Calculate the molality of a solution containing 141.5 g of glycine (NH2CH2COOH) dissolved in 4.456 kg of H2O

Answer:

pH = 8.72

Explanation:

This is like a titration of a weak acid and a strong base, in this case, we are at the equivalence point plus we have the same mmoles of acid and base. We have completely neutralized the acid.

CH₃COOH      +     OH⁻        ⇄    CH₃COO⁻   +   H₂O

0.1M . 15 mL      0.1M . 15 mL

We only have (0.1M . 15 mL) mmoles of acetate ion. → 1.5 mmoles

As this compound acts like a base, we propose this equilibrium:

CH₃COO⁻   +  H₂O  ⇄  CH₃COOH      +     OH⁻   Kb

We need to work with Kb and we know, that Kw = Ka. Kb so, Kb = Kw/Ka

Kb = 1×10⁻¹⁴ /1×10 ⁻⁵ = 5.55×10⁻¹⁰

Concentration of CH₃COO⁻ → 1.5 mmol / 30mL (volumes of the solutions are additive) = 0.05M

So: [CH₃COOH] . [OH⁻] / [CH₃COO⁻] = Kb

x²/ 0.05-x = 5.55×10⁻¹⁰

We can avoid the quadractic equation because Kb is so small

[OH⁻] = √(5.55×10⁻¹⁰ . 0.05) = 5.27×10⁻⁶

pOH = - log [OH⁻]  → 5.28

pH = 14 - pOH = 8.72

The pH of a solution made by mixing 15.00 mL of 0.10 M acetic acid should be 8.72.

Since the following equation should be used.

CH₃COOH      +     OH⁻        ⇄    CH₃COO⁻   +   H₂O

0.1M . 15 mL      0.1M . 15 mL

Now

(0.1M . 15 mL) mmoles of acetate ion. → 1.5 mmoles

So,

CH₃COO⁻   +  H₂O  ⇄  CH₃COOH      +     OH⁻   Kb

Now

Kw = Ka. Kb

Kb = Kw/Ka

And,

Kb = 1×10⁻¹⁴ /1×10 ⁻⁵

= 5.55×10⁻¹⁰

Now

[CH₃COOH] . [OH⁻] / [CH₃COO⁻] = Kb

x²/ 0.05-x = 5.55×10⁻¹⁰

Now

[OH⁻] = √(5.55×10⁻¹⁰ . 0.05) = 5.27×10⁻⁶

pOH = - log [OH⁻]  → 5.28

pH = 14 - pOH

= 8.72

Hence, The pH of a solution made by mixing 15.00 mL of 0.10 M acetic acid should be 8.72.

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Answer:

2

Explanation:

A single covalent bond is formed when two electrons are shared between the same two atoms, one electron from each atom.

Answer:

the answer is 2

Explanation:

Answer:

Fe(s) + Cu^2+(aq) ---> Fe^2+(aq) + Cu(s)

Explanation:

The ionic equation shows the actual reaction that took place. It excludes the spectator ions. Spectator ions are ions that do not really participate in the reaction even though they are present in the system.

For the reaction between iron and copper II nitrate, the molecular reaction equation is;

Fe(s) + Cu(NO3)2(aq)----> Fe(NO3)2(aq) +Cu(s)

Ionically;

Fe(s) + Cu^2+(aq) ---> Fe^2+(aq) + Cu(s)

Answer:

Boiling

Explanation:

When a liquid is heated, the vapor pressure rises steadily. When water attains a temperature of 100°C or 212°F its vapor pressure is now equal to the atmospheric pressure at sea level, this is what we mean by boiling.

When this occurs, water continues to evaporate untill the vapor pressure inside the bubbles becomes high enough to stop water bubbles from collapsing again from the pressure of the water around it so the bubbles rise and break the surface.

Answer:

Cu2 + 2Mg-> 2Cu+ Mg2

Explanation:

Balance the equation and make sure both the reactant and the products are the same

Hope it will be helpful

[tex]Cu^{+2} + 2Mg[/tex]  -> [tex]2Cu + Mg^+2[/tex]  is the correct half-reactions.

