A chemistry student needs 60.00 g of butanoic acid for an experiment. He has available 120. g of a 36.9% w/w solution of butanoic acid in ethanol. Calculate the mass of solution the student should use. If there's not enough solution, press the "No solution" button.

Answer : The correct option is, (7) 332 mL

Explanation : Given,

Density of solution of sulfuric acid = [tex]1.29g/cm^3=1.29g/mL[/tex]

38.1 % acid by mass that means 38.1 grams of sulfuric acid present in 100 grams of solution of sulfuric acid.

Now we have to calculate the mass of solution of sulfuric acid.

As, 38.1 grams of sulfuric acid present in 100 grams of solution of sulfuric acid

So, 163 grams of sulfuric acid present in [tex]\frac{163}{38.1}\times 100=428.95[/tex] grams of solution of sulfuric acid

Now we have to calculate the volume of sulfuric acid solution.

[tex]Density=\frac{Mass}{Volume}[/tex]

[tex]1.29g/mL=\frac{428.95g}{Volume}[/tex]

[tex]Volume=332mL[/tex]

Therefore, the volume of sulfuric acid solution needed is 332 mL.

Answer:

11.6 mL

Explanation:

First we need to calculate the number of moles of Zn2+ present in the solution:

[tex]n=V*C\\n_{Zn^{2+}}=0.65*0.012=7.8x10^{-3}moles[/tex]

As the charge of ion zinc is 2+ and the charge of hydroxide is 1-, we need double moles of NaOH:

[tex]n_{NaOH}=0.0156moles[/tex]

As we have the concentration and the moles, we can calculate the volume:

[tex]V=\frac{n}{C} \\\\V=0.01164L=11.6mL[/tex]

Answer:

b) Solvent B was not a good mobile phase because it could not dissolve the ink.

Explanation:

The base for chromatography is that the mixture to analyze is soluble in the solvent, if this happen, there will be separation. The only thing that may affect the separation, is that the mixture is not soluble in the solvent.

Answer: Option B

Explanation:

The paper chromatography can be defined as the technique in which there are two phase. One is stationary phase that lies in the cellulose fiber of the paper.

The mobile phase is the solvent that moves on the stationary phase. This is the usually solvent.

The interaction between the mobile phase and stationary phase must be good so as to carry the stationary phase along with the mobile phase.

If the stationary phase is not carried along with the mobile phase then the interaction between the stationary phase and mobile phase is not strong.

Answer:

c) kg

Explanation:

Kilograms stands alone. It has to be hooked up to another unit for it to be a derived unit.

I am joyous to assist you anytime.

Answer:

The answer to your question is c) kg

Explanation:

Derived units is when to units are combined to get a new one, For example, combining length and length gives lenght² and then we measured area, this is a derived unit.

a) cm³ this answer is wrong because we are combining length x length x length so we get the units of volume, this is a derived unit.

b) g/mL this answer is wrong because we are combining mass and volume so we can measure density, this units are derived.

c)kg this answer is correct because it measures mass and only that, then it's a fundamental unit.

The mass of oxygen, silicon, ruthenium, and rhodium in the earth's crust is 2.274 × 1031 g, 1.360 × 1031 g, 4.998 × 1021 g, and 4.998 × 1021 g respectively.

To find the total mass of each of these elements in Earth's crust, we must first calculate the total volume of the crust. This is done by multiplying the surface area of the Earth by the mean thickness of the crust: 5.10 × 108 km2 * 35 km = 1.785 × 1010 km3.

Next, we convert this volume into cm3, because the density is given in g/cm3: 1.785 × 1010 km3 * (105)3 = 1.785 × 1025 cm3.

Then, we can calculate the total mass of the crust by multiplying the volume by the density: 1.785 × 1025 cm3 * 2.8 g/cm3 = 4.998 × 1025 g.

From there, we can calculate the mass of each element by multiplying the total mass by the abundance of each element. For oxygen and silicon, this would be: 4.998 × 1025 g * 4.55 × 105 g/t = 2.274 × 1031 g for oxygen and 4.998 × 1025 g * 2.72 × 105 g/t = 1.360 × 1031 g for silicon. For the rare elements ruthenium and rhodium, this would be: 4.998 × 1025 g * 1 × 10-4 g/t = 4.998 × 1021 g for both elements.

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Final answer:

To exceed the FDA's assumed sodium intake limit of 2,300 mg per day for an adult, one would have to consume approximately 44.23 liters of softened water with a sodium concentration of 5.2×10⁻²% by mass.

Explanation:

The question pertains to the consumption of softened water and its sodium concentration relative to the FDA recommendations for sodium intake. The FDA's recommended limit is not specified in the question, so for this explanation, we will assume a commonly referenced guideline limit of 2,300 mg sodium per day for an adult. The concentration of sodium in the softened water is given as 5.2×10⁻²% by mass. Given this and the density of water (1.0 g/mL), we can calculate the volume of softened water consumed that would exceed the FDA's recommended sodium intake limit.

