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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

Where 1 mole of C₂₁H₄₄ produce 3 moles of ethene, C₂H₄.

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:

Moles Ethene -Molar mass: 28.05g/mol-

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

Moles C₂₁H₄₄:

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

Mass C₂₁H₄₄:

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

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Alveolar air (a mixture of nitrogen, oxygen, and carbon dioxide) has a total pressure of 0.998 atm. If the partial pressure of o

inn [45]

Answer:

The partial pressure of carbon dioxide is 22.8 mmHg

Explanation:

Dalton's Law is a gas law that relates the partial pressures of the gases in a mixture. This law says that the pressure of a gas mixture is equal to the sum of the partial pressures of all the gases present.

In this case:

Ptotal=Pnitrogen + Poxygen + Pcarbondioxide

You know that:

Ptotal= 0.998 atm
Pnitrogen= 0.770 atm
Poxygen= 0.198 atm
Pcarbondioxide= ?

Replacing:

0.998 atm=0.770 atm + 0.198 atm + Pcarbondioxide

Solving:

Pcarbondioxide= 0.998 atm - 0.770 atm - 0.198 atm

Pcarbondioxide= 0.03 atm

Now you apply the following rule of three: if 1 atm equals 760 mmHg, 0.03 atm how many mmHg equals?

$Pcarbondioxide=\frac{0.03 atm*760 mmHg}{1 atm}$

Pcarbondioxide= 22.8 mmHg

The partial pressure of carbon dioxide is 22.8 mmHg

6 0

2 years ago

HA and HB are two strong monobasic acids. 25.0cm3 of 6.0mol/dm3 HA is mixed with 45.0cm3 of 3.0mol/dm3 HB.

Lostsunrise [7]

Answer:

The H+ (aq) concentration of the resulting solution is 4.1 mol/dm³

(Option C)

Explanation:

Given;

concentration of HA, $C_A$ = 6.0mol/dm³

volume of HA, $V_A$ = 25.0cm³, = 0.025dm³

Concentration of HB, $C_B$ = 3.0mol/dm³

volume of HB, $V_B$ = 45.0cm³ = 0.045dm³

To determine the H+ (aq) concentration in mol/dm³ in the resulting solution, we apply concentration formula;

$C_iVi = C_fV_f$

where;

$C_i$ is initial concentration

$V_i$ is initial volume

$C_f$ is final concentration of the solution

$V_f$ is final volume of the solution

$C_iV_i = C_fV_f\\\\Based \ on \ this\ question, we \ can \ apply\ the \ formula\ as;\\\\C_A_iV_A_i + C_B_iV_B_i = C_fV_f\\\\C_A_iV_A_i + C_B_iV_B_i = C_f(V_A_i\ +V_B_i)\\\\6*0.025 \ + 3*0.045 = C_f(0.025 + 0.045)\\\\0.285 = C_f(0.07)\\\\C_f = \frac{0.285}{0.07} = 4.07 = 4.1 \ mol/dm^3$

Therefore, the H+ (aq) concentration of the resulting solution is 4.1 mol/dm³

7 0

2 years ago

The average concentration of bromde ion in seawater is 65 mg of bromide ion per kg of seawater. what is the molarity of the brom

allsm [11]

Usually concentrations are expressed as molarity, or moles of solute per liter solution. First, convert the mass of bromide ion to moles. The molar mass of bromine is 79.904 g/mol.

Moles of bromine = 65 mg * 1 g/1000 mg * 1 mol/79.904 g = 8.135×10⁻⁴ moles

Next, convert the mass of seawater to volume using the density.

Volume of seawater = 1 kg * 1 m³/ 1,025 kg * 1000 L/1 m³ = 0.976 L

Thus,
Molarity = 8.135×10⁻⁴ moles/0.976 L = 8.335×10⁻⁴ M

5 0

2 years ago

A gas occupies a volume at 34.2 mL at a temperature of 15.0 C and a pressure of 800.0 torr. What will be the volume of this gas

Grace [21]

The answer is 34.1 mL.
Solution:
Assuming ideal behavior of gases, we can use the universal gas law equation
P1V1/T1 = P2V2/T2
The terms with subscripts of one represent the given initial values while for terms with subscripts of two represent the standard states which is the final condition.
At STP, P2 is 760.0torr and T2 is 0°C or 273.15K. Substituting the values to the ideal gas expression, we can now calculate for the volume V2 of the gas at STP:
(800.0torr * 34.2mL) / 288.15K = (760.0torr * V2) / 273.15K
V2 = (800.0torr * 34.2mL * 273.15K) / (288.15K * 760.0torr)
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2 years ago

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Using complete subshell notation (not abbreviations, 1s 22s 22p 6 , and so forth), predict the electron configuration of each of

dybincka [34]

Answer: The electronic configuration of the elements are written below.

Explanation:

Electronic configuration is defined as the representation of electrons around the nucleus of an atom.

Number of electrons in an atom is determined by the atomic number of that atom.

For the given options:

Option a: Carbon (C)

Carbon is the 6th element of the periodic table. The number of electrons in carbon atom are 6.

The electronic configuration of carbon is $1s^22s^22p^2$

Option b: Phosphorus (P)

Phosphorus is the 15th element of the periodic table. The number of electrons in phosphorus atom are 15.

The electronic configuration of phosphorus is $1s^22s^22p^63s^23p^3$

Option c: Vanadium (V)

Vanadium is the 23rd element of the periodic table. The number of electrons in vanadium atom are 23.

The electronic configuration of vanadium is $1s^22s^22p^63s^23p^64s^23d^3$

Option d: Antimony (Sb)

Antimony is the 51st element of the periodic table. The number of electrons in antimony atom are 51.

The electronic configuration of antimony is $1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^3$

Option e: Samarium (Sm)

Samarium is the 62nd element of the periodic table. The number of electrons in samarium atom are 62.

The electronic configuration of samarium is $1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^66s^24f^6$

Hence, the electronic configuration of the elements are written above.

4 0

2 years ago