A balanced equation is an equation for a chemical reaction in which the number of atoms for each element in the reaction and the total oxidation numbers is the same for both the reactants and the products.

[tex]Cu^{+2} + 2Mg[/tex]  -> [tex]2Cu + Mg^+2[/tex] is the correct half-reactions.

Magnesium is oxidized because its oxidation state increased from 0 to +2 while Cu is reduced because its oxidation state decreased from +2 to 0.

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Answer:

A. Ka = [CO2] / [C] [O2]^1/2

B. Kb = [CO2] / [CO] [O2]^1/2

Explanation:

Equilibrium constant is simply defined as the ratio of the concentration of the products raised to their coefficient to the concentration of the reactants raised to their coefficient.

Now, we shall obtain the expression for the equilibrium constant for the reaction as follow:

A. Determination of the expression for equilibrium constant Ka.

This is illustrated below:

C(s) + 1/2 O2(g) <==> CO(g)

Ka = [CO2] / [C] [O2]^1/2

B. Determination of the expression for equilibrium constant Kb.

This is illustrated below:

CO(g) + 1/2 O2(g) <==> CO2(g)

Kb = [CO2] / [CO] [O2]^1/2

Answer:

Companies know that people will be willing to spend more to get an in-demand product.

Explanation:

When a product is really in demand, many customers are willing to part with more money order to purchase the product, as a result of that, many companies may take advantage of the increasing demand for the product to hike it's price.

Hence, the increase in price may not really have a negative impact on the quantity demanded because the demand for the product is high and customers are willing to spend more money in order to purchase an in-demand product, hence the answer above.

Prices increase when demand is high because companies know that people will be willing to spend more to get in-demand products.

Prices generally increase with higher demand for goods because the higher demand creates pressure for the supply to meet up.

Manufacturing companies can either increase their production to meet up with demand at the same price or capitalize on the situation to make more money by increasing the price without increasing the supply.

Since there is a buying pressure on the product in the market already, people would still be open to buying even at higher prices.

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Answer:

D

Explanation:

Addition of 0.05 M HCl, will react with all of the C2H3O2- from LiAc which will give 0.05 M more HAc. So there will be no Acetate ion left to make the solution buffer. Hence, the correct option for the this question is d, which is adding 0.050 moles of HCl.

The action that destroys the buffer is option c. adding 0.050 moles of HCl.

It is a solution of a weak acid and salt.

Here, The buffer will destroy at the time when either HC2H3O2 or NaC2H3O2 should not be present in the solution.

The addition of equal moles of HCl finishly reacts with equal moles of NaC2H3O2. Due to this,  there will be only acid in the solution.

Since

moles of HC2H3O2 = 1*0.250 = 0.250

moles of NaC2H3O2 = 1*0.050 = 0.050.

moles of HCl is added = 0.050

Now

The reaction between HCl and NaC2H3O2

[tex]HCl + NaC_2H_3O_2 \rightarrow HC_2H_3O_2 + NaCl[/tex]

Now

BCA table is

            NaC2H3O2  HCl       HC2H3O2

Before 0.050 0.050 0.250

Change -0.050 -0.050 +0.050

After 0 0 0.300

Now, the solution contains the acid (HC2H3O2 ) only.

Therefore addition of 0.050 moles of HCl will destroy the buffer.

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Answer:

16G = 1 mole ; then 4G = how many moles? 4/16 = 0.25 mole; That means 4 grams of oxygen is 0.25 moles.

Explanation:

A mole of water molecules contains 2 moles of hydrogen atoms and 1 mole of oxygen atoms.

Answer:

CO2 will diffuse more rapidly.

Explanation:

From Graham's law of diffusion, we understood that the rate of diffusion of a gas is inversely proportional to the square root of its density as shown below:

Rate (R) & 1/√Density (d)

R & 1/√d

But, the density of a gas is directly proportional to the relative molecular mass (M) of the gas.