If the softened water contains 5.2×10⁻²% sodium by mass, this means there are 0.052 grams of sodium in every 100 grams of the softened water. Since 1 mL of water weighs 1 gram, this also means there are 0.052 grams of sodium per 100 mL of water.

To find the volume (V) of water that contains 2,300 mg (2.3 grams) of sodium, we set up the following equation:

V (in mL) × 0.052 g/100 mL = 2.3 g.

Solving for V, we get:

V = (2.3 g × 100 mL) / 0.052 g

V = 44,230.77 mL.

Therefore, one would have to consume approximately 44.23 liters of this softened water to exceed the FDA recommendation for sodium intake.

Answer:

26.18 g

Explanation:

The molar volume of oxygen at 1 atm and 0ºC is 22.4 L/mol.

If you want to react 7 L it means you will use

7/22.4 = 0.3125 moles of oxygen

the balanced equation is 3 Fe + 2 O2 = Fe3O4

which means that 2 moles of oxygen reacts with 3 moles of iron

If you have 0.3125 moles of oxygen, 0.468 moles of iron will be needed.

Iron molecular weight is 55.84 and than, 0.3125 moles corresponds to a mass of iron equal to 26.18

To determine the mass of iron required to react with 7.0 L of oxygen, we need to use the balanced equation and the molar ratios between iron and oxygen.

To determine the mass of iron required to react with 7.0 L of oxygen, we need to use the balanced equation and the molar ratios between iron and oxygen. The balanced equation for the reaction is:

4Fe + 3O2 → 2Fe2O3

From the equation, we can see that it takes 4 moles of iron to react with 3 moles of oxygen to produce 2 moles of Fe2O3. First, we need to convert the volume of oxygen to moles by using the ideal gas law. At 0 degrees Celsius and 1 atm, the volume can be calculated as follows:

V = nRT/P

V = (7.0 L)(0.0821 L/mol·K)(273 K) / (1 atm)

V = 17.109 moles

Next, we need to use the molar ratios to determine the moles of iron required. Since the ratio is 4 moles of iron to 3 moles of oxygen, we can set up a proportion:

4 moles Fe / 3 moles O2 = x moles Fe / 17.109 moles O2

Solving for x:

x = (4 moles Fe)(17.109 moles O2) / 3 moles O2

x = 22.812 moles Fe

Lastly, we can convert the moles of iron into grams using the molar mass of iron:

(22.812 moles)(55.85 g/mol) = 1276.06 grams

Therefore, approximately 1276.06 grams of iron is required to react with 7.0 L of oxygen to produce Fe3O4.

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Answer:

The answer to your question is: 85.458 amu

Explanation:

data

Rubidium-85   A = 84.9118 amu   abundance = 72.15%  

Rubidium - 87  A = 86.9092 amu abundance = 27.85%

Atomic weight = ?

Atomic weight = 84.9118(0.7215) + 86.9092(0.2785)

Atomic weight = 61.2538 + 24.2042

Atomic weight = 85.458 amu

Rubidium has two naturally occurring isotopes, rubidium -85 ( atomic mass = 84.9118 amu; abundance = 72.15%) and rubidium -87 (atomic mass = 86.9092; abundance = 27.85%). The atomic weight of rubidium is 85.46 amu.

To calculate the average atomic mass of element use this formula

Average Atomic Mass = f₁M₁ + f₂M₂

where,

f = Fraction of natural abundance of isotope

M = Mass number of isotope

Isotope Rubidium -85 (Atomic mass = 84.9118 amu and Abundance = 72.15)

Abundance = [tex]\frac{72.15}{100}[/tex]

                    = 0.7215

Isotope Rubidium- 87 (Atomic mass = 86.9092 amu and Abundance = 27.85%)

Abundance = [tex]\frac{27.85}{100}[/tex]

                    = 0.2785

Now put the value in above formula, we get:

Average Atomic Mass = f₁M₁ + f₂M₂

                                     = (84.9118 × 0.7215) + (86.9092 × 0.2785)

                                     = 61.26 + 24.20

                                     = 85.46 amu

Thus from the above conclusion we can say that Rubidium has two naturally occurring isotopes, rubidium -85 ( atomic mass = 84.9118 amu; abundance = 72.15%) and rubidium -87 (atomic mass = 86.9092; abundance = 27.85%). The atomic weight of rubidium is 85.46 amu.

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The names have been correctly sorted as follows;

Molecule 1 - 3-chloro-4-iodo-2-methyl hexane

Molecule 2 - 4-chloro-3-iodo-2-methyl hexane

Molecule 4 - 3-chloro-2-iodo-4-methyl hexane

The IUPAC naming system for organic compounds

To name organic compounds, identify the longest continuous chain of carbon atoms, the "parent chain". Look up the root name of this chain based on its length, and identify each substituent and its position on the chain with a number.

Combine the root name, substituent prefixes with their positions, and functional group suffixes to form the complete name. Lastly, use commas and hyphens to separate and connect different parts of the name.

