How many moles of atoms are in 150 g S

Answer:

Explanation:

Here, we want to know what is meant by quantitative transfer

In a quantitative transfer of analytes, no analytes must be lost. What this simply means is that every part or grain of the substance to be weighed must be transferred

Thus, the correct answer choice here is the option B

The following balanced chemical equation may be used to describe the interaction between aluminum (Al) and calcium nitrate (Ca(NO₃)₂):

2 Al + 3 Ca(NO₃)₂ → 2 Al(NO₃)3 + 3 Ca

Reactants are the chemicals that begin a chemical reaction, while products are the compounds that are created as a result of the reaction.

The substances that initiate a chemical reaction. Products are the substances that are created during the reaction. Compounds or elements can act as reactants and products.

Aluminium and calcium nitrate interact in this reaction to form aluminium nitrate (Al(NO₃)₃) and calcium (Ca), which are the end products.

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The enthalpy of combustion for 1kg of urea is -1223525.84 J/mol.

Urea is a compound that is used in fertilizers and in some plastics.The enthalpy of combustion for urea is the amount of energy that is released when urea is burned. In order to calculate the enthalpy of combustion for 1kg of urea, we need to use the information that is provided to us in the question. Let us start by writing down the balanced equation for the combustion of urea: CO(NH2)2 + 3/2 O2 → CO2 + 2H2O + N2
The balanced equation shows that 1 mole of urea reacts with 1.5 moles of oxygen gas to produce 1 mole of carbon dioxide, 2 moles of water, and 1 mole of nitrogen gas. The enthalpy change for this reaction is equal to the amount of energy that is released when 1 mole of urea is burned.
The heat of combustion (ΔHc) of urea is -632.6 kJ/mol. This means that 632.6 kJ of energy is released when 1 mole of urea is burned. We know that 12g of urea raised the temperature of water by 30°C. We can use this information to calculate the amount of energy that was released when 12g of urea was burned.
The specific heat capacity of water is 4.18 J/g°C. This means that it takes 4.18 J of energy to raise the temperature of 1 gram of water by 1°C. Therefore, it takes 4.18 x 1000 = 4180 J of energy to raise the temperature of 1 kg of water by 1°C.
We know that 12g of urea raised the temperature of water by 30°C. Therefore, the amount of energy that was released when 12g of urea was burned is:
Energy = mass x specific heat capacity x temperature change
Energy = 0.012 kg x 4180 J/kg°C x 30°C
Energy = 1497.6 J
We can now use this information to calculate the enthalpy of combustion for 1kg of urea:
Enthalpy of combustion = energy released / moles of urea burned
Enthalpy of combustion = 1497.6 J / (0.012 kg / 60.06 g/mol)
Enthalpy of combustion = - 1223525.84 J/mol
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Answer:

lungs

Explanation:

Answer:

molarity of Cl⁻ ions = 5.0 M

Explanation:

The molarity of a solution expresses the amount in moles of a solution present in a litre of solvent. Its unit is mol/L or mol/dm³ written as M.

MgCl₂ when dissolved to form an aqueous solution dissociates into positive and negative ions, Mg²⁺and Cl⁻ ions.

Equation of the dissociation of MgCl₂ in aqueous solution is given below:

MgCl₂(aq) -----> Mg²⁺(aq) + 2Cl⁻(aq)

1 mole of aqueous MgCl₂ produces 2 moles of chloride ions, Cl⁻

Molar concentration of MgCl₂(aq) = 2.5 M

Molarity of chloride ions = number of moles of Cl⁻ ions * molarity of MgCl₂

molarity of Cl⁻ ions = 2 * 2.5 M

molarity of Cl⁻ ions = 5.0 M

The electron configuration for the element Iodine will be:

[Kr] 4d10 5s2 5p5, now we have every electron represented within their orbitals

Answer:

A large absorption at 2950 cm⁻¹ indicates that there is likely an sp³C-H functional group somewhere in the molecule. The "sp³" designation means that this is a single bond because these atoms have 4 orbitals available (s + p + p + p) for hybridizing, meaning only single bonds are possible.

The standard reduction potential of the Cr³+/Cr electrode is -0.46V. None of the option is correct.

To determine the standard reduction potential of the Cr³+/Cr electrode, we can use the Nernst equation, which relates the standard reduction potential to the cell potential under non-standard conditions. The Nernst equation is given by:

E = E° - (0.0592/n) * log(Q)

where E is the cell potential, E° is the standard reduction potential, n is the number of electrons transferred in the half-reaction, and Q is the reaction quotient.

In this case, we have the standard potential of the cell as −0.60V. We know that the standard reduction potential of the Sn/Sn²+ electrode is -0.14V. Therefore, the reduction potential of the Cr³+/Cr electrode can be calculated as:

E = -0.60V - (-0.14V)

E = -0.60V + 0.14V

E = -0.46V

Therefore, the standard reduction potential of the Cr³+/Cr electrode is -0.46V.

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Answer:

5l NO

2

at STP

No. of molecules=

22.4

5

mol=

22.4

5

×N

A

molecules

A) 5ℊ of H

2

(g)

No. of moles=

2

5

mol=

2

5

×N

A

molecules

B) 5l of CH

4

(g)

No. of moles of CH

4

=

22.4

5

mol=

22.4

5

N

A

molecules

C) 5 mol of O

2

=5N

A

O

2

molecules

D) 5×10

23

molecules of CO

2

(g)

Molecules of 5l NO

2

(g) at STP=5l of CH

4

(g) molecules at STP

Therefore, option B is correct.

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Answer:

28 meters

Explanation:

If a reaction starts with 30 grams of material, the same amount of mass 30 grams of material should be present at the end, according to the Law of Conservation of Mass.

The law of conservation of mass states that no atoms are misplaced or made in a chemical reaction. as an alternative, the atoms be a part of collectively in distinctive methods to form merchandise.

The law of conservation of mass states that during a chemical reaction mass is neither created nor destroyed. for example, the carbon atom in coal turns into carbon dioxide when it's miles burned. The carbon atom changes from a stable structure to gas but its mass does not trade.

The law of conservation of mass states that depend cannot be created or destroyed in a chemical response. for example, when wooden burns, the mass of the soot, ashes, and gases equals the unique mass of the charcoal and the oxygen when it first reacted. So the mass of the product equals the mass of the reactant.

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Answer:

No because their children shouln't be born into this world as if they were a customized doll. It could also be seen as sexualizing them because they would most likely make them to their liking which is inappropriate because you dont know what the parents intentions could be with it.

The molar solubility of CuS is 7.8 x 10⁻¹⁹ M.

The solubility product expression for CuS is:

Since CuS dissociates to form one Cu²⁺ ion and one S²⁻ ion, the molar solubility of CuS can be represented as x. Therefore, at equilibrium:

Substituting these values in the Ksp expression:

Solving for x, we get:

Therefore, the molar solubility of CuS is 7.8 x 10⁻¹⁹ M.

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The complete question is:

Calculate the molar solubility, S , of CuS if it has a Ksp value of 6×10⁻³⁷.

Answer: oxide

Explanation:..

Answer:

weight is the measure of force put onto an object

like if you put a heavier object on something elastic, it would go lower because weight creates force and friction

The partial pressure of hydrogen gas in the mixture is:0.53334 atm.

What is partial pressure?
partial pressure. noun. the pressure that a gas would have if it took up the entire volume that the mixture of gases currently occupies.

P total,and the total number of moles in the mixture would be,

n total=nH2+nO2+nN2

This means that you could write

P total.V=ntotal.RT

p total=n total.RT/V

This is equivalent to

p total=[nH2+nO2+nN2].RT/V

therefore,

P total =PH2+PO2+pN2

now, to get the partial pressure of,let's Hydrogen gas

RT/V=Ptotal/nH2+nO2+nN2

therefore

PN2=XN2.p total

similarly,

PO2=XO2.p total

PN2=XN.p total

The total number of moles will be

\(ntotal\) = 0.40+0.15+0.50=1.50 moles.

\(PH2\)=0.40/1.50*2.0 atm =0.5334 atm.

\(PO2\)=0.15/1.50*2.0 atm =0.2 atm.

\(PN2\)=0.50/1.50*2.0 atm =0.66 atm.

Hence, the partial pressure of hydrogen gas in the mixture is:0.53334 atm.

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Answer: The molar mass is the mass of a given chemical element or chemical compound (g) divided by the amount of substance (mol).

Explanation: The molar mass of a compound can be calculated by adding the standard atomic masses (in g/mol) of the constituent atoms.

Equivalent of sulfate ions = 4.00 EqSO4^2−

1) equivalent weight of sulphate = 96.0636/2

= 48.0328 g/equivalent

2) 2 mol will have a weight of = 2*96.0636 = 192.1272 g

3) so the equivalent of sulphate = 192.1272/48.0328

= 3.999, round it up to 4.00

Sulfur and oxygen atoms make up the majority of the sulphate ion's mass. Four oxygen atoms that are placed at equal intervals around the centre sulphur atom in this structure. In terms of bonding, two oxygen atoms make S=O bonds, while the other two form S-O bonds.

Two double bonds and two single bonds with O make up the more stable structure of SO42-, however the single bonds have a formal charge of -1.

An insoluble precipitate of BaSO4 forms when aqueous barium chloride is introduced to an acidified solution containing sulphate ions. This precipitate, which is white in colour, proves that sulphate is present.

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Answer:

c

Explanation:

mass is like matter it cant be created or destroyed so mass stays the same.

Answer:

C) It is equal to the total mass of the products.

