The enthalpy of fusion of methanol (CH3OH) is 3.16 kJ/mol. How much heat would be absorbed or released upon freezing 25.6 grams of methanol?

Answer: only H2O is a compound

Explanation:

Answer:

1. Reduce

2. Oxidize

3. Oxidize

4. Reduce

5. Oxidize

6. Reduce

Explanation:

Answer:

1. Reduce, Oxidize

2.Oxidize, reduce

3.Reduce, Oxidize

Explanation:

Answer:

224 grams of O₂

Explanation:

This is the reaction:

2 C₂H₆ (g) + 7 O₂(g) → 4 CO₂ (g) + 6 H₂O (g)

2 moles of ethene react with 7 moles of oxygen.

Let's convert the ethene's mass into moles (mass / molar mass)

60 g / 30 g/m = 2 mol

So, if 2 moles of ethene must react with 7 moles of O₂ and we have 2 moles, obviously we would need 7 moles of oyxgen.

Let's convert the moles to mass ( mol . molar mass)

7 m . 32 g/m = 224 grams

Answer: Mass of O2 is 224g

Explanation:

From the equation of the reaction, Ethane to oxygen is 2 to 7

While the molar masses respectively gives 30 and 32 g/mol

2 moles of methane gives 7 moles of O2;

60g/30=2moles for methane gives x/32 for O2;

Cross multiplying yields

7 x 2 = 2 x mass of O2/32

14 = 2x/32

Final answer gives 224 grams of oxygen

Answer:

The correct answer is option b.

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

[tex]\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i[/tex] .....(1)

We are given:

Mass of isotope X-28 = 27.979 amu

Percentage abundance of isotope X-28 = 92.21%

Fractional abundance of isotope X-28 = 0.9221

Mass of isotope X-29 = 28.976 amu

Percentage abundance of isotope X-29 = 4.70%

Fractional abundance of isotope X-29 = 0.047

Mass of isotope X-30= 29.974 amu

Percentage abundance of isotope X-30 = 3.09%

Fractional abundance of isotope X-30 = 0.0309

Putting values in equation 1, we get:

[tex]\text{Average atomic mass of X}=\sum[(27.979 amu\times 0.9221)+(28.976 amu\times 0.047)+(29.974 amu\times 0.0309)][/tex]

[tex]\text{Average atomic mass of X}=28.09 amu[/tex]

The atomic weight of X is 28.09 amu.

The atomic weight of element X is calculated by considering the masses and abundances of its isotopes. By multiplying each isotope's mass by its abundance and summing up the products, the atomic weight of X can be determined to be 28.985 amu. Closest option to answer is option a.

The atomic weight of an element is calculated by taking into account the masses and abundances of its isotopes. In this case, element X exists in three isotopic forms with different masses and abundances. To calculate the atomic weight of X, we multiply the mass of each isotope by its abundance and sum up the products.

(27.979 amu * 0.9221) = 25.699 amu

(28.976 amu * 0.047) = 1.361 amu

(29.974 amu * 0.0309) = 0.925 amu

Summing up the products:

25.699 amu + 1.361 amu + 0.925 amu = 28.985 amu

Therefore, the atomic weight of X is 28.985 amu.

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Answer:

a. b + s = 26

b. b = 26 — s

Explanation:

Answer:

Explanation:

Because most of the volume occupied by the substance is empty space.

Answer:

He have higher pressure at room temperature.

Explanation:

It is given that both the gases are kept in 5 L chambers.

Therefore, volume is constant.

Also, they both are at room temperature, so temperature is also constant.

Now, number of moles of [tex]O_2[/tex] = [tex]\dfrac{Given\ mass}{Molecular \ mass}=\dfrac{5}{32}=0.16\ mol.[/tex]

Also, number of moles of He =[tex]\dfrac{Given\ mass}{Molecular \ mass}=\dfrac{5}{4}=1.25\ mol.[/tex]

Now, according to GAS LAW,

[tex]PV=nRT[/tex]  ( all terms have their usual meaning).

In this case, V, R and T are constant.

So, pressure is directly proportional to n i.e number of moles.