When a chemical reaction take up energy then this type of reaction is called endothermic reaction. Therefore,  heat is being transferred from the surroundings to the system.

In chemistry there are various type of reaction out of which the two main types are the exothermic reaction and endothermic reaction.

Exothermic reaction is the one in which energy releases in form of heat respiration reaction is an example of exothermic reaction. In respiration food that we eat are broken down in glucose with release of energy.

Endothermic reaction is the one in which energy is taken out during the reaction. Photosynthesis is an example of endothermic reaction where sunlight energy is taken by the plants to make food.

H[tex]_2[/tex]+ Br[tex]_2[/tex]→ 2 HBr  ΔH = 72.80 kJ.

72.80 kJ of heat should be transferred when 38.2g of liquid brownie reacts with excess hydrogen gas to form hydrogen bromine.  Heat is being transferred from the surroundings to the system.

Therefore,  heat is being transferred from the surroundings to the system.

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The question contains a typo and seems to refer to a chemical reaction between hydrogen gas and bromine to form hydrogen bromide. Without specific data on the enthalpy change, one cannot conclusively say how much heat is transferred or definitively whether the reaction is exothermic or endothermic.

The original question seems to have a typo referring to 'liquid brownie' reacting with hydrogen gas, which is non-sensical in a chemistry context. Assuming the question pertains to the reaction of hydrogen gas (H₂) with bromine (Br₂) to form hydrogen bromide (HBr), as per the given example:

H₂(g) + Br₂(g) → 2 HBr(g)

To determine how much heat is transferred during this reaction and whether the reaction is exothermic or endothermic, we need additional data such as the enthalpy change (ΔH) for the reaction, which is not provided in the question. In general, if ΔH is negative, the reaction is exothermic, meaning heat is transferred from the system to the surroundings. Conversely, if ΔH is positive, the reaction is endothermic, and heat is transferred from the surroundings to the system.

Given the typical nature of reactions between halogens and hydrogen, one could infer that the reaction is likely exothermic, but without specific data on enthalpy change, this is an educated guess.

Answer:

The answer to your question is: the last option ( 2 mole Fe : 3 mol Cl2)

Explanation:

The mole ratio in a chemical reaction is expressed by the integers before each element, for example if we have

                                                           2H2SO4 + 3NaOH,

The mole ratio between, H2SO4 and NaOH is 2H2SO4 : 3 NaOH

In your question, the question is the ratio between Fe and Cl2, so we look at their coefficients and then:

                    2 mole Fe : 3 mol Cl2

Answer:

1)

Explanation:

Fe+2 + 2e -> Fe

The element that is reduced gains electrons. In this case Fe is gaining electrons

Answer: The half reaction that describes the reduction reaction is [tex]Fe^{2+}(aq.)+2e^-\rightarrow Fe(s)[/tex]

Explanation:

Oxidation reaction is defined as the reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.

[tex]X\rightarrow X^{n+}+ne^-[/tex]

Reduction reaction is defined as the reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

[tex]X^{n+}+ne^-\rightarrow X[/tex]

For the given chemical reaction:

[tex]Mg(s)+Fe^{2+}(aq.)\rightarrow Mg^{2+}(aq.)+Fe(s)[/tex]

The half reactions follows:

Oxidation half reaction:  [tex]Mg(s)\rightarrow Mg^{2+}(aq.)+2e^-[/tex]

Reduction half reaction:  [tex]Fe^{2+}(aq.)+2e^-\rightarrow Fe(s)[/tex]

Hence, the half reaction that describes the reduction reaction is [tex]Fe^{2+}(aq.)+2e^-\rightarrow Fe(s)[/tex]

Cheddar cheese, iced tea with no ice, chicken noodle soup, vanilla ice cream, antifreeze, rocky road ice cream, and trail mix are examples of homogeneous and heterogeneous mixtures.

Homogeneous mixtures:

Heterogeneous mixtures:

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Homogeneous mixtures are uniform throughout; examples include iced tea with no ice and antifreeze. Heterogeneous mixtures aren't uniform and individual components can be seen; examples include cheddar cheese, chicken noodle soup, bread pudding, vanilla ice cream, rocky road ice cream, and trail mix.

The question pertains to the classification of various items as either homogeneous or heterogeneous mixtures. A homogeneous mixture is a combination of substances with a composition that is uniform throughout. In such a mixture, you cannot see the individual components, and there's no variation from point to point. On the other hand, a heterogeneous mixture is a collection of substances whose composition can vary from point to point and where the individual components can be seen.

From the provided list, the examples of homogeneous mixtures are Iced tea with no ice and Antifreeze because their compositions are the same throughout. Meanwhile, Cheddar Cheese, Chicken noodle soup, Bread pudding, Vanilla ice cream, Rocky road ice cream, and Trail mix could be considered examples of heterogeneous mixtures as their compositions are not uniform and the different components can be seen distinctly.

