Which chemical equation is balanced as written?

PLS ANSWER FAST

Explanation:

Intermolecular forces hold multiple molecules together and determine many of a substance's properties. All of the attractive forces between neutral atoms and molecules are known as van der Waals forces, although they are usually referred to more informally as intermolecular attraction.

Intermolecular forces are the forces of attraction or repulsion which act between neighboring particles (atoms, molecules, or ions ). These forces are weak compared to the intramolecular forces, such as the covalent or ionic bonds between atoms in a molecule.

Answer:

AFAIK

Explanation:

uric acid is much less toxic than ammonia, hence bigger concentrations of it are tolerated in the body. This means you can excrete it while excreting very little water - beneficial wherever water's not abundant.

There's a tradeoff though, uric acid requires more energy to synthesize than ammonia, so pretty much all fish, say, excrete ammonia rather than uric acid - it's no problem to dilute ammonia since there's no water shortage.

Answer:

https://www.chegg.com/homework-help/questions-and-answers/1-10-points-solution-15-percent-mass-kcl-benzene-new-boiling-point--901-c-b-921-c-c-821-c--q63751186

Explanation: Thats your answer

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45° Celsius

Formula (113°F − 32) × 5/9 = 45°C

Answer:

Explanation:

The correct answer is

C. -112.0 J/mol K

Given :

For a chemical reaction, the activation energy for the forward reaction is +181 kJ and the activation energy for the backward reaction is +62 kJ.

To Find :

The overall energy change for the forward reaction.

Solution :

The overall energy change for the forward reaction is :

[tex]\Delta E_f = E_f - E_b\\\\\Delta E_f = 181 - 62 \ Kj\\\\\Delta E_f = 119 \ Kj[/tex]

Therefore, the overall energy change for the forward reaction is 119 Kj.

The sun is the source, and the change is damage

Answer:

it would be Convection.

Answer:

Examples of chemical properties include flammability, toxicity, acidity, reactivity (many types), and heat of combustion.

Answer:

W = 615.91 kJ

Explanation:

We need to use the following expression:

W = ΔG° * mol   (1)

But in order to determine the ΔG° we need the following expression:

ΔG° = -n * F * E°   (2)

Where F is a constant and is 96,500 J/V mol

n is the number of transferred electrons in the reaction. As we are passing from Al to Al³⁺ we can say that the number of electrons are 3.

Finally to get the moles, we need the the atomic weight of aluminum which is 26.98 g/mol, so the moles:

moles = m/MM   (3)

Let's calculate the moles of aluminum:

moles = 50 / 26.98 = 1.85 moles of aluminum

Now let's calculate the gibbs energy using (2):

ΔG° = -3 * 96,500 * 1.15

ΔG° = -332,925 J or simply -332.925 kJ/mol

Finally, using (1) we can determine the work done:

W = 332.925 * 1.85

Hope this helps

The maximum work available from 50.0 g of aluminum in the following cell when the cell potential is 1.15 V will be 615.91 kJ

The cell potential, Ecell, is the measure of the potential difference between two half cells in an electrochemical cell.

Expression for work done ;

W = ΔG° x mol   (1)

But, to determine the ΔG°

ΔG° = -n x F x E°   (2)

Where,

(As there is variation from Al to Al³⁺ the change in number of electrons are 3)

To get the moles,

moles = m/MM  (3)

moles = 50 / 26.98 = 1.85 moles of aluminum

Now,

ΔG° = -3 x 96,500 x 1.15

ΔG° = -332,925 J  

       = -332.925 kJ/mol

Now, using (1) we can determine the work done:

W = 332.925 x 1.85

W = 615.91 kJ

Hence, The maximum work available from 50.0 g of aluminum in the following cell when the cell potential is 1.15 V will be 615.91 kJ

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#SPJ5

Answer:

400k

Explanation:

volume is constant

P1/T1=P2/T2

1.5/100=6/T2

T2=(100×6)/1.5

T2=400k

Answer:

[Your answer would be in this document of mine below] Thanks :D

Explanation:

Answer:

22.67 L of PH₃

Explanation:

The balanced equation is:

[tex]P_4 (s) + 6H_2(g) \to 4PH_3(g)[/tex]

From the equation:

[tex]34 L \times \dfrac{1 \ mol \ of H_2 }{22.4 \ L \ H_2} \times \dfrac{4 \ mol \ of \ PH_3}{6 \ mol \ H_2} \times \dfrac{22.4 \ L \ PH_3}{1 \ mol \ PH_3}[/tex]

= 22.67 L of PH₃

Answer:

just awser

eeww

Explanation:

Answer:

Please find the solution in the attached file.

Explanation:

The two chairs of the confirmation of the alpha glucose were defined in the attached file.  