So, moles of He is more than moles of [tex]O_2[/tex].

Therefore, He have higher pressure at room temperature.

Hence , this is the required solution.

The gas that has a higher pressure is He.

Number of moles of oxygen gas = 5.00 g /32 g/mol = 0.156 moles

From PV = nRT

P = ?

V = 5 L

n =  0.156 moles

T = 25 + 273 = 298 K

R = 0.082 atmLK-1mol-1

P = nRT/V

P =  0.156 moles × 0.082 atmLK-1mol-1 × 298 K/5 L

P = 0.76 atm

Number of moles of He = 5/4 g/mol = 1.25 moles

P = ?

V = 5 L

n = 1.25 moles

T = 25 + 273 = 298 K

R = 0.082 atmLK-1mol-1

P = nRT/V

P =1.25 moles × 0.082 atmLK-1mol-1 × 298 K/5 L

P = 6.11 atm

The gas that has a higher pressure is He.

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Answer:

[C₆H₁₂O₆] = 0.17M

Explanation:

3% by mass, means 3 g of solute in 100 g of solution.

Density is mass / volume. This data always refers to solution.

Solution density = Solution mass / Solution volume

1.0097 g/mL = 100 g / Solution volume

Solution volume = 100 g / 1.0097 g/mL → 99.03 mL

Let's convert the mL to L, for molarity (mol/L)

99.03 mL = 0.09903 L

Now we have to find out the moles. Let's calculate them with the molar mass

(mass / molar mass)

3 g / 180 g/mol = 0.0166 mol

Molarity is mol/L → 0.0166 mol/0.09903 L → 0.17 M

The molarity of dextrose in a 3.00% by mass aqueous solution, with a density of 1.0097 g/mL, is calculated to be approximately 0.168 M by dividing the number of moles of dextrose (0.01665 mol) by the volume of the solution in liters (0.09903 L).

To calculate the molarity of dextrose in the solution, we'll first need to understand that molarity is defined as the number of moles of solute per liter of solution. Here, the solution is 3.00% by mass dextrose. This means that in 100 grams of the solution, there are 3 grams of dextrose.

Using the density of the solution (1.0097 g/mL), we'll calculate the volume that 100 grams of the solution occupies:

100 grams of solution contains 3 grams of dextrose. With the molar mass of dextrose (C6H12O6) being 180.16 g/mol, we calculate the number of moles in those 3 grams:

To find the molarity, we need the volume in liters:

Thus, the molarity (M) is:

The molarity of dextrose in the solution is approximately 0.168 M.

Answer:

Proton: m= 1.6726x10⁻²⁷ kg

Neutron: m= 1.6749x10⁻²⁷ kg

Electron: m= 9.1164x10⁻³¹ kg

Explanation:

We can calculate the mass of a proton, neutron, and electron using the following data:

mass of proton: 1.00728 amu

mass of neutron: 1.00867 amu

mass of electron: 5.49x10⁻⁴ amu

1 amu = 1.66054x10⁻²⁷ kg

Now, the mass of a proton, neutron, and electron in kilograms can be calculated using the next relation:

[tex] m = 1.66054 \cdot 10^{-27} \frac{kg}{amu} \cdot particle's mass (amu) [/tex]

For the proton:

[tex] m = 1.66054 \cdot 10^{-27} \frac{kg}{amu} \cdot 1.00728 amu = 1.6726 \cdot 10^{-27} kg [/tex]

For the neutron:    

[tex] m = 1.66054 \cdot 10^{-27} \frac{kg}{amu} \cdot 1.00867 amu = 1.6749 \cdot 10^{-27} kg [/tex]

For the electron:

[tex] m = 1.66054 \cdot 10^{-27} \frac{kg}{amu} \cdot 5.49x10⁻⁴ amu = 9.1164 \cdot 10^{-31} kg [/tex]    

I hope it helps you!

The masses of proton, neutron, and electron are approximately 1.6726 x 10-27 kg, 1.6749 x 10-27 kg and 9.109 x 10-31 kg, respectively. This is calculated by multiplying each particle's mass in Atomic Mass Units by the factor 1.66054 × 10-27 kg/amu.

To calculate the mass of a proton, neutron, and electron in kilograms, we must know their masses in Atomic Mass Units (amu). We know that 1 amu = 1.66054 × 10

-27 kg. Using this we can get the masses as:

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If an evaluating committee places the burden of proof of the safety of a new chemical on the manufacturer of the chemical, then the committee is using the precautionary principle

Explanation:

This is a method that is used to handle the problems that are related in affecting any person. This principle mainly aims in determining the safety of any new thing before it is released to the public. This will test whether the new thing does not make harm or destroy anything before it is introduced to public.

For instance let us take an example of a medicine introduction. Any new tablets when it is discovered must be tested. Without any testing, releasing it to public will only have adverse effect on them. The given scenario also relates with the precautionary principle since, the safety is considered as an important one by the evaluating committee.  

Answer:

The value of Kp is 4

Explanation:

Step 1: Data given

The initial pressure of N2 = 1 atm

The initial pressure of  H2 = 2 atm

Temperature = 200 °C

At equilibnrium, the toal pressure is 2 atm

Step 2: The balanced equation

N2(g) + 3H2(g) → 2NH3(g)

Step 3: The initial pressures

N2 = 1.0 atm

H2 = 2.0 atm

NH3 = 0 atm

Step 4: The partial pressures at equilibrium

For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3

There will react X moles of N2,

There will react 3X moles of H2

There will be produced 2X moles NH3

The partial pressure of N2 at the equilibrium is (1.0 -X)atm

The partial pressure of H2 at the equilibrium is (2.0 - 3X)atm

The partial pressure of NH3 at the equilibrium is 2X

The total pressure at the equilibrium = 2.0 atm

The total pressure = pN2 + pH2 +pNH3 = 2.0 atm

2.0 atm = (1.0-X) + (2.0 - 3X) + 2X

2.0 =3.0 -2X

X = 0.50

The partial pressure of N2 at the equilibrium is (1.0 -0.50) = 0.50 atm

The partial pressure of H2 at the equilibrium is (2.0 - 3*0.5) = 0.50 atm

The partial pressure of NH3 at the equilibrium is 2*0.5 = 1.0 atm

Step 5: Calculate Kp

Kp = pNH3 / (pN2)*(pH2)

Kp = 1.0 / (0.5*0.5)

Kp = 4

The value of Kp is 4

Answer:

The answer to your question is a heterogeneous mixture

Explanation:

The Matter is classified as Pure substances and Mixtures

Pure substances are elements or compounds which are not mixed with more substances.

Mixtures are several elements or compounds together, mixtures can be homogeneous or heterogeneous.

Homogeneous mixtures are when the different components can be identified.

Heterogeneous mixtures are when the different components can be identified just by looking at the mixture.

In concrete we can identify the different components like gravel, sand, crushed rocks, etc, so concrete is a heterogeneous mixture.

Answer: The final pressure in the vessel will be 0.965 atm

Explanation:

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]      .....(1)

Given mass of phosphorus trichloride = 20.0 g

Molar mass of phosphorus trichloride = 137.3 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of phosphorus trichloride}=\frac{20.0g}{137.3g/mol}=0.146mol[/tex]

Given mass of oxygen gas = 3.15 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of oxygen gas}=\frac{3.15g}{32g/mol}=0.098mol[/tex]

The chemical equation for the reaction of phosphorus trichloride and oxygen gas follows:

[tex]2PCl_3+O_2\rightarrow 2POCl_3[/tex]

By Stoichiometry of the reaction:

2 moles of phosphorus trichloride reacts with 1 mole of oxygen gas

So, 0.146 moles of phosphorus trichloride will react with = [tex]\frac{1}{2}\times 0.146=0.073mol[/tex] of oxygen gas

As, given amount of oxygen gas is more than the required amount. So, it is considered as an excess reagent.