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Answer:

The answer to your question is letter A.

Explanation:

Isomers are molecules that have the same molecular formula but have a different structure. The molecule from which are looking an isomer has 5 carbons and 1 double bond. Then we need to look for another molecule with these components.

A.- This molecule has 5 carbons and 1 double bond, This structure is an isomer of the first one.

B.- This molecule has 3 carbons and 1 double bond, it's not an isomer of the first structure.

C. This molecule has 4 carbons and 1 triple bonds, it's not an isomer of the first structure.

D. This molecule has 5 carbons but it doesn't have any double bond, then it's not an isomer of the first structure.

Answer: 85.47u

Explanation:

The average atomic mass of an element is determined by taking the weighted average of the atomic masses of its naturally occurring isotopes.

Now, weighted average simply means that each isotope contributes to the average atomic mass of the element proportionally to its percent abundance.

avg. atomic mass=∑i(isotopei×abundancei)

The more abundant an isotope is, the more its atomic mass will influence the average atomic mass of the element.

In your case, you know that rubidium has two stable isotopes

85Rb → 84.91 u, 72.16% percent abundance

87Rb →86.91 u, 27.84% percent abundance

When you calculate the average atomic mass, make sure that you use decimal abundance, which is simply percent abundance divided by

100.

So, plug in your values to get

avg. atomic mass = 84.91 u × 0.7216 + 86.91 u × 0.2784

avg. atomic mass = 85.4668 u

Rounded to four sig figs, the answer will be

avg. atomic mass =85.47 u

Answer:

b

Explanation:

Answer:

The answer to your question is p O2 = 17.2 kPa, letter b

Explanation:

Remember that in a mixture of gases, total pressure is equal to the sum of the pressure of individual gases.

Then

          Pt = p He + p CO2 + p O2

Data

Pt = 101.3 kPa

pHe = 84kPa

pCO2 = 0.1 kPa

p O2 = ?

       Substitution

       101.3 = 84 + 0.1 + pO2

       pO2 = 101.3 - 84 - 0.1

        pO2 = 17.2 kPa

Answer:

mouth, stomach

Explanation:

The carbohydrates start the digestion in the mouth, with the enzymes dissolved in the saliva, and during the rest of the digestive system it continues until the intestine.

Most of the digestion of lipids is in the intestine, with the lipid enzymes that come from the bile. But it starts in the mouth with the absorption of the small ones.

The chemical digestion of the proteins begins in the stomach because the peptide enzyme it turns active with the acidic environment of the stomach.

Answer:

Explanation:

I could have set up an experiment to determine the density of the metal in the jar. When I derive the density of the metal, I will compare it with that of every other metal to see if it properly fits any.

Density is the mass per unit volume of substance. It is the amount of substace contained per unit volume.

To find the density of the metal, the mass of the metal is obtained by directly weighing it. The volume is found by immersing the metal in water as it will sink. The volume of the water displaced is the volume of the metal. A measuring cylinder or an overflow is used to determine the volume of the liquid displaced.

     Then:

             Density = [tex]\frac{mass}{volume}[/tex]

Final answer:

To confirm the identity of the metal as pure zinc, perform an experiment with hydrochloric acid, observe reactions, record properties, and differentiate based on reactivity and density.

Explanation:

Experiment:

Place a few drops of zinc metal in a test tube and cover with dilute hydrochloric acid.

Heat the test tube containing the mixture to observe reactions and record your observations.

Wait for the product to cool, break the test tube, and examine the product to verify if it matches properties of pure zinc.

Observations:Zinc's reactivity with hydrochloric acid, its density compared to water, and its behavior in displacement reactions can be used to differentiate and confirm its identity as pure zinc.

Chemical Reactions:When zinc metal reacts with sulfuric acid, it produces zinc sulfate and hydrogen gas; whereas zinc oxide reacts with sulfuric acid to form zinc sulfate only.

Answer:

Concentration of A at equilibrium = 1 - 0.5 = 0.5 M

Explanation:

[tex]A \leftrightharpoons B + C[/tex]

Equilibrium constant = 0.5

Initial concentration of A = 1 M

             [tex]A \leftrightharpoons B + C[/tex]

Initial      1            0     0

At equi.  1-x         x      x

Equilibrium constant = [tex]\frac{[B][C]}{[A]}[/tex]

[tex]0.5 = \frac{x \times x}{1-x} \\0.5(1-x) = x^2\\0.5 -0.5x = x^2\\x^2+0.5x - 0.5 = 0[/tex]

on solving,

x = 0.5 M

Concentration of A at equilibrium = 1 - 0.5 = 0.5 M

Final answer:

Use the equilibrium constant and an ICE table to set up the equation Keq = [B][C]/[A] and solve for x to determine the equilibrium concentration of A after adding more of the substrate to the mixture.