Thus, we can say that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. This can be represented mathematically as:

Rate (R) & 1/√Molar mass (M)

R & 1/√M

From the above illustration, we can say that the lighter the gas, the faster the rate of diffusion and the heavier the gas, the slower the rate of diffusion.

Now, to answer the question given above,let us determine the molar mass of Cl2 and CO2.

This is illustrated below:

Molar mass of Cl2 = 2 x 35.5 = 71 g/mol

Molar mass of CO2 = 12 + (2x16) = 12 + 32 = 44 g/mol

Summary

Gas >>>>>> Molar mass

Cl2 >>>>>> 71 g/mol

CO2 >>>>> 44 g/mol

From the illustration above, we can see that CO2 is lighter than Cl2.

Therefore, CO2 will diffuse more rapidly.

Answer: CO2

Explanation:

Answer:

The correct answer is - 1.02 V

Explanation:

From the reduction-oxidation reaction:

Ni²⁺(aq) + 2 Fe²⁺(aq) → Ni(s) + 2 Fe³⁺(aq)

Ni²⁺ is reduced to Ni(s) while Fe²⁺ is oxidized to Fe³⁺. Thus, the half reactions are:

Reduction (cathode) : Ni²⁺(aq) + 2 e‑ → Ni(s)                    Eº= ‑0.25 V

Oxidation (anode) :  2 x (Fe²⁺ → Fe³⁺ + e-)(aq)                Eº= -0.77 V

                                -------------------------------------

                     Ni²⁺(aq) + 2 Fe²⁺(aq) → Ni(s) + 2 Fe³⁺(aq)

In order to calculate the Eºcell, we have to add the reduction potential of the reaction in cathode (reduction) to the oxidation potential of the anode (oxidation):

Eºcell= Eºr + Eºo= (-0.25 V) + (-0.77 V) = - 1.02 V

Answer:

2.81 g of H2O.

Explanation:

We'll begin by calculating mass of O2 that contains 1.25×10²³ molecules O2.

This can be obtained as follow:

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.022×10²³ molecules. This implies that 1 mole of O2 also contains 6.022×10²³ molecules.

1 mole of O2 = 16x2 = 32 g.

Thus 6.022×10²³ molecules is present in 32 g of O2,

Therefore, 1.25×10²³ molecules will be present in =

(1.25×10²³ × 32) / 6.022×10²³ = 6.64 g of O2.

Therefore, 1.25×10²³ molecules present in 6.64 g of O2.

Next, the balanced equation for the reaction. This is given below:

CH4(g) + 2O2(g) —> CO2(g) + 2H2O(g)

Next, we shall determine the masses of CH4 and O2 that reacted and the mass of H2O produced from the balanced equation.

This can be obtained as follow:

Molar mass of CH4 = 12 + (4x1) = 16 g/mol.

Mass of CH4 from the balanced equation = 1 x 16 = 16 g

Molar mass of O2 = 16x2 = 32 g/mol.

Mass of O2 from the balanced equation = 2 x 32 = 64 g

Molar mass of H2O = (2x1) + 16 = 18 g/mol.

Mass of H2O from the balanced equation = 2 x 18 = 36 g

From the balanced equation above,

16 g of CH4 reacted with 64 g of O2 to produce 36 g if H2O.

Next, we shall determine the limiting reactant.

This can be obtained as follow:

From the balanced equation above,

16 g of CH4 reacted with 64 g of O2.

Therefore, 1.25 g of CH4 will react with = (1.25 x 64)/16 = 5 g of O2.

From the above calculations, we can see that only 5 g out of 6.64 g of O2 is needed to react completely with 1.25 g of CH4.

Therefore, CH4 is the limiting reactant.

Finally, we shall determine the mass of H2O produced from the reaction.

In this case, the limiting reactant will be used because it will give the maximum yield of H2O.

