Is A and C correct?

I'm a little unsure of this question, but is it right?

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The combustion of ethane (C2H6) produces carbon dioxide and steam. 2C2H6(g)+7O2(g)⟶4CO2(g)+6H2O(g) How many moles of CO2 are produced when 5.90 mol of ethane is burned in an excess of oxygen?

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Final answer:

The combustion of ethane yields carbon dioxide, and with 5.90 moles of ethane being reacted, it results in the production of 11.8 moles of CO2.

Explanation:

The question pertains to the concept of stoichiometry in chemistry, and the chemical reaction in question is a combustion reaction involving ethane (C2H6). From the balanced reaction, it is evident that 2 moles of ethane (C2H6) produce 4 moles of carbon dioxide (CO2). Therefore, if we have 5.90 moles of ethane reacting, it's a straightforward calculation to determine that this would yield twice that many moles of CO2. We simply multiply the moles of ethane by the stoichiometric ratio (4/2) to get the moles of CO2.

Example Calculation: 5.90 moles of ethane x (4 moles CO2 / 2 moles C2H6) = 11.8 moles CO2

So, when 5.90 moles of ethane are burned in an excess of oxygen, 11.8 moles of CO2 are produced.

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Final answer:

In the combustion of ethane, for every mole of ethane burned, two moles of carbon dioxide are produced. Hence, when 5.90 moles of ethane are burned, 11.8 moles of carbon dioxide are produced.

Explanation:

The chemical reaction given, 2C2H6(g) + 7O2(g) ⟶ 4CO2(g) + 6H2O(g), states that 2 moles of ethane (C2H6) produce 4 moles of carbon dioxide (CO2). Thus, the mole-to-mole ratio of ethane to carbon dioxide is 2:4, or simplified, 1:2. So, for every mole of ethane burned, two moles of carbon dioxide are produced.

Given that 5.90 moles of ethane are burned, we can calculate the quantity of carbon dioxide produced by multiplying 5.90 moles by 2. Hence, when 5.90 moles of ethane are burned in an excess of oxygen, 11.8 moles of carbon dioxide are produced.

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Which of the following electron configurations represents an excited state of the indicated atom? Group of answer choices a.Na: 1s2 2s2 2p6 3s2 3p2 3s1
b.Ne: 1s2 2s2 2p6
c.N: 1s2 2s2 2p3
d.P: 1s2 2s2 2p6 3s2 3p2 4s1
e.He: 1s2

Answers

Answer: The electronic configuration of the atom that represents excited state is P: $1s^22s^22p^63s^23p^24s^1$

Explanation:

There are 2 states classified under energy levels:

Ground state: This is the lower energy state which is termed as stable state.
Excited state: This is the upper energy state and is termed as the unstable state. All the electrons which are present in this state always come back to the ground state.

For the given options:

For a: Sodium

The atomic number of sodium element is 11. The ground state electronic configuration of this element is $1s^22s^22p^63s^1$

For b: Neon

The atomic number of neon element is 10. The ground state electronic configuration of this element is $1s^22s^22p^6$

For c: Nitrogen

The atomic number of nitrogen element is 7. The ground state electronic configuration of this element is $1s^22s^22p^3$

For d: Phosphorus

The atomic number of phosphorus element is 15. The ground state electronic configuration of this element is $1s^22s^22p^63s^23p^3$

One electron from the valence shell jumps into outer shell and the excited state electronic configuration becomes $1s^22s^22p^63s^23p^24s^1$

For e: Helium

The atomic number of helium element is 2. The ground state electronic configuration of this element is $1s^2$

Hence, the electronic configuration of the atom that represents excited state is P: $1s^22s^22p^63s^23p^24s^1$

Final answer:

The sodium atom with the electron configuration 1s2 2s2 2p6 3s2 3p2 3s1 is in an excited state because other sodium atom stages are not completely filled.

Explanation:

An atom is in an excited state when one or more electrons have moved to a higher energy level. Normal electron configurations have the electrons in the lowest possible energy states (or orbitals). In this case, the answer choice is (a) sodium (Na) with the electron configuration 1s2 2s2 2p6 3s2 3p2 3s1. Sodium normally has the 3s state fully occupied, so the presence of an electron in a higher energy state (3p) and the vacancy in the lower energy state (3s) indicates an excited state.