Explanation:

To calculate the concentration of A when the reaction reaches equilibrium after adding 0.50 moles of A, we use an ICE table for the equilibrium process and the equilibrium constant given as Keq = 0.5. Initially, we have 1.0 M of A without B. Upon addition, we have 1.50 M of A. If we let x be the change in concentration of A as it reacts to form B and C, at equilibrium, [A] = 1.50 - x, [B] = x, and [C] = x because the mole ratio of A:B:C is 1:1:1. Setting up the equilibrium expression as Keq = [B][C]/[A], and substituting the terms with the expressions in terms of x, we get 0.5 = x²/(1.50 - x). We must solve this quadratic equation to find the value of x and, consequently, the equilibrium concentration of A.

Equilibrium constant = ([B][C])/([A])

0.5 = (x * x)/(1-x) \n0.5(1-x) = x^2\n0.5 -0.5x = x^2\nx^2+0.5x - 0.5 = 0

on solving,

x = 0.5 M

Concentration of A at equilibrium = 1 - 0.5 = 0.5 M

Answer:

43.668 kg

Explanation:

First we set the equation:

[tex]Al_{2}O_{3} + 6NaOH + 12HF \longrightarrow 2Na_{3}AlF_{6}+9H_{2}O[/tex]

Now, we need to now the kmoles for each reactant:

[tex]M_{Al_{2}O_{3}}=101.96kg/kmol\\M_{NaOH}=40kg/kmol\\M_{HF}=20.01kg/kmol\\n_{Al_{2}O_{3}}=0.104kmol\\n_{NaOH}=1.285kmol\\n_{HF}=2.57kmol[/tex]

With this, we can see that the limit reactant is the aluminum oxide, so, with the equation for the reaction we know that 1 kmol of aluminum oxide, produces 2 kmol of cryolite, so we set a rule of three and see that 0.208 kmoles of cryolite are produced, the we proceed to calculate the mass:

[tex]M_{Na_{3}AlF_{6}}=209.94kg/kmol\\n_{Na_{3}AlF_{6}}=0.208kmol\\m_{Na_{3}AlF_{6}}=43.668kg[/tex]

Answer:

The answer is: Law of multiple proportions        

Explanation:

The law of multiple proportions is a law of chemical combination given by Dalton in 1803.

According to this law, if more than one chemical compound is formed by combining two elements, then the mass of an element that combines with the fixed mass of other element is represented in the form of small whole number ratio.

Therefore, is an illustration of the law of the law of multiple proportions.

Methane and ethane are both made up of carbon and hydrogen. In methane, there are 12.0 g of carbon for every 4.00 g of hydrogen, a ratio of 3:1 by mass. In ethane, there are 24.0 g of carbon for every 6.00 g of hydrogen, a ratio of 4:1 by mass. This is an illustration of the law of multiple proportions.

The law of multiple proportions states that when two elements form more than one compound, the masses of one element that combine with a fixed mass of the other element are in a ratio of small whole numbers.

 In the given example, methane [tex]CH_3[/tex] and ethane [tex]C_2H_6[/tex] are compounds formed by carbon (C) and hydrogen (H). The mass ratios of carbon to hydrogen in these compounds are as follows:

For methane :

- Mass of carbon (C) = 12.0 g

- Mass of hydrogen (H) = 4.00 g

- Mass ratio of C to H = 12.0 g / 4.00 g = 3:1

 For ethane

- Mass of carbon (C) = 24.0 g

- Mass of hydrogen (H) = 6.00 g

- Mass ratio of C to H = 24.0 g / 6.00 g = 4:1

The ratios of the masses of carbon that combine with a fixed mass of hydrogen (12.0 g of carbon for methane and 24.0 g of carbon for ethane) are in a simple whole number ratio of 1:2 (or 3:6 if we consider the hydrogen masses as well). This exemplifies the law of multiple proportions, as the mass of carbon combining with a fixed mass of hydrogen increases in a small whole number ratio from one compound to the other.

Answer:

How many grams of CCL4 were formed? 116.9 g

How many grams of Cl2 reacted with the CH4? 243.8 g

Explanation:

First we need to know the molar mass for every element or compound in the reaction:

[tex]M_{CH_{4}}=16 g/mol\\M_{CH_{3}Cl}=50.49g/mol\\M_{CH_{2}Cl_{2}}=84.93g/mol\\M_{CHCl_{3}}=119.38g/mol\\M_{CCl_{4}}=153.82g/mol[/tex]

Now we proceed to calculate the amount of moles produced, per product:

[tex]n_{CH_{4}}=2.4\\n_{CH_{3}Cl}=0.18\\n_{CH_{2}Cl_{2}}=0.55\\n_{CHCl_{3}}=0.91\\n_{CCl_{4}}=n_{CH_{4}}-(n_{CH_{3}Cl}+n_{CH_{2}Cl_{2}}+n_{CHCl_{3}})\\n_{CCl_{4}}=0.76mol\\m_{CCl_{4}}=n_{CCl_{4}}*M_{CCl_{4}}\\m_{CCl_{4}}=116.9g[/tex]

To calculate the mass of chlorine we just need to make a mass balance:

[tex]m_{CH_{4}}+m_{Cl_{2}}=m_{CH_{3}Cl}+m_{CH_{2}Cl_{2}}+m_{CHCl_{3}}+m_{CCl_{4}}\\m_{Cl_{2}}=m_{CH_{3}Cl}+m_{CH_{2}Cl_{2}}+m_{CHCl_{3}}+m_{CCl_{4}}-m_{CH_{4}}\\m_{Cl_{2}}=243.8g[/tex]

Final answer:

To find the grams of CCl4 formed, we need to calculate the grams of Cl2 reacted with CH4 and then convert it to grams of CCl4 using the stoichiometry of the balanced equation. 2.4 mol of Cl2 reacted with 38.4 g of CH4. Therefore, 369.17 grams of CCl4 were formed.

Explanation:

To find the grams of CCl4 formed, we need to calculate the grams of Cl2 reacted with CH4 and then convert it to grams of CCl4 using the stoichiometry of the balanced equation. We start by calculating the molar mass of CH4, which is 16.04 g/mol. Since all the CH4 reacted, we can use its mass to calculate the moles of CH4. 38.4 g CH4 / 16.04 g/mol = 2.4 mol CH4.

Next, we use the balanced equation to determine the moles of Cl2 that reacted with CH4. From the equation, we know that 1 mole of CH4 reacts with 4 moles of Cl2. So, 2.4 mol CH4 x (4 mol Cl2 / 1 mol CH4) = 9.6 mol Cl2.

Now, we can convert the moles of Cl2 to grams using its molar mass of 70.90 g/mol. 9.6 mol Cl2 x 70.90 g/mol = 681.84 g Cl2 reacted. This is the answer to the second part of the question.

Finally, we use the stoichiometry of the balanced equation to calculate the grams of CCl4 formed. From the equation, we know that 1 mole of CH4 reacts to produce 1 mole of CCl4. So, the moles of CCl4 formed is equal to the moles of CH4 reacted, which is 2.4 mol.

Finally, we convert the moles of CCl4 to grams using its molar mass of 153.82 g/mol. 2.4 mol CCl4 x 153.82 g/mol = 369.17 g CCl4. Therefore, 369.17 grams of CCl4 were formed.

Calculate the volume of air in two different rooms based on their dimensions in feet. The volume of air in each room is converted to liters using the given conversion factor. The first room has approximately 24,900 liters of air, and the second room has around 34,300 liters of air.

The volume of air in a room can be calculated using the formula:

Volume = Length x Width x Height

A) For a room measuring 10.0 ft x 11.0 ft x 8.00 ft:

Volume = 10.0 ft x 11.0 ft x 8.00 ft = 880 ft³ = 24,900 L

B) For a room measuring 11.0 ft x 11.0 ft x 10.0 ft:

Volume = 11.0 ft x 11.0 ft x 10.0 ft = 1,210 ft³ = 34,300 L

Answer: The empirical formula for the given compound is [tex]C_3H_3O_2[/tex]

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

[tex]C_xH_yO_z+O_2\rightarrow CO_2+H_2O[/tex]

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of [tex]CO_2=0.9267g[/tex]

Mass of [tex]H_2O=0.1897g[/tex]

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 0.9267 g of carbon dioxide, [tex]\frac{12}{44}\times 0.9267=0.2527g[/tex] of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 0.1897 g of water, [tex]\frac{2}{18}\times 0.1897=0.021g[/tex] of hydrogen will be contained.

Mass of oxygen in the compound = (0.4987) - (0.2527 + 0.021) = 0.225 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =[tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.2527g}{12g/mole}=0.021moles[/tex]

Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.021g}{1g/mole}=0.021moles[/tex]

Moles of Oxygen = [tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.225g}{16g/mole}=0.014moles[/tex]

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.485 moles.

For Carbon = [tex]\frac{0.021}{0.014}=1.5[/tex]

For Hydrogen  = [tex]\frac{0.021}{0.014}=1.5[/tex]

For Oxygen  = [tex]\frac{0.014}{0.014}=1[/tex]

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1.5 : 1.5 : 1

To make in whole number we multiply the ratio by 2, we get:

The ratio of C : H : O = 3 : 3 : 2

The empirical formula for the given compound is [tex]C_3H_3O_2[/tex]

Thus, the empirical formula for the given compound is [tex]C_3H_3O_2[/tex]

The temperature outside on the hot summer day was approximately 37.18 °C.