The limiting reactant is CH4 and the mass of H2O produced from the reaction can be obtained as follow:

From the balanced equation above,

16 g of CH4 reacted to produce produce 36 g if H2O.

Therefore, 1.25 g of CH4 will react to produce = (1.25 x 36)/16 = 2.81 g of H2O.

Therefore, 2.81 g of H2O were obtained from the reaction.

The mass in grams of H₂O which can be formed when 1.25g CH₄ are combined with 1.25×10²³ molecules O₂ is 2.8 grams.

Stoichiometry of any reaction tells about the amount of species present before and after the completion of the reaction.

Given chemical reaction is:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

Moles of CH₄ will b calculate as:

n = W/M, where

W = given mass = 1.25g

M = molar mass = 16g/mol

n = 1.25/16 = 0.078 moles

Molecues of CH₄ in 0.078 moles = 0.078×6.022×10²³ = 0.46×10²³

Given molecules of O₂ = 1.25×10²³

Required molecules of CH₄ is less as compared to the molecules of O₂, so here CH₄ is the limiting reagent and formation of water is depends on it only.

From the stoichiometry of the reaction it is clear that:

1 mole of CH₄ = will produce 2 moles of H₂O

0.078 moles of CH₄ = will produce 2×0.078=0.156 moles of H₂O

Mass of H₂O will be calculated by using its moles as:

W = (0.156)(18) = 2.8g

Hence required mass of H₂O is 2.8g.

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Answer:

Explanation:

A tertiary alcohol is a compound (an alcohol) in which the carbon atom that has the hydroxyl group (-OH) is also bonded (saturated) to three different carbon atoms.

Based on the question, the only tertiary alcohol that can result from C₆H₁₄O that have a 4-carbon chain is

2-hydroxy-2,3-dimethylbutane

     H  OH   H    H

      |     |       |      |

H - C - C -   C  - C - H

      |     |       |      |

     H  CH₃  CH₃ H

From the above, we can see that the carbon atom having the hydroxyl group is also bonded to three other carbon atoms. And since we aren't considering stereochemistry, this is the only tertiary alcohol we can have with a 4-carbon chain

Answer:

OPTION C is correct

3.86 x 10-19 J

Explanation:

Energy of the line can be calculated using below formula

E= h ν.................(1)

Where E= energy

h= plank constant= 6.626 10-34 J s

c=speed of light=3 x 108 m/s

But we know that Velocity V= = c / λ

Then substitute into equation (1) we have

E = h c / λ.............(2)

We can calculate our( hc ) in nm for unit consistency

h c =( 6.626 ×10^-34)x(3×108)

h c = (1.986 x 10-16 )

hc = 1.986 x 10-16 J nm then since our (hc) and λ are in the same unit , were good to go then substitute into equation(2)

E = h c / λ = (1.986 x 10-16) / 515

E = 3.86 x 10-19 J

Therefore, the Energy is 3.86 x 10-19 J

Answer:

Reaction of Chlorine with Hydrogen Chlorine and Hydrogen mixed together explodes when exposed to sunlight, which produces Hydrogen Chloride. In the dark away from sunlight, no reaction occurs, so light energy is required for a reaction. Cl2 + H2 = 2 HCl Reaction of Chlorine with Non-Metals Chlorine directly combines with most non-metals.

Explanation:

I hope this helps bro

Answer:

While balancing a chemical equation, we change the coefficient  to balance the number of atoms on each side of the equation

Explanation:

While balancing a chemical equation, we change the coefficient to balance the number of atoms on each side of the equation.

To summarize in chemistry terms, a chemical equation depicts the initial chemicals, or reactants, on the left-hand side and the final compounds, or products, just on right-hand side, divided by an arrow. In the chemical equation, the number of atoms in each element as well as the total charge are the same on opposite of the equation's sides.

Chemical equations are used in chemistry to depict chemical processes by writing the reactants and products in terms of their corresponding chemical formulas. While balancing a chemical equation, we change the coefficient to balance the number of atoms on each side of the equation.