Thus, phosphorus trichloride is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of phosphorus trichloride produces 2 moles of [tex]POCl_3[/tex]

So, 0.146 moles of phosphorus trichloride will produce = [tex]\frac{2}{2}\times 0.146=0.146mol[/tex] of [tex]POCl_3[/tex]

To calculate the pressure of the vessel, we use the equation given by ideal gas follows:

[tex]PV=nRT[/tex]

where,

P = pressure of the vessel = ?

V = Volume of the vessel = 6.00 L

T = Temperature of the vessel = [tex]210^oC=[210+273]K=483K[/tex]

R = Gas constant = [tex]0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]

n = number of moles = 0.146 moles

Putting values in above equation, we get:

[tex]P\times 6.00L=0.146mol\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 483K\\\\P=\frac{0.146\times 0.0821\times 483}{6.00}=0.965atm[/tex]

Hence, the final pressure in the vessel will be 0.965 atm

Answer:

The final pressure in the vessel is 1.13 atm

Explanation:

Step 1: Data given

Volume of the vessel = 6.00 L

Mass of PCl3 = 20.0 grams

Mass of O2 = 3.15 grams

Temperature = 15.0 °C

The vessel is heated to 210°C

Molar mass of PCl3 = 137.33 g/mol

Step 2: The balanced equation

2PCl3 + O2 → 2POCl3

Step 3: Calculate moles PCl3

MolesPCl3 = mass PCl3 / molar mass PCl3

Moles PCl3 = 20.0 grams / 137.33 g/mol

Moles PCl3 =0.146 moles PCl3

Step 4: Calculate moles O2

Moles O2 = 3.15 grams/ 32.0 g/mol

Moles O2 = 0.0984 moles O2

Step 5: Calculate the limiting reactant

PCl3 is the limiting reactant. It will completely be consumed(0.146 moles). So at completion there is no PCl3 remaining.

O2 is in excess. There will react 0.146/2 = 0.073 moles. There will remain 0.0984 - 0.073 = 0.0254 moles O2

Step 6: Calculate moles POCl3

For 2 moles PCl3 we need 1 mol O2 to produce 2 moles POCl3

For 0.146 moles PCl3 we'll have 0.146 moles POCl3

Step 7: Calculate final pressure

p*V = n*R*T

p = (n*R*T)/V

⇒ with n = the number of moles = 0.146 moles of POCl3 produced + 0.0254 moles O2 remaining = 0.1714 moles gas

⇒ with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temperature = 210 +273 = 483 Kelvin

⇒ with V = the volume = 6.00 L

p = (0.1714 *0.08206 * 483) / 6.00

p = 1.13 atm

The final pressure in the vessel is 1.13 atm

Answer:

Mg₃N₂ + 6H₂O  →  2NH₃  +  3Mg(OH)₂

Coefficient of water is 6 (option 3)

Explanation:

The reaction is:

Mg₃N₂ + H₂O  →  NH₃  +  Mg(OH)₂

Let's balance the reaction.

In reactant side we have 3 Mg, therefore in product side, we add 3 Mg to the hydroxide.

This change, modified the hydroxide, so now we have 6 O and 6 H, but we have in total 9 H (6 from the hydroxide + 3 from the ammonia)

As we have 2N, in reactant side, we must add 2 N to the ammonia, so now

we have 12 H in product side . We must complete with 6, the water so the H are ballanced.

In reactant side we have 6 O, therefore in product we must have 6 O (two O, in the OH but we have 3 moles, so in total we have 6 O) - BALANCED

The balance reaction is:

Mg₃N₂ + 6H₂O  →  2NH₃  +  3Mg(OH)₂

The chemical reaction of magnesium nitride with water to form ammonia and magnesium hydroxide has a coefficient of 6 for water in the balanced equation.

Preparing ammonia through the reaction of magnesium nitride with water yields ammonia and magnesium hydroxide. The balanced chemical equation for this reaction is: Mg3N2 + 6H2O → 3NH3 + 3Mg(OH)2. Thus, the coefficient of water in the balanced chemical equation is 6, which corresponds to option number 3 in your question.

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Answer:

C₃H₈  +  5O₂  →  3CO₂  +  4H₂O

Option D.