Charles's law, which states that for constant pressure and volume of a gas, the volume of a gas is proportional to its absolute temperature (measured in Kelvin), can be used to resolve this:

Calculating the temperature outside ([tex]\rm T_2[/tex]) in Kelvin using the information provided:

Initial temperature ([tex]\rm T_1[/tex]) = 19.0 °C + 273.15 K = 292.15 K

Initial diameter ([tex]\rm d_1[/tex]) = 50.0 cm

Final diameter ([tex]\rm d_2[/tex]) = 51.0 cm

[tex]\rm (T_1 / T_2) = (d_1^3 / d_2^3)[/tex]

[tex]\rm (292.15 K / T_2) = (50.0^3 / 51.0^3)[/tex]

Calculating the right-hand side of the equation:

[tex](50.0^3 / 51.0^3)[/tex] ≈ 0.94149

Solving for [tex]\rm T_2[/tex]:

[tex]\rm T_2[/tex] = 292.15 K / 0.94149 ≈ 310.33 K

Converting [tex]\rm T_2[/tex] back to Celsius:

Temperature outside (in Celsius) = [tex]\rm T_2[/tex] - 273.15

Temperature outside (in Celsius) = 310.33 K - 273.15 K ≈ 37.18 °C

Therefore, the temperature outside on the hot summer day was approximately 37.18 °C.

Learn more about temperature, here:

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The problem is based on Charles's Law which states the volume of a gas is directly proportional to its temperature at constant pressure. The initial and final conditions of the balloon are used to calculate the outside temperature to be approximately 38.35°C.

To solve this problem, we need to use the concepts of gas laws, specifically Charles's Law. Charles's Law states that volume of a gas is directly proportional to its temperature if pressure and the amount of gas remain constant. In this case, we assume that the pressure inside the balloon remains constant because it is exposed to constant external atmospheric pressure both inside the air-conditioned room and outside.

Let's express the initial and final conditions as follows:

According to Charles's Law, V1/T1=V2/T2. So, we can find the final temperature as T2 = V2*(T1/V1). Substituting our values, we find T2 = 69813*(292.15/65450) = 311.5 K. Finally, convert it back to Celsius by subtracting 273.15: T2 = 311.5 - 273.15 = 38.35°C. So, it was approximately 38.35°C outside.

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Answer: The mass of bromine reacted is 160.6 grams.

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 aluminium = 18.1 g

Molar mass of aluminium = 27 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of aluminium}=\frac{18.1g}{27g/mol}=0.670mol[/tex]

The chemical equation for the reaction of aluminium and bromide follows:

[tex]2Al+3Br_2\rightarrow 2AlBr_3[/tex]

By Stoichiometry of the reaction:

2 moles of aluminium reacts with 3 moles of bromine gas

So, 0.670 moles of aluminium will react with = [tex]\frac{3}{2}\times 0.670=1.005mol[/tex] of bromine gas.

Now, calculating the mass of bromine gas, we use equation 1:

Moles of bromine gas = 1.005 moles

Molar mass of bromine gas = 159.81 g/mol

Putting values in equation 1, we get:

[tex]1.005mol=\frac{\text{Mass of bromine}}{159.81g/mol}\\\\\text{Mass of bromine}=(1.005mol\times 159.81g/mol)=160.6g[/tex]

Hence, the mass of bromine reacted is 160.6 grams.

Final answer:

To find out how many grams of bromine react with 18.1g of aluminum, we first determine the mass ratio from a previous reaction and then use that to calculate the mass of bromine. The proportional relationship indicates that approximately 159.8 grams of bromine would react with 18.1 grams of aluminum.

Explanation:

The student asks how many grams of bromine react with 18.1g of aluminum. The chemical reaction is similar in behavior to the reaction of aluminum with chlorine where aluminum bromide is formed. Given a previous reaction scale of 27g aluminum yielding 266.7g aluminum bromide, we use stoichiometry to find the grams of bromine reacting with 18.1g aluminum.

First, determine the molar mass of aluminum (Al) and aluminum bromide (AlBr3). Aluminum has a molar mass of approximately 27g/mol and aluminum bromide has a molar mass of roughly 267g/mol based on the atomic weights of aluminum (approximately 27) and bromine (approximately 80 per bromine atom, with three bromine atoms per molecule). Find the mole ratio of Al to AlBr3 from the balanced chemical equation, which is 2:2 (or 1:1 for simplicity). From the initial reaction scale, you can calculate the ratio of aluminum to bromine involved in the reaction.

Next, set up a proportional relationship:

To solve for x (the mass of aluminum bromide produced from 18.1g of aluminum), cross-multiply and divide:

27g Al * x g AlBr3 = 18.1g Al * 266.7g AlBr3

x = (18.1g Al * 266.7g AlBr3) / 27g Al

x ≈ 177.7g AlBr3

Finally, determine the mass of bromine involved. Since 27g of Al produces 266.7g of AlBr3, the mass of bromine can be calculated by subtracting the mass of aluminum from the total mass of the product:

266.7g AlBr3 - 27g Al = 239.7g Br2

Now, calculate the mass of bromine that would react with 18.1g of Al using the same proportion:

Cross-multiply and solve for y:

y = (18.1g Al * 239.7g Br2) / 27g Al

y ≈ 159.8g Br2

Therefore, approximately 159.8 grams of bromine would react with 18.1 grams of aluminum.