Answer:

₰₮₤₳V₤

Explanation:

First you must calculate the number of moles in this solution, by rearranging the equation. No. Moles (mol) = Molarity (M) x Volume (L) = 0.5 x 2. = 1 mol.

For NaCl, the molar mass is 58.44 g/mol. Now we can use the rearranged equation. Mass (g) = No. Moles (mol) x Molar Mass (g/mol) = 1 x 58.44. = 58.44 g.

Answer: Salt is an ionic compound.

Explanation:

A compound formed by transfer of electrons from one atom to another is called an ionic compound.

For example, chemical formula of salt is NaCl (also called table salt).

Salt is formed when sodium (atomic no. 11) donates its one valence electron to chlorine (atomic no. 17). As a result, sodium ion and chlorine ion chemically combine together and form the compound NaCl.

Thus, we can conclude that salt is an ionic compound.

Answer: Anode: [tex]Pb+SO_4^{2-}\rightarrow PbSO_4+2e^-[/tex]

Cathode: [tex]PbO_2+4H^++SO_4^{2-}+2e^-\rightarrow PbSO_4+2H_2O[/tex]

Overall cell reaction : [tex]Pb+2SO_4^{2-}+PbO_2+4H^+\rightarrow 2PbSO_4+2H_2O[/tex]

Explanation:

Lead storage battery is a secondary cell used in automobiles and invertors. The anode is made up of lead and undergoes oxidation during discharging and cathode is made up of lead oxide and acts as cathode during discharging. The electrolyte used is dilute .

Lead storage battery acts as electrochemical cell while discharging.

Discharging reaction for  Anode:

Discharging reaction for  Cathode: [tex]PbO_2+4H^++SO_4^{2-}+2e^-\rightarrow PbSO_4+2H_2O[/tex]

Overall cell reaction : [tex]Pb+2SO_4^{2-}+PbO_2+4H^+\rightarrow 2PbSO_4+2H_2O[/tex]

A lead storage battery is an energy storage device. At the anode lead and sulfate ion reacts to produce Lead(II) sulfate and release two electrons.

In secondary cells or the recharge-discharge cells during recharge, the positive electrode is the anode, while during discharge cathode is the positive electrode.

The anode (lead) of the cell undergoes an oxidation reaction during the discharge, whereas the lead oxide or the cathode undergoes reduction.

The discharging reaction at the anode of the cell is given as,

[tex]\rm Pb + SO_{4}^{2-} \rightarrow PbSO_{4} + 2e^{-}[/tex]

The discharging reaction at the cathode of the cell is given as,

[tex]\rm PbO_{2} + 4H^{+} + SO_{4}^{2-} + 2 e^{-} \rightarrow PbSO_{4} + 2H_{2}O[/tex]

The overall cell reaction is given as,

[tex]\rm Pb + 2SO_{4}^{2-} + PbO_{2} + 4H^{+} \rightarrow 2PbSO_{4} + 2H_{2}O[/tex]

Therefore, the gain and loss of electrons are represented at the cathode and the anode of the cell.

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Answer: A chemical change occurs when the substance's composition is changed. When bonds are broken and new ones are formed a chemical change occurs.

Explanation:

As I think Option C is correct i.e. NH4+.

Answer:

number of moles of (N) = 0.794 moles

Explanation:

From the given information:

no of moles of nitrogen (N) = (unknown)???

mass of nitrogen = 13.5 g

molar mass of NH3 = 14 +( 1 × 3) = 17 g/mol

To calculate the no of moles of N, we have:

number of moles of (N) = mass of N/molar mass

number of moles of (N) = 13.5 g/17 g/mol

number of moles of (N) = 0.794 moles

Answer:

Option C, an increase of reaction rate with increase in temperature

Explanation:

Colligative properties are as follows

a) Decrease of  vapor pressure

b) Increase of boiling point

c) Reduction of freezing point

d) Increase of osmotic pressure

There is no impact on reaction rate and hence it is not a colligative property.

Thus, option c is the right choice

An increase of reaction rate with increase in temperature isn't an example

of colligative properties of solutions

Colligative properties of solutions depend on the ratio of the number of

solutes to that of the solvent(concentration) and not on the nature of the

substances involved.

Examples of colligative properties include vapor pressure lowering, boiling

point elevation, freezing point depression, and osmotic pressure. Increase of

reaction rate with increase in temperature is therefore not an example of

colligative properties of solutions.

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Answer:

Option C is the answer

Butene, also called Butylene, 4 isomeric compound belonging to the series of olefinic hydrocarbons. The chemical formula is C4H8, option c is correct.

Butene, also called Butylene, four isomeric Combinations belong to the series of olefinic hydrocarbons. The chemical formula is C4H8.

The isomeric forms are 1-butene, cis-2-butene, trans-2-butene, and isobutylene.

Thus, option "C" is correct. the chemical formula is C4H8.

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