Find the molarity of 194. 55 g of sugar (C12H22O11) in 250. ML of water

The reaction performed with the elephant toothpaste demonstration is known as a decomposition reaction.

The process of breaking down a chemical compound into smaller molecules, atoms, or ions is known as a decomposition reaction. It is also known as analysis or disintegration. A reaction in which a single substance is broken down into two or more simpler substances is known as a decomposition reaction. The elephant toothpaste demonstration is a simple chemical reaction in which hydrogen peroxide breaks down into oxygen gas and water in a matter of seconds.

The formula for hydrogen peroxide is H₂O₂. It is a pale blue liquid that contains hydrogen, oxygen, and water. When you add yeast, soap, and food coloring, the reaction is more exciting. The yeast acts as a catalyst, breaking down hydrogen peroxide into water and oxygen gas. The oxygen gas created causes the soap to foam up, creating the "elephant toothpaste" effect. The chemical reaction that takes place during the elephant toothpaste demonstration can be written as follows:

2H₂O₂(liquid) → 2H₂O (liquid) + O₂ (gas)

This reaction is an example of a decomposition reaction.

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The correct name for the given compound S4P2 is- option B( tetrasulfur diphosphide).    

When two or  further  rudiments chemically combine in a fixed  rate by mass, the  attained product is known as a  emulsion. Composites can be defined as substances  conforming of 2 or  further different types of  rudiments in a fixed  rate of their  tittles. When the  rudiments combine, some individual property of the  rudiments is lost and the  recently formed  emulsion has new  parcels. composites are represented by their chemical formula. A chemical formula is a emblematic  representation of the proportions of  tittles that constitute a particular chemical  emulsion.  The prefix" tetra-" indicates that there are four sulfur  tittles, and the suffix" ide" indicates that the  emulsion contains two phosphorus  tittles.

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Answer:

electrons

Explanation:

Electrons are the negatively charged atoms that orbit  around the nucleus of an atom

The average speed is approximately 19.0 kilometers per hour.

When we say the average speed is approximately 19.0 kilometers per hour, it means that over a given time period, the object or person in question traveled a total distance of 19.0 kilometers divided by the total time it took to cover that distance. In this case, the given speed is 19 kilometers per hour.

To calculate the average speed, we use the formula:

Average Speed = Total Distance / Total Time

In the given question, the speed is already mentioned as 19 kilometers per hour. So, we can conclude that the total distance covered is 19 kilometers.

However, to find the total time taken, we need more information. The question only provides the speed but not the duration of the journey. Without knowing the total time, we cannot determine the average speed accurately. Therefore, in the absence of further information about the time, we can only state that the average speed is approximately 19.0 kilometers per hour based on the given speed.

Average speed and how it is calculated by considering the total distance and total time traveled.

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The concentration of A decreases and then levels out. Option A

In many chemical reactions, a reactant is consumed as the reaction progresses, leading to a decrease in its concentration over time. The reactant molecules are transformed into products, and as the reaction proceeds, the concentration of the reactant gradually diminishes.

At equilibrium, the concentrations of both reactants and products remain relatively constant over time, although they can coexist.

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Answer:

1. A

2. B

3. D

4. C

Explanation:

Just trust me. Good luck on your test! (I go to K12 as well.)

Answer:

I dont think so. If you continue to heat them up they will shrink and dissapper r become ashes. Thats a very interesting question btw. :) hope this helps!

Explanation:

The question gives us the reaction to produce sodium sulfide (Na2S) from sodium sulfate (Na2SO4) and carbon (C) and provides the amount of reactants used (15.0 g of Na2SO4 and 7.50 g of C), asking the limiting reactant, the excess reactant and the amount of Na2S produced.

I) The first step for this type of question is checking if the given equation is balanced. For this case, we don't need to adjust the coefficients as the equation is already balanced.

II) Next, we need to calculate the molar mass of the compounds we'll be using. To calculate the molar mass, I'll be using the following atomic masses and considering the number of each atom in the molecules:

Na: 22.99 u

S: 32.07 u

O: 15.99 u

C: 12.01 u

Now, we can calculate the molar mass of Na2SO4, C and Na2S:

Na2SO4: molar mass = (2 * 22.99) + (1 * 32.07) + (4 * 15.99) = 142.01 g/mol

C: molar mass = (1 * 12.01) = 12.01 g/mol

Na2S: molar mass = (2 * 22.99) + (1 * 32.07) = 78.05 g/mol

iii) The third step is to convert the masses given for Na2SO4 and C into the correspondent number of moles using the molar mass of these compounds:

IV) On the forth step, we must define the limiting reactant for this reaction considering the amounts used of each one and the stoichiometric coefficients:

1 mol Na2SO4 reacts with 1 mol C

0.106 mol Na2SO4 reacts with...?

Solving this calculation, we have that we would need 0.106 mol of C to react with 15.0 g of Na2SO4. Since we there are 0.624 mol of C available to react, we can conclude that carbon is the reactant in excess (there is an excess of 0.518 mol) and sodium sulfate is the limiting reactant.

V) At last, we can calculate the amount of Na2S produced from the limiting reactant amount used (0.106 mol of Na2SO4) and the stoichiometric coefficients:

1 mol Na2SO4 ----------1 mol Na2S

0.106 mol Na2SO4 ---- y

Solving for y, we have that 0.106 mol of Na2S will be produced.

vWe can convert this amount into mass of Na2S using its molar mass:

In summary:

Na2SO4 is the limiting reactant and there is an excess of 0.518 mol or 6.22 g of C;

8.27 g of Na2S will be produced from 15.0 g of Na2SO4.

(e) (i) Number average (Mn) = 6,900g/mol

(ii) Mn will remain the same in this case as it does not depend on weight.= 6,900g/mol

(iii) The degree of polymerization (DP) is 4492.

(i) Number average (Mn)

The number average molecular weight (Mn) is the sum of the molecular weights of all polymer chains divided by the total number of polymer chains.

Weight average (Mw)

The weight average molecular weight (Mw) is the sum of the product of the molecular weight and the fraction of the total polymer chains that have that molecular weight.

Using the given formula, let's first calculate the Mn:

(2100 + 6600 + 12000) / 3 = 6,900g/mol

Let's now calculate the Mw:

[(2100 x 2100) + (6600 x 6600) + (12000 x 12000)] / (2100 + 6600 + 12000)= 9966.67g/mol

(ii) Equal weights of chains

Mn will remain the same in this case as it does not depend on weight.

Mw, on the other hand, will change.

The following formula will be used:

2100 x (1/3) + 6600 x (1/3) + 12000 x (1/3) = 6,900g/mol

(iii) The degree of polymerization (DP)

DP refers to the number of repeating units in the polymer chain.

We can calculate the DP using the following formula:

DP = (Mn / M) * NA where M is the molar mass of the repeating unit, and NA is Avogadro's number.

Using the Mn value from part (i) and the given molecular weights for the repeating units, we can calculate the DP:

DP = (6,900 / ((126 + 324 + 300) / 3)) * 6.02 × 1023= 4492.45 (approximately 4492)

Therefore, the degree of polymerization is 4492.

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The weight average molecular weight (Mw) of the polymer will be \($$M_w=\frac{1\times2100^2+1\times6600^2+1\times12000^2{1\times2100+1\times6600+1\times12000}=8,580\;g/mol$$\).

Weight average molecular weight (Mw) of the polymer will be \($$M_w=\frac{1.0587\times2100^2+1.0587\times6600^2+1.0587\times12000^2{1.0587\times2100+1.0587\times6600+1.0587\times12000}=8,825\;g/mol$$\).

The degree of polymerization of the polymer in the first question is 25.01.

Number average (Mn) and weight average (Mw) molecular weights of a polymer are calculated using the following formula:

\($$M_n = \frac{\sum N_iM_i}{\sum N_i}$$\)

and

\($$M_w = \frac{\sum N_iM_i^2}{\sum N_iM_i}$$\)

where Ni is the number of chains with molecular weight Mi.