Answer:

1. evaporation of ethanol

2. condensation of ethanol

Explanation:

The first and second process imply a change of phase, which is essentially a  physical change because the matter change from liquid to gas and from gas to liquid respectively and it can ve reversed.

The options 3 to 6 are related to a chemical reaction because the outcomes of the process are different compounds than the ones we had at the beginning. The main reaction that implies process 3 to 6 is represented by:

C6H12O6 +  impurities → 2 C2H5OH + 2 CO+ C

For option 5. burning of natural gas , we have:

Natural gas(Mainly a mix of CH4 and C2H10) +O2→COn+H2O

Therefore, justo 1 and 2 are a physical change

The evaporation and condensation of ethanol are physical changes as these do not alter the substance's chemical composition. The formation of carbon deposit, carbon dioxide gas, the burning of natural gas, and ethanol formation all involve chemical reactions and are thus considered chemical changes.

In the experiment described, both physical and chemical changes occur. Physical changes involve a change in physical state without altering the substance's chemical composition. Therefore, the evaporation of ethanol (#1), and the condensation of ethanol (#2) are physical changes. The substance remains ethanol in both cases; it merely changes from liquid to gas or from gas to liquid.

Chemical changes, on the other hand, involve a chemical reaction where a new substance is formed. Therefore, the formation of a carbon deposit inside the flask from decomposition of glucose (#3), the formation of carbon dioxide gas from glucose (#4), the burning of natural gas (#5), and the formation of ethanol from glucose by yeast (#6) are examples of chemical changes.

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Final answer:

The limiting reagent in this reaction is CuSO4.

Explanation:

The limiting reagent in this reaction is CuSO4.

To determine the limiting reagent, we need to compare the number of moles of each reactant with the stoichiometric ratio of the balanced equation. First, calculate the number of moles of Na2S and CuSO4 using their molar masses. Then, compare the moles of each reactant to the stoichiometric ratio in the balanced equation. The reactant that produces fewer moles of the product is the limiting reagent.

In this case, calculate the moles of Na2S and CuSO4. The stoichiometric ratio is 1:1 for Na2S and CuSO4. Since the moles of CuSO4 is smaller than the moles of Na2S, CuSO4 is the limiting reagent.

Answer: The value of n =2.

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.

To calculate the moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given molecules}}{\text {Avogadro's number}}=\frac{8.06\times 10^{20}}{6.023\times 10^{23}}=0.0013moles[/tex]

0.0013 moles of [tex]XeF_n[/tex] weigh = 0.227 grams

Thus 1 mole of [tex]XeF_n[/tex]  will weigh = [tex]\frac{0.227}{0.0013}\times 1=174.62[/tex]  grams

Molar mass of [tex]XeF_n[/tex] = [tex]1(131.3)+n(19)=174.62[/tex]

[tex]n=2[/tex]

Thus the value of n =2

The xenon fluoride compound is likely XeF2. This was deduced by relating the given mass and number of molecules of the compound to its molar mass and Avogadro's number. The calculated molar mass fits best with that of XeF2.

The subject of your question lies in the area of analytical chemistry, specifically in calculations involving the concept of the mole. Here we want to find the molecular formula for the xenon fluoride, XeFn using the data provided. We know the Avogadro's number which states that one mole of any substance contains 6.02 x 1023 molecules.

In this question, we're given that 8.06 x 1020 molecules of XeFn weigh 0.227 g. We can calculate the molar mass of this specific compound. To elaborate, if one mole weighs 'M' grams and contains 6.02 x 1023 molecules, then 0.227 g would have (0.227/M) moles or (0.227/M) x 6.02 x 1023 molecules. If we set this equal to 8.06 x 1020 and solve for 'M', the molar mass of the compound, we can calculate a value of approximately 169 g/mol.

Given that the molar mass of xenon (Xe) is about 131 g/mol and that of fluorine (F) is approximately 19 g/mol, we can reason that the molar mass of the compound XeFn is closest to that of XeF2, since 131 + 2(19) = 169 g/mol. Therefore, the xenon fluoride compound being referred to in the problem is XeF2.

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Answer:

The answer to your question is; H2O and CO.

Explanation:

Ionic bonding is when a metal donates its valence electrons to a nonmetal.

Covalent bonding is when two nonmetals share electrons to reach noble gases' configuration.

NaF   here we have a metal and a nonmetal, then the bonding is ionic.

MgO  here, there is also a metal and a nonmetal, the bonding is ionic.

H2O  here, there are two nonmetals, so this is a covalent bonding.

CO here, there are two nonmetals, the bonding is covalent.

Final answer:

H₂O and CO are the compounds most likely to have covalent bonds because they are composed of non-metallic elements. NaF and MgO are ionic compounds as they contain a metal and a non-metal.

Explanation:

Compounds that are most likely to have covalent bonds are usually formed between non-metallic elements. In the choices given, compounds H₂O (water) and CO (carbon monoxide) are most likely to have covalent bonds because they consist of non-metallic elements bonded together.

Let's evaluate the options further: