ratio between the number of atoms of these elements.

<h3>Step Two: arrange the elements in an appropriate order</h3>

Apply the Hill System to arrange these four elements in the empirical formula. In the Hill System:

If carbon, $\rm C$ , is present in this compound, then:

$\rm C$ (carbon) and then $\rm H$ (hydrogen) will be the first two elements listed in the formula (ignore the hydrogen if it is not in the compound.)
The other elements in this compound will be listed in alphabetical order.

If there is no carbon $\rm C$ in this compound, then list all the elements in this compound in alphabetical order.

Both $\rm C$ (carbon) and $\rm H$ (hydrogen) are found in this compound. Therefore, the first element in the list would be $\rm C\!$ . The second would be $\rm H\!$ , followed by $\rm N\!$ and then $\rm O\!$ .

Hence, the empirical formula of this compound would be $\rm C_2H_8N_2O_3$ .

6 0

2 years ago