Explanation:

This is a combustion reaction, where a compound reacts with oxygen to produce CO₂ and water.

You have to look at the reactants and the products.

C₃H₈  +  O₂  →  CO₂  +  H₂O

To balance you must have the same atoms of each elements in both sides.

Actually we have 3 C, 8 H and 2 O in reactant side and 1 C, 3 O and 2H in product side.

We can add 4 to water to have 8 H in product side to balance the H with reactants but we modified the amount of oxygens. Now, we have 4 O in water and 2 O from CO₂ (6 in total).

In reactant side, we have 3 C, therefore we add 3 to CO₂ and now, we have 3 C on both sides and 8 H in both sides, so, as in product side we have 10 O, 6 from CO₂ and 4 from water, we must add a 5 to balance in reactant side.  The balance equation will be:

C₃H₈  +  5O₂  →  3CO₂  +  4H₂O

Answer:

Option 5. 21.7 %

Explanation:

Let's think the reaction:

2Al (s) + 6HCl (l) → 3H₂ (g) + AlCl₃ (aq)

We have the data of hydrogen, we formed so let's apply the Ideal Gases Law equation to solve the amount of gas.

P . V = n . R . T

1.03 atm . 1.53L = n . 0.082 L.atm . 298K

(1.03 atm . 1.53L) / (0.082 L.atm . 298K) = n → 0.0645 moles of H₂

In the reaction, 3 moles of H₂ are produced by 2 moles of Al. Let's determine, how many moles of Al produced, the 0.0645 moles of H₂.

3 moles of H₂ were produced by 2 moles of Al

Then, 0.0645 moles of H₂ would be produced by (0.0645 .2)/3 = 0.043 moles.

If we convert the moles to mass, we can know the mass of Al in the alloy. (mol . molar mass)

0.043 m . 26.98 g/mol = 1.16 g

As the sample had a mass of 5.34 g, let's determine the % of Al.

(1.16 g / 5.34 g) . 100 = 21.7%

Answer: 2,625.3 g AlCl3

Explanation: solution attached:

First balance the chemical equation then do basic stoichiometry.

Answer:

(a). Biological fixation

(b). Fixation by lightning

Explanation:

Gaseous nitrogen in the atmosphere has to be converted or "fixed" into a suitable form before it can be utilised by living organisms.

There two main ways of nitrogen fixation are

(a). Biological fixation: Majority (approximately 90%) of nitrogen fixation is carried out by bacteria. Bacterias such as Cyanobacteria transforms nitrogen into ammonium and ammonia : N2 + 3 H2 → 2 NH3. The produced ammonia can then be taken in directly by plants and /or the conversion products of ammonium and ammonia may further react in the process of nitrification.

(b). Fixation by lightning: Lightning energy causes the combination nitrogen (N2) and water (H2O) forming nitrates (NO3) and ammonia (NH3) . Rain water dissolves the formed nitrates and ammonia and the solution is drained into the ground, where they can be reached by plants plant roots for consumption.

Biological nitrogen fixation and lightning are the major natural processes responsible for converting atmospheric nitrogen to useable forms.

Nitrogen gas goes into the soil from the atmosphere, and nitrogen fixing bacteria convert this nitrogen to ammonium ions (NH4+), which can be used by plants.

Whereas, Lightning converts atmospheric nitrogen into ammonia and nitrate (NO3) that comes to the soil with the help of rainfall.

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Answer: B

Double replacement reaction

Explanation:

Answer:

1.4moles

Explanation:

From the balanced equation below;

2H2 + O2 → 2H2O

2moles of hydrogen reacts with 1mole of oxygen to produce 2moles of water.

since we are only interested in comparing hydrogen and water, we can say 2moles of hydrogen will produce 2moles of water

Hence by comparism of moles;

1.4moles of hydrogen will produce 1.4moles of water

In the reaction 2H2 + O2 → 2H2O, the formation of water from hydrogen gas is a 1:1 ratio. Thus, 1.4 moles of hydrogen gas would produce 1.4 moles of water.

Analyzing balanced chemical equations is a fundamental approach to determine the stoichiometry of a chemical reaction. In this case, the equation 2H2 + O2 → 2H2O provides crucial information regarding the mole ratios of reactants and products. It signifies that 2 moles of hydrogen gas (H2) react completely with 1 mole of oxygen gas (O2) to yield 2 moles of water (H2O).

Therefore, when 1.4 moles of hydrogen gas are fully consumed in the reaction, the stoichiometry reveals that an equivalent amount, 1.4 moles, of water will be generated. This clear and direct correlation underscores the predictive power of balanced chemical equations in understanding the quantities of reactants and products in a chemical reaction.

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Answer:

HA +  KOH  →  KA  +  H₂O

Explanation:

The unknown solid acid in water can release its proton as this:

HA  +  H₂O  →  H₃O⁺  +  A⁻

As we have the anion A⁻, when it bonded to the cation K⁺, salt can be generated, so the reaction of HA and KOH must be a neutralization one, where you form water and a salt

HA +  KOH  →  KA  +  H₂O

It is a neutralization reaction because H⁺ from the acid and OH⁻ from the base can be neutralized as water

Answer:

Partia pressure N₂ → 0.975 atm

Explanation:

Let's analyse the moles fractions:

N₂ → 0.25

O₂ → 0.65

He → 0.1

Partial pressure / Total pressure = Mole fraction

Partial pressure N₂ / 3.9 atm = Mole fraction N₂

Partial pressure N₂ / 3.9 atm = 0.25

Partial pressure N₂ = 3.9 atm . 0.25 → 0.975 atm

The pressure of nitrogen in the mixture has been 0.975 atm.

Partial pressure has been defined as the pressure exerted by the gas molecules in the mixture. The partial pressure ([tex]P_A[/tex])  has been expressed as:

[tex]P_A=X_A\;\times\;P[/tex]

Where,  Mole fraction of the element A, [tex]X_{N_2}=0.25\;\rm atm[/tex]

The total pressure of the mixture, [tex]P=3.9\;\rm atm[/tex]

Substituting the values, the pressure of nitrogen ([tex]P_{\rm N_2}[/tex]),

[tex]P_{\rm N_2}=0.25\;\times\;3.9\;\rm atm\\ \textit P_{N_2}=0.975\;atm[/tex]

The pressure of nitrogen in the mixture has been 0.975 atm.

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Answer: B. Rusting of the metal is a chemical change.

Explanation: Rusting is considered a chemical change since it involves a change in the composition of iron through oxidation. Since there is a presence of oxygen and moisture it weakens the bonds of iron molecules to react and form iron oxide another substance in the chemical process.

Answer:

covalent bond

Explanation:

Covalent bond -

It is the type of interaction observed between two species , which share the electrons in order to attain stability , is referred to as covalent bond.

The shared electrons are referred to as the bonding pairs or the shared pairs .

Stability and completion of the octet is the driving force for the formation covalent bond.

The molecules of the organic compound usually shows this type bonding .

Like the bonds between - carbon , oxygen and hydrogen are covalent bonds.

Answer:

A. increases

Explanation:

According to the Gay-Lussac's law:-

Thus, at constant volume and number of moles, Pressure of the gas is directly proportional to the temperature of  the gas.

P ∝ T

Also,  it can be written as:-

[tex]\frac {P_1}{T_1}=\frac {P_2}{T_2}[/tex]

Thus, if the temperature is increased, the pressure exerted by the gas also increases.

Answer:

The answer is D

Explanation:

Law of Definite Proportions states that a given compound always contain its component elements in fixed ratio (by mass) and does not depend on its source and/or method of preparation.  

If one is to explain the law of definite proportions properly, then Dalton's fourth atomic theory comes to mind which states that atoms chemically combine with other atoms in fixed, whole number ratios.

The law of definite proportions suggests that compounds always contain its component elements in fixed ratio no matter the source, for example, oxygen makes up 8/9 of the mass of the sample of pure water regardless of the source while hydrogen makes up the remaining 1/9 of the mass. This generally shows that when oxygen and hydrogen combine in the ratio 8:1, water molecule will be formed as suggested by the Dalton Atomic Theory.

The Dalton's atomic theory postulate explaining the Law of Definite Proportions is 'Atoms chemically combine with other atoms in fixed, whole-number ratios.' This principle underlies consistent elemental proportions in compounds, such as the 1:8 ratio of hydrogen to oxygen in water.

The postulate of Dalton's atomic theory that is vital to explain the observations summarized by the Law of Definite Proportions is: 'Atoms chemically combine with other atoms in fixed, whole-number ratios.'

This is crucial because the Law of Definite Proportions states that a given compound, regardless of the source or preparation method, will always consist of the same elements in the same proportion by mass.

For example, water (H2O) is always composed of 2 hydrogen atoms and 1 oxygen atom (fixed, whole-number ratio) regardless of the source. In water, the ratio of the mass of hydrogen to oxygen is always 1:8, which corroborates the Law of Definite Proportions.

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HF and CH3OH exhibit dipole-dipole forces when dissolved in water because they both can form hydrogen bonds. CaCl2 and FeBr3 display ion-dipole forces in water because they are ionic compounds.

When these substances are added to water, they exhibit different solute-solvent interactions due to their different properties.

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Answer:

Diamond is burnt

Diamond (Carbon)+ Oxygen → CO2

Water electrolysed

2H2O → 2H2 + O2

Explanation:

Dalton atomic theory states that in a chemical reaction, the atoms of the elements join together or combine in simple whole number ratios

Hence when, at very high temperatures, diamond which consists of carbon is burnt we have

Diamond (Carbon)+ Oxygen → CO2

Also when water is electrolysed it decomposes as follows

2H2O → 2H2 + O2

Answer:

B. 0.075 m (NH4)3PO4

Explanation:

Our strategy here is to recall the van´t Hoff factor, i, for the colligative properties of electrolyte solutions which appears as the consequence that electrolytes disociate completely in their solutions in water.

Thus in this problem we need to determine i and then realize the one with the lowest freezing point will have the biggest  i ( all the concentrations are equal) since

ΔTf = i m Kf

Substance  van´t Hoff factor

Li I                             2

(NH4)3PO4              4

NaIO4                       2

KCN                          2

KNO2                       2

The correct answer is B. 0.075 m (NH4)3PO4

The aqueous solution with the lowest freezing point is 0.075 m (NH4)3PO4 because it dissociates into the highest number of particles, leading to the greatest freezing point depression.

To find the aqueous solution with the lowest freezing point, we need to determine the effective concentration of solute particles after dissociation. Freezing point depression is larger for solutions with a higher number of dissolved particles. Therefore, we must consider the van't Hoff factor (i), which is the number of particles a compound dissociates into in solution. For example, LiI (i=2), NaIO4 (i=1), and KCN (i=2) will produce fewer particles than (NH4)3PO4, which will dissociate into four particles (i=4).

Given that the molal concentrations (m) are the same, the solution with the highest van't Hoff factor will have the highest concentration of particles and therefore, the lowest freezing point. Hence, solution B, 0.075 m (NH4)3PO4, with an effective concentration of 0.3 m (0.075 m × 4), will have the lowest freezing point.

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Answer:

Serine will be on the exterior of the globular protein while leucine on the interior of the globular proteins

Explanation:

The nature or solubility of the side cham determines the poition of amino acid on the globular protein and it is either hydrophilic or hydrophobic.

Serine is an hydrophilic amino acid and so it is position on the surface of the globular protein (Exterior)

While Leucine side chain is hydrophobic in nature is positioned on the interior of the globular protein.

Leucine, a non-polar amino acid, tends to be in the interior of globular proteins while serine, a polar amino acid, is often found on the surface.

In an aqueous environment of globular proteins, the position of an amino acid is influenced by the polar or non-polar nature of its R group or side chain. Leucine, with a non-polar R group, prefers to stay in the interior part of the protein, away from the water, because non-polar R groups are hydrophobic. On the other hand, serine, with a polar R group, tends to stay on the surface, close to the water, due to its hydrophilic or water-attracting nature.

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