Explanation:

The law of conservation of mass states that in a chemical reaction, the total mass of reactants is equal to the total mass of products.

- profparis

According to what is known about bases, the correct answer to this question is B. They are slippery.

Answer:

a. TFinal = -6.57°C

b. Tfinal = 101.80°C

Explanation:

When a solute is added to a solvent producing an ideal solution, the freezing point of the solution decreases with regard to pure solvent. Also, boiling point increases with regard to pure solvent.

The formulas are:

Freezing point:

ΔT = Kf×m×i

Where Kf is freezeing point depression constant of water (1.86°C/m), m is molality of solution and i is van't Hoff factor (1 for ethylene glycol).

Boiling point:

ΔT = Kb×m×i

Where K is freezeing point depression constant of water (0.51°C/m), m is molality of solution and i is van't Hoff factor (1 for ethylene glycol).

Moles of 21.4g of ethylene glycol (Molar mass: 62.07g/mol) are:

21.4g C₂H₆O₂ ₓ (1mol / 62.07g) = 0.345 moles

And kg of 97.6mL of water = 97.6g are 0.0976kg. Molality of the solution is:

0.345mol / 0.0976kg = 3.5325m

Replacing in the formulas:

a. Freezing point:

ΔT = 1.86C/m×3.5325m×1

ΔT = 6.57°C

0°C - Tfinal = 6.57°C

b. Boiling point:

ΔT = 0.51°C/m×3.5325m×1

ΔT = 1.80°C

Tfinal - 100°C = 1.80°C

Answer:

M= ml - 45301 - 11.59M.

Explanation:

In 250 mL of volumetric flask add 0.975875 grams of \(CaF_2\) and dissolve it in the 250 mL of water.

Given:

The solid of calcium fluoride.

To prepare:

The 250 mL solution of 0.100 M of fluoride ions from solid calcium fluoride.

Method:

Molarity of the fluoride ion solution needed = M = 0.100 M

The volume of the fluoride ion solution needed = V = 250 mL

\(1 mL = 0.001L\\V=250 mL=250\times 0.001 L=0.250 L\)

The moles of fluoride ion needed = n

According to the definition of molarity:

\(M=\frac{n}{V}\\0.100M=\frac{n}{0.250 L}\\n=0.100M\times 0.250 L=0.025 mol\)

Moles of fluoride ion = 0.025 mol

We know that solid calcium fluoride dissolves in water to give calcium ions and fluoride ions.

\(CaF_2(s)\rightarrow Ca^{2+}(aq)+2F^-(aq)\)

According to reaction, 2 moles of fluoride ions are obtained from 1 mole of calcium fluoride, then 0.025 moles of fluoride ions will be obtained from:

\(=\frac{1}{2}\times 0.025 mol=0.0125 \text{mol of } CaF_2\)

Moles of calcium fluoride = 0.0125 mol

Mass of calcium fluoride needed to prepare the solution :

\(=0.0125 mol\times 78.07 g/mol=0.975875 g\)

Preparation:

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Answer:

Alkali metals have the lowest, and just in case, halogens have the highest

Explanation:

Report if wrong

Answer:

Alkali Metals

Explanation:

Most noble gases don't form compounds.

Answer: the concentration of the solution after the dllution is 0.050 M

Explanation:

The question requires us to determine the concentration of a solution after a dilution process.

The following information was provided by the question:

Concentration of initial solution = C1 = 0.30 M

Volume taken from initial solution = V1 = 25.0 mL

Volume of final solution = V2 = 150 mL

When a solution is diluted, the number of moles of the substance remains the same. Thus, we can say that:

where (1) is the solution before dilution and (2) refers to the solution after dilution.

The number of moles of a substance can be obtained from the definition of molarity (or molar concentration), where the number of moles is divided by the volume of the solution:

Thus, we can write that for a dillution:

From the expression above we can determine the concentration of the final solution. Applying the values provided by the question to the expression above, we'll have:

Therefore, the concentration of the solution after the dilution is 0.050 M.

There are 288 possible isomers for 3-chlorobutan-2-ol.

There are 7200 possible isomers for 4-chloropentane-2,3-diol.

3-chlorobutan-2-ol:

Total number of isomers = number of possible arrangements of carbon atoms x number of possible arrangements of other atoms

= 24 x 12

= 288

Therefore, there are 288 possible isomers for 3-chlorobutan-2-ol.

4-chloropentane-2,3-diol:

Total number of isomers = number of possible arrangements of carbon atoms x number of possible arrangements of other atoms

= 120 x 60

= 7200

Therefore, there are 7200 possible isomers for 4-chloropentane-2,3-diol.

Isomers are molecules with the same molecular formula (the same amount and types of atoms) but different atomic arrangements in space. This means that despite having the same molecular formula, isomers have distinct physical and chemical properties.

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Answer:

Use a scientific calculator