(i) When the number of chains with molecular weights of 2100, 6600, and 12000 are mixed together:

Molecular weight (M) Number of chains (N) 2100 1 6600 1 12000 1

Total 3

The number average molecular weight (Mn) of the polymer will be:

\($$M_n=\frac{1\times2100+1\times6600+1\times12000}{1+1+1}= 6,900\;g/mol$$\)

The weight average molecular weight (Mw) of the polymer will be:

\($$M_w=\frac{1\times2100^2+1\times6600^2+1\times12000^2}{1\times2100+1\times6600+1\times12000}=8,580\;g/mol$$\)

(ii) When the equal weights of chains are mixed together:

The total weight of the chains is:

2100 + 6600 + 12000 = 20,700 g

Number of chains with molecular weight 2100 = 2100 x n1

Number of chains with molecular weight 6600 = 6600 x n2

Number of chains with molecular weight 12000 = 12000 x n3

So, the total weight of each group of chains will be

n1M1 + n2M2 + n3M3.

Now, we can calculate the values of n1, n2, and n3 as follows:

n1 = 6600 x n2n2 = 12000 x n3M1 + M2 + M3 = 2100 + 6600 + 12000 = 20700n1M1 + n2M2 + n3M3 = 20700

Equating the value of n1 in terms of n2 and n3 and substituting it in the equation above:

n1 = 6600 x n2

\($$\frac{6600}{n1}=n2$$\)

\($$\frac{12000}{n2}=n3$$\)

\($$n1=\frac{20700}{2100+6600+12000}=0.1739$$\)

n2 = 0.0208, n3 = 0.0052

Therefore, the number of chains with molecular weight 2100 = 0.1739 x 2100 / 6600 x 0.1739 x 6600 = 0.0208 x 12000 / 6600 x 0.0208 x 6600 = 0.0052 x 2100 / 6600 x 0.0052 x 12000 ≈ 1.0587

Number average molecular weight (Mn) will be:

\($$M_n=\frac{1.0587\times2100+1.0587\times6600+1.0587\times12000}{1.0587+1.0587+1.0587}= 7,170\;g/mol$$\)

Weight average molecular weight (Mw) of the polymer will be:

\($$M_w=\frac{1.0587\times2100^2+1.0587\times6600^2+1.0587\times12000^2}{1.0587\times2100+1.0587\times6600+1.0587\times12000}=8,825\;g/mol$$\)

(iii) The degree of polymerization (DP) of the polymer in the first question will be:

Number of chains with molecular weight 2100 = 1

Number of chains with molecular weight 6600 = 1

Number of chains with molecular weight 12000 = 1

The molecular weight of the repeating units of the three different chains have molecular weights of 126, 324, and 300 respectively.

Therefore, the degree of polymerization (DP) of the polymer in the first question will be:For 2100 chain,

\($$n_1=\frac{2100}{126}=16.67$$\)

For 6600 chain,

\($$n_2=\frac{6600}{324}=20.37$$\)

For 12000 chain,

\($$n_3=\frac{12000}{300}=40.00$$\)

So, the average degree of polymerization (DP) is:

\($$DP=\frac{1\times16.67+1\times20.37+1\times40.00}{1+1+1}=25.01$$\)

Therefore, the degree of polymerization of the polymer in the first question is 25.01.

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Answer:

\(\large \boxed{\text{-92 $^{\circ}$C}}\)

Explanation:

We can use the Ideal Gas Law and solve for T.

pV = nRT

Data  

p = 1.25 atm

V = 25.0 L

n = 2.10 mol

R = 0.082 06 L·atm·K⁻¹mol⁻¹

Calculations

1. Temperature in kelvins

\(\begin{array} {rcl}pV & = & nRT\\\text{1.25 atm} \times \text{25.0 L} & = & \rm\text{2.10 mol} \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times T\\31.25&=&0.09847T\text{ K}^{-1}\\T& = &\dfrac{31.25}{\text{0.098 47 K}^{-1}}\\\\& = &\text{181 K}\end{array}\)

2. Temperature in degrees Celsius

\(\begin{array} {rcl}T & = & (181 - 273.15) \, ^{\circ}\text{C}\\& = & -92 \, ^{\circ}\text{C}\\\end{array}\\\text{The temperature of the gas is $\large \boxed{\mathbf{-92 \, ^{\circ}}\textbf{C}}$}\)

If the freezing point depression for a solution is 2.5°c and kf = 4.5°c/m, the molality of the solution is

m= 0.56m

While molecular weight (MW) and molecular weight (MW) may be used to measure concentration, molecular weight (MW) and molecular weight (MW) is used to measure molecular weight and concentration, respectively.