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If a molecule diffusing through extracellular fluid travels 1 mm in 1 sec, how long will it take that molecule to diffuse 1 cm

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given that the rate of diffusion is 1 mm per 1 second. then the time it travels in 1 cm can be solve using the formula
t = d / r
where d is the distance
r is the rate

first, 1 cm is equal to 10 mm

t = 10 mm / ( 1 mm / s )
t = 10 s

A concentration cell is constructed using two Ni electrodes with Ni2+ concentrations of 1.0 M and 1.00 � 10�4 M in the two half-cells. The reduction potential of Ni2+ is �0.23 V. Calculate the potential of the cell at 25�C if the more dilute Ni2+ solution is in the anode compartment.

Answers

Answer: The cell potential of the cell is +0.118 V

Explanation:

The half reactions for the cell is:

Oxidation half reaction (anode): $Ni(s)\rightarrow Ni^(2+)+2e^-$

Reduction half reaction (cathode): $Ni^(2+)+2e^-\rightarrow Ni(s)$

In this case, the cathode and anode both are same. So, $E^o_(cell)$ will be equal to zero.

To calculate cell potential of the cell, we use the equation given by Nernst, which is:

$E_(cell)=E^o_(cell)-(0.0592)/(n)\log ([Ni^(2+)_(diluted)])/([Ni^(2+)_(concentrated)])$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_(cell)$ = ?

$[Ni^(2+)_(diluted)]$ = $1.00* 10^(-4)M$

$[Ni^(2+)_(concentrated)]$ = 1.0 M

Putting values in above equation, we get:

$E_(cell)=0-(0.0592)/(2)\log (1.00* 10^(-4)M)/(1.0M)$

$E_(cell)=0.118V$

Hence, the cell potential of the cell is +0.118 V

The cell potential for the cell as calculated is 0.118 V.

What is the Nernst equation?

The Nernst equation can be used to obtain the cell potential of a cell under non- standard conditions. The standard cell potential in this case is zero owing to the fact that both cathode and anode are made of nickel.

Hence;

Ecell = E°cell - 0.0592/nlog Q

Ecell = 0 - 0.0592/2 log (1 00 * 10^-4/1)

Ecell = 0.118 V

The cell potential for the cell as calculated is 0.118 V.

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Compute the repeat unit molecular weight of polystyrene. (b) Compute the number-average molecular weight for a polystyrene for which the degree of polymerization is 23000.

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Answer :

(a) The repeat unit is, styrene

(b) The number-average molecular weight is, 2392000 g/mol

Explanation :

First we have to calculate the repeat unit molecular weight of polystyrene.

As, the repeat unit is, styrene having chemical formula $C_8H_8$

Molecular weight of repeat unit = 8 × C + 8 × H

Molecular weight of repeat unit = 8 × 12 g/mol + 8 × 1 g/mol

Molecular weight of repeat unit = 104 g/mol

Now we have to calculate the number-average molecular weight.

Number-average molecular weight = Average repeat molecular weight × Degree of polymerization

Number-average molecular weight = (104 g/mol) × (23000)

Number-average molecular weight = 2392000 g/mol

Thus, the number-average molecular weight is, 2392000 g/mol

Final answer:

The repeat unit molecular weight of polystyrene is 104.15 g/mol. The average molecular weight of polystyrene with a polymerization degree of 23000 is approximately 2,395,450 g/mol.

Explanation:

To answer this question, we first need to understand that the repeating unit in polystyrene is the styrene monomer, which is C8H8. The molecular weight of this unit can be calculated by adding up the atomic weights of all the atoms in the monomer. The atomic weights of carbon (C), hydrogen (H), and styrene-based on the periodic table are approximately 12.01 amu, 1.01 amu, and 104.15 g/mol respectively. This gives a total of 104.15 g/mol for the repeat unit molecular weight of polystyrene.

Given that the degree of polymerization is 23000, we can calculate the number-average molecular weight by multiplying the repeat unit molecular weight (104.15 g/mol) by the degree of polymerization (23000). This gives a total of approximately 2,395,450 g/mol for the number-average molecular weight.

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The heat capacity of chloroform (trichloromethane,CHCl3)in the range 240K to 330K is given
byCpm/(JK-1mol-1) = 91.47
+7.5x10-2(T/K). In a particular experiment,
1.0molCHCl3 is heated from 273K to 300K. Calculate the
changein molar entropy of the sample.