Therefore, while balancing a chemical equation, we change the coefficient to balance the number of atoms on each side of the equation.

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Answer:

A divergent plate boundary  

Explanation:

At a divergent boundary, the plates pull away from each other and generate new crust.

Answer:

Because the ocean becomes larger, this is a divergent plate boundary. Divergent plates cause the ocean floor to expand, making the ocean larger.

Explanation:

PLATO ANSWER

Answer:

The correct answer is CaO > LiBr > KI.

Explanation:

Lattice energy is directly proportional to the charge and is inversely proportional to the size. The compound LiBr comprises Li+ and Br- ions, KI comprises K+ and I- ions, and CaO comprise Ca²⁺ and O²⁻ ions.  

With the increase in the charge, there will be an increase in lattice energy. In the given case, the lattice energy of CaO will be the highest due to the presence of +2 and -2 ions. K⁺ ions are larger than Li⁺ ion, and I⁻ ions are larger than Br⁻ ion.  

The distance between Li⁺ and Br⁻ ions in LiBr is less in comparison to the distance between K⁺ and I⁻ ions in KI. As a consequence, the lattice energy of LiBr is greater than KI. Therefore, CaO exhibits the largest lattice energy, while KI the smallest.  

Arranging the chemical compounds in order of decreasing magnitude of lattice energy, we have:

c. CaO.

a. LiBr

b. KI

Lattice energy can be defined as a measure of the energy required to dissociate one (1) mole of an ionic compound into its constituent anions and cations, in the gaseous state.

Hence, it is typically used to measure the bond strength of ionic compounds.

Generally, lattice energy is inversely proportional to the size of the ions and directly proportional to their electric charges.

Lithium bromide (LiBr) comprises the following ions:

Potassium iodide (KI) comprises the following ions:

Calcium oxide (CaO) comprises the following ions:

From the above, we can deduce that there is an increase in the charge possessed by the ionic chemical compounds and as such this would result in an increase in the lattice energy.

In order of decreasing magnitude of lattice energy, the chemical compounds are arranged as:

I. CaO.

II. KI.

III. LiBr.

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Answer:

B) geminal diol

Explanation:

Hello,

In this case, considering the attached picture, you can see that the substance resulting from the hydrolysis of an aldehyde or a ketone is a geminal diol since the two hydroxyl groups are in the same carbon. Such hydrolysis could be carried out in either acidic or basic conditions depending upon the equilibrium constant.

Regards.

Answer:

The concentration in mol/L =  4.342 mol/L

Explanation:

Given that :

mass of sodium chloride = 25.4 grams

Volume of the volumetric flask = 100 mL

We all know that the molar mass of sodium chloride NaCl = 58.5 g/mol

and number of moles = mass/molar mass

The number of moles of sodium chloride = 25.4 g/58.5 g/mol

The number of moles of sodium chloride = 0.434188 mol

The concentration in mol/L = number of mol/ volume of the solution

The concentration in mol/L = 0.434188 mol/ 100 × 10⁻³ L

The concentration in mol/L =  4.34188 mol/L

The concentration in mol/L =  4.342 mol/L

Answer:

(a) when a reaction system reaches a state of equilibrium, the concentration of the products is equal to the concentration of the reactants

Determining whether the statements about equilibrium is True or False

A) The concentration of the products is equal to the concentration of the reactants at equilibrium : TRUE

B) When a system is at equilibrium, Keq = 1 : TRUE

C) The rates of the forward reaction and the reverse reaction are equal at equilibrium :  TRUE

D) Adding a catalyst to a reaction system will shift the position of equilibrium to the right : FALSE

In a chemical reaction at equilibrium the value of Keq will be equal to 1 because the concentration of the products is equal to the concentration of the reactants in the chemica reaction. Also at equilibrium the rate of forward reaction is same as the rate of reverse reaction.

A catalyst can only affect the rate of reaction and not the amount of product ( yield of reaction).

Hence we can conclude that the answers to your questions are as listed above.

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Explanation:

An atom undergoes alpha decay by losing a helium atom.

So when bismuth undergoes alpha decay, we have;

²¹⁰₈₃Bi --> ⁴₂He + X

Mass number;

210 = 4 + x

x = 206

Atomic number;

83 = 2 + x

x = 81

The element is Thallium. The symbol is Ti.

For the second part;

X --> ⁴₂He + ²³⁴₉₀Th

Mass number;

x = 4 + 234 = 238

Atomic Number;

x = 2 + 90 = 92

The balanced nuclear equation is;

²³⁸₉₂U --> ⁴₂He + ²³⁴₉₀Th

Answer:

We can have: Calcium, strontium, or barium

Explanation:

In this case, we have to remember the solubility rules for sulfate [tex]SO_4~^-^2[/tex] and the chloride [tex]Cl^-[/tex]:

Sulfate

All sulfate salts are SOLUBLE-EXCEPT those also containing: Calcium, silver, mercury (I), strontium, barium or lead.([tex]Ca^+^2~,Ag^+~,Hg_2^+^2~,Sr^+2~,Ba^+^2~,Pb^+^2[/tex]), which are NOT soluble.

Chloride

All chloride salts as SOLUBLE-EXCEPT those also containing: lead, silver, or mercury (I). ([tex]Pb^+^2~,Ag^+~,Hg_2~^+^2[/tex]), which are NOT soluble.

If we the salt formed a precipitated with the sulfate anion, we will have as possibilities "Calcium, silver, mercury (I), strontium, barium or lead". If We dont have any precipitated with the Chloride anion we can discard "Silver, mercury (I),  lead" and our possibilities are:

"Calcium, strontium, or barium".

I hope it helps!

Explanation:

To calculate [H3O+] in the solution we must first find the pH from the [ OH-]

That's

pH + pOH = 14

pH = 14 - pOH

To calculate the pOH we use the formula

pOH = - log [OH-]

And [OH-] = 5.5 × 10^-5 M

So we have

pOH = - log 5.5 × 10^ - 5

pOH = 4.26

Since we've found the pOH we can now find the pH

That's

pH = 14 - 4.26

pH = 9.74

Now we can find the concentration of H3O+ in the solution using the formula

pH = - log H3O+

9.74 = - log H3O+

Find the antilog of both sides

The solution is basic since it's pH lies in the basic region.

Hope this helps you

Answer:

d) a solution that is 0.025M in HCOOH and 0.025M in HCOONa

Explanation:

The reaction of a weak acid (HOOH) with NaOH is as follows:

HCOOH + NaOH → HCOONa + H₂O

Based on the reaction, 1 mole of the acid reacts with 1 mole of the base (Ratio 1:1).

The initial moles of both species are:

HCOOH: 0.050L × (0.1mol / L) = 0.0050 moles of HCOOH

NaOH: 0.050L × (0.05 mol / L) = 0.0025 moles NaOH

After the reaction, all NaOH reacts with HCOOH producing HCOONa (Because moles of NaOH < moles HCOOH).

Final moles:

HCOOH: 0.0050 moles - 0.0025 moles (After reaction) = 0.0025 moles

HCOONa: Moles HCOONa = Initial Moles NaOH: 0.0025 moles

As volume of the mixture is 100mL (50 from the acid + 50 from NaOH), molarity of both HCOOH and HCOONa is:

0.0025 moles / 0.100L = 0.025M of both HCOOH and HCOONa

Thus, the initial mixture is equivalent to:

Answer:

6.5 mg/L.

Explanation:

Step one: write out and Balance the chemical reaction in the Question above:

NiCl2 + 2AgNO3 =====> 2AgCl + Ni(NO3)2.

Step two: Calculate or determine the number of moles of AgCl.

So, we are given that the mass of AgCl = 3.6 mg = 3.6 × 10^-3 g. Therefore, the number of moles of AgCl can be calculated as below:

Number of moles AgCl = mass/molar mass = 3.6 × 10^-3 g / 143.32. = 2.5118 × 10^-5 moles.

Step three: Calculate or determine the number of moles of NiCl2.

Thus, the number of moles of NiCl2 = 2.5118 × 10^-5/ 2 = 1.2559 × 10^-5 moles.

Step four: detemine the mass of NiCl2.

Therefore, the mass of NiCl2 = number of moles × molar mass = 1.2559 × 10^-5 moles × 129.6 = 1.6 × 10^-3 g.

Step five: finally, determine the concentration of NiCl2.

1000/ 250 × 1.6 × 10^-3 g. = 6.5 mg/L.

Answer:

A) 3.17 g of Zn

Explanation:

Let's consider the reduction of Zn(II) that occurs in an electrolysis bath.

Zn⁺²(aq) + 2e⁻ → Zn(s)

We can establish the following relations:

The mass of Zn deposited under these conditions is:

[tex]24min \times \frac{60s}{1min} \times \frac{6.5C}{s} \times \frac{1mol\ e^{-} }{96,468C} \times \frac{1molZn}{2mol\ e^{-}} \times \frac{65.38g}{1molZn} = 3.17 g[/tex]

Answer:

A.) 3.17

Explanation:

I got it right in class!

Hope this Helps!! :))

Answer:

0 g.

Explanation:

Hello,

In this case, since the reaction between methane and oxygen is:

[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]

If 0.963 g of methane react with 7.5 g of oxygen the first step is to identify the limiting reactant for which we compute the available moles of methane and the moles of methane consumed by the 7.5 g of oxygen:

[tex]n_{CH_4}=0.963gCH_4*\frac{1molCH_4}{16gCH_4}=0.0602molCH_4\\ \\n_{CH_4}^{consumed}=7.5gO_2*\frac{1molO_2}{32gO_2}*\frac{1molCH_4}{2molO_2} =0.117molCH_4[/tex]

Thus, since oxygen theoretically consumes more methane than the available, we conclude the methane is the limiting reactant, for which it will be completely consumed, therefore, no remaining methane will be left over.

[tex]left\ over=0g[/tex]

Regards.

Answer:

We can then determine that a reaction will shift to the right if Q<K

Explanation:

Comparing Q with K allows to find out the status and evolution of the system:

In this case, we can then determine that a reaction will shift to the right if Q<K

Answer:

[tex]\large \boxed{\text{0.57 g/mL; 3.7 mL; 6.6 g; 0.366 g/mL}}[/tex]

Explanation:

1. Density from mass and volume

[tex]\text{Density} = \dfrac{\text{mass}}{\text{volume}}\\\\\rho = \dfrac{m}{V}\\\\\rho = \dfrac{\text{8.6 g}}{\text{15 mL}} = \text{0.57 g/mL}\\\text{The density is $\large \boxed{\textbf{0.57 g/mL}}$}[/tex]

2. Volume from density and mass

[tex]V = \text{9.7 g}\times\dfrac{\text{1 mL}}{\text{2.6 g}} = \text{3.7 mL}\\\\\text{The volume is $\large \boxed{\textbf{3.7 mL}}$}[/tex]

3. Mass from density and volume

[tex]\text{Mass} = \text{4.1 cm}^{3} \times \dfrac{\text{1.6 g}}{\text{1 cm}^{3}} = \textbf{6.6 g}\\\\\text{The mass is $\large \boxed{\textbf{6.6 g}}$}[/tex]

4. Density by displacement

Volume of water + object = 24.6 mL

Volume of water                = 12.8 mL

Volume of object               = 11.8 mL

[tex]\rho = \dfrac{\text{4.3 g}}{\text{11.8 mL}} = \text{0.36 g/mL}\\\text{The density is $\large \boxed{\textbf{0.36 g/mL}}$}[/tex]

Your drawing showing water displacement using a graduated cylinder should resemble the figure below.