Generally, the equation for  of freezing point of depression with molality is  mathematically given as

\(\Delta T_{f} = k_{f} *m\)

Therefore

m = 2.5/4.5

m= 0.56m

In conclusion,  the molality of the solution  is 0.56m

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Explanation:

I Would Determine Mass,Volume And Weight

The reaction of acetyl chloride with methanol is an example of an acyl substitution reaction. The mechanism of this reaction can be described as follows:

Step 1: Protonation of Acetyl Chloride

Acetyl chloride (CH3COCl) reacts with a proton (H+) from a proton source, such as HCl, to form the acylium ion (CH3CO+).

CH3COCl + H+ → CH3CO+ + Cl-

Step 2: Nucleophilic Attack by Methanol

Methanol (CH3OH) acts as a nucleophile and attacks the acylium ion at the carbonyl carbon atom, leading to the formation of a tetrahedral intermediate.

CH3CO+ + CH3OH → CH3COCH3OH+

Step 3: Loss of Protonated Alcohol

The tetrahedral intermediate formed in step 2 is unstable and undergoes elimination of the protonated alcohol to form the acetylated methanol product (CH3COOCH3) and a hydronium ion (H3O+).

CH3COCH3OH+ → CH3COOCH3 + H3O+

Overall, the reaction can be summarized as follows:

CH3COCl + CH3OH → CH3COOCH3 + HCl

In this reaction, acetyl chloride acts as the acylating agent and methanol acts as the nucleophile. The reaction proceeds through an intermediate and the final product is an ester, acetylated methanol. This reaction is widely used in organic synthesis for the preparation of esters

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The yield of P₄O₁₀ is 4.14 g. This means that 4.14 grams of P₄O₁₀ are obtained in the reaction, given the conditions and percent yield provided.

To calculate the yield of P₄O₁₀, we need to use the percent yield and the given mass of phosphorus (P₄) burned. The balanced chemical equation for the reaction is:

4P + 5O₂ → P₄O₁₀

From the equation, we can see that the molar ratio between P₄O₁₀ and P₄ is 1:1. This means that the theoretical yield of P₄O₁₀ is equal to the mass of P₄ burned.

Given that 6.20 g of phosphorus (P₄) is burned, we can assume that the theoretical yield of P₄O₁₀ is also 6.20 g.

Now, we can use the percent yield formula to calculate the actual yield of P₄O₁₀:

Percent Yield = (Actual Yield / Theoretical Yield) × 100

Rearranging the formula, we can solve for the actual yield:

Actual Yield = (Percent Yield / 100) × Theoretical Yield

Substituting the given values, we have:

Actual Yield = (67.0 / 100) × 6.20 g = 4.14 g

Therefore, the yield of P₄O₁₀ is 4.14 g.

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Answer:

Explanation:

The density of a substance can be found by using the formula

\(density = \frac{mass}{volume} \\\)

From the question

mass = 4.85 kg

volume = 280 cm³

We have

\(density = \frac{4.85}{280} \\ = 0.01732...\)

We have the final answer as

Hope this helps you

This is defined as a variant form of a gene. We were told that allele for  white wool is dominant to the allele for black wool and two heterozygous sheep mated.

When we cross Ww and Ww we get WW, Ww, Ww, ww. Since white is dominant then the percentage is 3/4 = 75 percent.

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Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

Force = 50 × 5

We have the final answer as

Hope this helps you

Answer:

However, when she looks at a sample using a magnifying lens, she can see small droplets of liquid surrounded by another liquid. Which conclusion best fits her observations? The mixture is a solution because it has the same appearance throughout the sample

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The empirical formula of oxide of iron is Fe2O2 and its name is iron oxide.

Given:

Solution:

Now, calculate the relative mass of iron in iron oxide = Percentage of iron by mass / At. mass of iron

Relative mass of iron = 77.75/55.85 = 1.39

Relative mass of oxygen = Percentage of oxygen by mass / At. mass of iron = 22.25/16 = 1.39

Now the calculate the molar ratio of iron to oxygen = 1.39 : 1.39 = 2:2

Thus, the empirical formula of iron oxide is Fe2O2 and its name is iron oxide.

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Answer:

Alkanes

Explanation:

Explanation:

The answer is Alkanes.Alkanes are compound that consists entirely of atoms of carbon and hydrogen joined to one another by single bonds

Answer:

b) no change; an increase

Explanation:

Kinetic energy does not change but potential energy increases at melting point.

Answer:

Viscosity

Explanation:

Sijui kuexpalain lakini

Answer:

Explanation:

apart from boiling point and flammability viscosity density and solubility are those qualities of hydrocarbon which are effected by their size

The answer is given below

the specific heat is a weighted sum of these specific heats:

\(cv = \frac{n1}{n1 + n2} r + \frac{n2}{n1 + n2} r\)

If expression did not consist the divisor n1 + n2 in each term,then it did not consist the correct units. R has the units of a molar

specific heat (J/mol K), but nR is a heat capacity (J/K).

hence, In the limit n1 → 0 the expression must reduce to the specific heat for a diatomic gas, and indeed we find CV. Likewise, for n1 → 0 the expression reduces appropriately to the specific heat for a monatomic gas.

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The standard cell potential, E°(cell), for the given reaction is +0.23 V. To calculate the standard cell potential, E°(cell), for the given reaction, we need to use the reduction potentials of the half-reactions involved.

The half-reactions involved in the reaction are: \(Fe_{3+}\) (aq) + e- → \(Fe_{2+}\) (aq) (Reduction half-reaction), \(I_{2}\) (aq) + 2 e- → 2 I- (aq) (Oxidation half-reaction)

The reduction potential for \(Fe_{3+}\) (aq) + e- → \(Fe_{2+}\) (aq) is given as +0.77 V, and the reduction potential for \(I_{2}\) (aq) + 2 e- → 2 I- (aq) is given as +0.54 V.

To calculate E°(cell), we need to subtract the reduction potential of the oxidation half-reaction from the reduction potential of the reduction half-reaction.

E°(cell) = E°(reduction) - E°(oxidation) = (+0.77 V) - (+0.54 V) = +0.23 V. Therefore, the standard cell potential, E°(cell), for the given reaction is +0.23 V.

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A person with a history of chemical dependency on controlled substances who desires to obtain A License. A license to practice Nursing students may be required to show proof of current sobriety and fitness to practice.

This rule is from the Texas Board of Nursing’s (BON) Administrative Rules. It is stated that any nurse who has a chemical dependency diagnosis or a history of controlled substance abuse must demonstrate the following through "objective, verifiable evidence" of current sobriety and fitness to practice to get a license.

This objective, verifiable evidence could be obtained through AA records, negative drug tests, an assessment by a specialist in addiction, and testimonies from group members, colleagues, and other close friends.

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The standard cell potential is 1.7629 V.

The standard cell potential is a measure of the maximum potential difference between the electrodes of an electrochemical cell under standard conditions of temperature, pressure, and concentration. It is calculated as the difference between the standard reduction potential of the cathode and the standard oxidation potential of the anode. In this case, the standard cell potential is calculated by subtracting the standard oxidation potential of silver from the standard reduction potential of nitrate.

The anode reaction is the oxidation of silver to silver ions, and its standard oxidation potential is -0.7990 V. The cathode reaction is the reduction of nitrate to nitrogen monoxide, and its standard reduction potential is 0.9639 V. The standard cell potential is calculated as the difference between the two potentials, which is 1.7629 V. This problem involves the application of electrochemistry principles and the use of standard reduction potentials to determine the standard cell potential of an electrochemical cell.

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