In addition to NF3, two other fluoro derivatives of nitrogen are known: N2F4 and N2F2. What shapes do you predict for these two molecules?

Answer:

the final concentration is C=0.235 mg/L

Explanation:

doing a mass balance

ammonium mass outflow = ammonium inflow from river+ effluent discharge of ammonium  = 12000 L/s* 0.015 mg/L  + 1500 L /s * 2 mg/L = 3180 mg/s

outflow rate = 12000 L/s+ 1500 L /s  = 13500 L/s

then the final concentration is

C= ammonium mass outflow/outflow rate = 3180 mg/s/13500 L/s= 0.235 mg/L

Final answer:

To find the concentration of the mixture after becoming perfectly mixed downstream, we can use the principle of mass conservation.

Explanation:

To find the concentration of the mixture after becoming perfectly mixed downstream, we can use the principle of mass conservation. The total mass of ammonium-N in the effluent discharge is given by the product of its concentration and flow rate, while the total mass of ammonium-N in the river is given by the product of its concentration and flow rate. When the effluent discharge and river mix together, the total mass of ammonium-N remains constant.

We can use the equation:

(flow rate of effluent discharge * concentration of effluent discharge) + (flow rate of river * concentration of river) = (flow rate of mixture * concentration of mixture)

Plugging in the given values:

(1.5 m³/s * 2 mg/L) + (12 m³/s * 0.015 mg/L) = (flow rate of mixture * concentration of mixture)

Solving this equation will give us the concentration of the mixture after they have become perfectly mixed downstream.

Answer: The Cu-63 isotope has the highest percent natural abundance

Explanation:

Average atomic mass 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]

We are given:

Two isotopes of Copper, which are Cu-63 and Cu-65

Average atomic mass of copper = 63.54 amu

As, the average atomic mass of copper is closer to the mass of Cu-63 isotope. This means that the relative abundance of this isotope is the highest as compared to the other isotope.

Percentage abundance of Cu-63 isotope = 69.2%

Percentage abundance of Cu-65 isotope = 30.8 %

Hence, the Cu-63 isotope has the highest percent natural abundance

Answer:

The answer will be that if there is a linear relationship exist between CO2 and the global temperature, then if one variable change, the other also will change correspondingly.

The amount of hydrogen chloride that can be made is 1064 g

Why?

The two reactions are:

2H₂O → 2H₂ + O₂ 75.3 % yield

H₂ + Cl₂ → 2HCl 69.8% yield

We have to apply a big conversion factor to go from grams of water (The limiting reactant), to grams of HCl, the final product. We have to be very careful with the coefficients and percentage yields!

[tex]500.0gH_2O*\frac{1moleH_2O}{18.01 gH_2O}*\frac{2 moles H_2}{2 moles H_2O}*\frac{2.015g H_2}{1 mole H_2}*\frac{75.3 actual g}{100 theoretical g}=42.12 g H_2[/tex]

[tex]42.12H_2*\frac{1 mole H_2}{2.015gH_2}*\frac{2 moles HCl}{1 mole H_2}*\frac{36.46g}{1 mole HCl}*\frac{69.8 actualg}{100 theoreticalg} =1064gHCl[/tex]

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The amount of hydrogen chloride yield in the second reaction is 1065.7 g.

The two given reactions

The amount of hydrogen gas yield in the first reaction is calculated as follows;

[tex]\frac{500 \ g\ H_2O}{18 \ g \ H_2O} \times (2 \ mol\ H_2) \times 0.753= 41.83 \ g \ H_2[/tex]

The amount of hydrogen chloride yield in the second reaction is calculated as follows;

[tex]41.83 \ g \times (36.5 \ HCl) \times 0.698 = 1065.7 \ g \ HCl[/tex]

Thus, the amount of hydrogen chloride yield in the second reaction is 1065.7 g.

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Answer:

3.8 cm

Explanation:

Given data

Volume of a sphere: V = 4/3 × π × r³ = 4/3 × π × (2.5 cm)³ = 65 cm³

We can find the final volume using the combined gas law.

[tex]\frac{P_{1}\times V_{1}}{T_{1}} =\frac{P_{2}\times V_{2}}{T_{2}}\\\frac{3.2atm\times 65cm^{3} }{278.6K} =\frac{1.0atm\times V_{2}}{298.2K}\\V_{2}=223cm^{3}[/tex]

The final radius of the bubble is:

V = 4/3 × π × r³

223 cm³ = 4/3 × π × r³

r = 3.8 cm

Explanation:

A)

We know, each mole contains [tex]N_A=[/tex] [tex]6.023 \times 10^{23}[/tex] atoms.

It is given that mass of one oxygen atom is m= [tex]2.66\times 10^{-23}\ g[/tex].

Therefore, mass of one mole of oxygen, [tex]M=m\times N_A[/tex].

Putting value of n and [tex]N_A[/tex],

[tex]M=2.66\times 10^{-23}\times 6.023\times 10^{23} \ gm\\M=16.0\ gm[/tex]

B)

Given,

Mass of water in glass=0.050 kg = 50 gm.

From above part mass of one mole of oxygen atoms = 16.0 gm.

Therefore, number of mole of oxygen equivalent to 50 gm oxygen[tex]=\dfrac{50}{16}=3.1 \ moles.[/tex]

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Avogadro's number

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The mass of 1 mole of oxygen atoms is 16.00 grams. There are 3125 moles of oxygen atoms with a mass equal to that of a glass of water.

To find the mass of 1 mole of oxygen atoms, we need to use the molar mass of oxygen. The molar mass is the mass of one mole of a substance, and it is equal to the atomic mass in grams. The atomic mass of oxygen is 16.00 g/mol, so the mass of 1 mole of oxygen atoms is 16.00 grams.

To determine how many moles of oxygen atoms have a mass equal to the mass of a glass of water, we need to use the given mass of the water and the molar mass of oxygen. The molar mass of oxygen is 16.00 g/mol, and the mass of a glass of water is 0.050 kg (or 50,000 grams). Using the molar mass, we can set up a proportion to find the number of moles of oxygen atoms.

Moles of oxygen atoms = (Mass of water / Molar mass of oxygen) = (50,000 g / 16.00 g/mol) = 3125 moles of oxygen atoms.

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Answer:

When the neuron is at rest, what is primarily responsible for moving potassium ions OUT of the cell?​

The answer is "a concentration gradient"

Explanation:

A neuron is the main component of nervous tissue and it transmits information by electro-chemical signalling. For the nervous system to function, neurons must be able to send and receive signals.

A neuron at rest is negatively charged. The negative charge within the cell is created by the cell membrane being more permeable to potassium ion movement than sodium ion movement. At rest, there is a high concentration of potassium ions (K+) inside the cell compared to the extracellular fluid due to a net movement with the concentration gradient.

A concentration gradient acts on K+ (potassium ions). High number of potassium ions reside inside the cell, a chemical gradient occurs and pushes potassium out of the cell. The neuron membrane is more permeable to potassium ions than to other ions allowing it to selectively move out of the cell taking a positive charge with it down its concentration gradient.

Answer: 6.63

Explanation:Please see attachment for explanation

Final answer:

The pKa value for the weak acid HA can be calculated using the equation pKa = -log(Ka). Given the equilibrium concentrations, we can use the equation Ka = ([H+][A-])/[HA] to solve for Ka. After finding Ka, we can calculate pKa using the formula pKa = -log(Ka).

Explanation:

The pKa value for the weak acid HA can be calculated using the equation:

pKa = -log(Ka)

Given the equilibrium concentrations of [HA] = 0.170 M, [H+] = 2.00 × 10^-4 M, and [A-] = 2.00 × 10^-4 M, we can use the equation:

Ka = ([H+][A-])/[HA]

Plugging in the values gives us:

Ka = (2.00 × 10^-4)(2.00 × 10^-4)/(0.170)

After solving for Ka, we can calculate pKa using the formula pKa = -log(Ka).

Answer: The p-function of [tex]Zn^{2+}[/tex] and [tex]NO_3^{-}[/tex] ions are 2.51 and 2.14 respectively.

Explanation:

p-function is defined as the negative logarithm of any concentration.

We are given:

Millimolar concentration of zinc nitrate = 3.1 mM

Millimolar concentration of calcium nitrate = 4.2 mM

Converting this into molar concentration, we use the conversion factor:

1 M = 1000 mM

1 mole of zinc nitrate produces 1 mole of zinc ions and 2 moles of nitrate ions

Concentration of zinc ions = 0.0031 M

Concentration of nitrate ions in zinc nitrate, [tex]M_1=(2\times 0.0031)=0.0062M[/tex]

1 mole of calcium nitrate produces 1 mole of calcium ions and 2 moles of nitrate ions

Concentration of calcium ions = 0.0042 M

Concentration of nitrate ions in calcium nitrate, [tex]M_2=(2\times 0.0042)=0.0084M[/tex]

To calculate the concentration of nitrate ions in the solution, we use the equation:

[tex]M=\frac{M_1V_1+M_2V_2}{V_1+V_2}[/tex]

Putting values in above equation, we get:

[tex]M=\frac{(0.0062\times 1)+(0.0084\times 1)}{1+1}\\\\M=0.0073M[/tex]

Calculating the p-function of zinc ions and nitrate ions in the solution:

[tex]\text{p-function of }Zn^{2+}\text{ ions}=-\log[Zn^{2+}][/tex]

[tex]\text{p-function of }Zn^{2+}\text{ ions}=-\log(0.0031)\\\\\text{p-function of }Zn^{2+}\text{ ions}=2.51[/tex]

[tex]\text{p-function of }NO_3^{-}\text{ ions}=-\log[NO_3^{-}][/tex]

[tex]\text{p-function of }NO_3^{-}\text{ ions}=-\log(0.0073)\\\\\text{p-function of }NO_3^{-}\text{ ions}=2.14[/tex]

Hence, the p-function of [tex]Zn^{2+}[/tex] and [tex]NO_3^{-}[/tex] ions are 2.51 and 2.14 respectively.

The p-function values for Zn2+ and NO3- are 9.6 and 15.5, respectively.

The p-function for Zn2+ is 9.6, and the p-function for NO3- is 15.5.

Answer:

51:49 is the ratio of female participants to male participants.

Explanation:

Percentage of females in the race = 51 %

Percentage of males in the race = 100% - 51 % = 49%

Total participants in race = 400

Number of women participants = [tex]51% of 400=\frac{51}{100}\times 400=204[/tex]

Number of men participants  = [tex]49% of 400=\frac{49}{100}\times 400=196[/tex]

Ratio of female participants to male participants :

[tex]\frac{204}{196}=\frac{51}{49}=51:49[/tex]

Final answer:

To determine the ratio of female to male participants in a race with 400 participants and 51% females, we calculate 204 females and 196 males, resulting in a simplified ratio of 51:49.

Explanation:

To find the ratio of female to male participants in a local Thanksgiving 5K race with four hundred participants, where 51% are female, we first calculate the number of female and male participants. Since 51% of 400 participants are female, there are 204 female participants (0.51 × 400). The remaining participants are male, which accounts for 196 participants (400 - 204).

The ratio of female to male participants can be expressed as 204:196. To simplify this ratio, we divide both numbers by their greatest common divisor, which is 4. Therefore, the simplified ratio of female to male participants is 51:49.

Answer:

0,034 moles

Explanation:

Answer : The one part nitrogen gas combines with one part oxygen gas to form one part of dinitrogen monoxide.

Explanation :

Balanced chemical reaction : It is defined as the reaction in which the number of atoms of individual elements present on reactant side must be equal to the product side.

When nitrogen gas combines with oxygen gas then it react to give dinitrogen monoxide or nitrous oxide.

The balance chemical reaction will be:

[tex]2N_2(g)+O_2(g)\rightarrow 2N_2O(g)[/tex]

By the stoichiometry we can say that, 2 parts of nitrogen gas combines with 1 part of oxygen gas to give 2 parts of dinitrogen monoxide or nitrous oxide.

First we have to determine the limiting reagent.

From the reaction we conclude that,

As, 2 moles of nitrogen gas combine with 1 mole of oxygen gas

So, 1 moles of nitrogen gas combine with 0.5 mole of oxygen gas

It means that, oxygen gas is an excess reagent because the given moles are greater than the required moles and nitrogen gas is a limiting reagent and it limits the formation of product.

Now we have to determine the moles of dinitrogen monoxide.

As, 2 moles of nitrogen gas combine to give 2 mole of dinitrogen monoxide

So, 1 mole of nitrogen gas combine to give 1 mole of dinitrogen monoxide

Thus, the one part nitrogen gas combines with one part oxygen gas to form one part of dinitrogen monoxide.

Final answer:

One part nitrogen gas (N₂) combines with one part oxygen gas (O₂) to produce two parts of nitric oxide (NO).

Explanation:

When nitrogen gas (N₂) combines with oxygen gas (O₂), they can form several different nitrogen oxides, depending on the conditions and proportions in which they react. For the specific formation of nitric oxide (NO), we need to understand that one volume of nitrogen gas will combine with one volume of oxygen gas to form two volumes of nitric oxide. This is evident from the balanced chemical equation for this reaction:

N₂(g) + O₂(g) → 2NO(g)

This means 1 part of nitrogen gas combines with 1 part of oxygen gas to form 2 parts of nitric oxide (NO). The balancing of the equation indicates that two molecules of NO are formed from one molecule of nitrogen and one molecule of oxygen, due to the conservation of mass and volume in chemical reactions according to the Law of Combining Volumes.

Answer:

Xenon

Explanation:

Avogadro’s number represent the number of the constituent particles which are present in one mole of the substance. It is named after scientist Amedeo Avogadro and is denoted by [tex]N_0[/tex].

Avogadro constant:-

[tex]N_a=6.023\times 10^{23}[/tex]

Let the molar mass of the element is x g/mol

So,

[tex]6.023\times 10^{23}[/tex] atoms have a mass of x g

Also,

[tex]3.89\times 10^{24}[/tex] atoms have a mass of [tex]\frac{x}{6.023\times 10^{23}}\times 3.89\times 10^{24}[/tex] g

This mass is equal to 848 g

So,

[tex]\frac{x}{6.023\times 10^{23}}\times 3.89\times 10^{24}=848[/tex]

x= 131.3 g/mol

This mass correspond to xenon.

Explanation:

For the given reaction, the temperature of liquid will rise from 298 K to 339 K. Hence, heat energy required will be calculated as follows.

             [tex]Q_{1} = mC_{1} \Delta T_{1}[/tex]

Putting the given values into the above equation as follows.

           [tex]Q_{1} = mC_{1} \Delta T_{1}[/tex]

                      = [tex]27.3 g \times 1.70 J/g K \times 41[/tex]

                      = 1902.81 J

Now, conversion of liquid to vapor at the boiling point (339 K) is calculated as follows.

           [tex]Q_{2}[/tex] = energy required = [tex]mL_{v}[/tex]

    [tex]L_{v}[/tex] = latent heat of vaporization

Therefor, calculate the value of energy required as follows.

             [tex]Q_{2}[/tex] = [tex]mL_{v}[/tex]

                         = [tex]27.3 \times 444[/tex]

                         = 12121.2 J

Therefore, rise in temperature of vapor from 339 K to 373 K is calculated as follows.

            [tex]Q_{3} = mC_{2} \Delta T_{2}[/tex]

Value of [tex]C_{2}[/tex] = 1.06 J/g,    [tex]\Delta T_{2}[/tex] = (373 -339) K = 34 K

Hence, putting the given values into the above formula as follows.

             [tex]Q_{3} = mC_{2} \Delta T_{2}[/tex]

                       = [tex]27.3 g \times 1.06 J/g \times 34 K[/tex]

                       = 983.892 J

Therefore, net heat required will be calculated as follows.

            Q = [tex]Q_{1} + Q_{2} + Q_{3}[/tex]

                = 1902.81 J + 12121.2 J + 983.892 J

                = 15007.902 J

Thus, we can conclude that total energy (q) required to convert 27.3 g of THF at 298 K to a vapor at 373 K is 15007.902 J.

Answer: sulfur pentafluoride

Explanation:

The rules for naming of binary molecular compound :

In the given formula, the lower group number element is written first in the name and keep its element name and the higher group number is written second.

First element i.e. sulphur in the formula is named first and keep its element name.

1) Gets a prefix if there is a subscript on it such as mono for 1, di for 2, tri for 3 and so on.

Second element i.e. fluorine is named second.

1) Use the root of the element name, if it is an anion then use suffix (-ide).

2) Always use a prefix on the second element such as mono for 1, di for 2, tri for 3 and so on.

Therefore, the chemical name of compound [tex]SF_5[/tex] is sulfur pentafluoride

The name of the molecular compound SF₅ is sulfur pentafluoride. Therefore, option A is correct.

A molecular compound is a compound composed of two or more nonmetallic elements. In molecular compounds, atoms are joined together by covalent bonds, which involve the sharing of electrons between atoms.

Examples of molecular compounds include water (H₂O), carbon dioxide (CO₂), methane (CH₄), and ammonia (NH₃). Molecular compounds often have specific naming conventions based on the elements present and their respective ratios, such as using prefixes to indicate the number of atoms for each element.

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Final answer:

The balanced half-reaction for the oxidation of gaseous nitrogen dioxide to nitrate ion in acidic aqueous solution is 4NO2(g) + O2(g) + 4H+(aq) -> 4NO3-(aq).

Explanation:

The balanced half-reaction for the oxidation of gaseous nitrogen dioxide to nitrate ion in acidic aqueous solution is:

4NO2(g) + O2(g) + 4H+(aq) → 4NO3-(aq)

In this reaction, nitrogen dioxide (NO2) is oxidized to nitrate ion (NO3-), and oxygen gas (O2) is reduced to water (H2O). The reaction takes place in acidic surroundings, indicated by the presence of hydrogen ions (H+).

Answer: 40mL

Explanation: What is required for this question is to obtain the number of moles of NaOH solute in the required solution.

Number of moles of solute in solution = (Concentration of solution in mol/L) × (Volume of solution in L)

Number of moles of NaOH in the required solution = 0.2 × (300/1000) = 0.06 moles

This number of moles has to be in the volume of stock solution used to make the required solution.

Volume of stock solution needed = (Number of moles of NaOH in the stock solution that has to match that in the required solution in L)/(concentration of stock solution in mol/L)

Volume of stock solution required = 0.06/1.5 = 0.04L = 40mL

QED.

To make 300 mL of a 0.2000 M NaOH solution from a 1.5 M stock solution, you need 40 mL of the stock solution, as calculated using the dilution equation.

To find out how many milliliters of a 1.5 M stock solution is needed to make 300 mL of a 0.2000 M dilute NaOH solution, you can use the equation of dilution:
C1 times V1 = C2 times V2 where C1 and V1 are the concentration and volume of the stock solution, and C2 and V2 are the concentration and volume of the dilute solution, respectively.

The formula reorganized to solve for V1 is:
V1 = (C2  imes V2) / C1
Substituting the given values, we get:
V1 = (0.2000 M times 300 mL ) / 1.5 M
V1 = 40 mL

Therefore, you need 40 mL of the stock solution to prepare the required dilution.

Answer:  The work energy lost by the system is -6078 Joules

Explanation:

According to first law of thermodynamics:

[tex]\Delta E=q+w[/tex]

[tex]\Delta E[/tex]=Change in internal energy

q = heat absorbed or released

w = work done or by the system

w = work done by the system=[tex]-P\Delta V[/tex]  {Work is done by the system as the final volume is greater than initial volume and is negative}

where P = pressure = 3 atm

[tex]\Delta V[/tex] = change in volume = (55-35) L = 20 L

w =[tex]-3atm\times (20)L=-60Latm=-6078Joules[/tex]  

{1Latm=101.3J}

Thus work energy lost by the system is -6078 Joules

Answer:

Atomic mass of 35.5 g/mol is of chlorine.

Atomic mass of 89.02 g/mol is of Yttrium.

Ytterium(III) chloride is the correct name for [tex]YCl_3[/tex].

1.835 grams of [tex]YCl_3[/tex]can be prepared.

Explanation:

[tex]2M+3X_2\rightarrow 2MX_3[/tex]

Moles of [tex]X_2[/tex] =n

Number of moleules of [tex]X_2=8.92\times 10^{20} molecules[/tex]

1 mole = [tex]6.022\times 10^{23} molecules[/tex]

[tex]n=\frac{8.92\times 10^{20} molecules}{6.022\times 10^{23} molecules}[/tex]

n = 0.001481 mole

Mass of [tex]X_2=0.105 g[/tex]

Molar mass of [tex]X_2=m[/tex]

[tex]n=\frac{Mass}{\text{Molar mass}}[/tex]

[tex]0.001481 mol=\frac{0.105 g}{m}[/tex]

m = 71 g/mol

Atomic mass of X = [tex]\frac{71 g/mol}{2}=35.5 g/mol[/tex]

Atomic mass of 35.5 g/mol is of chlorine.

The compound MX3 consists of 54.47% X by mass:

Molar mass of compound = M'

Percentage of element in compound :

[tex]=\frac{\text{number of atoms}\times text{Atomic mass}}{\text{molar mas of compound}}\times 100[/tex]

X:

[tex]54.47\%=\frac{3\times 35.5 g/mol}{M'}\times 100[/tex]

M' = 195.52 g/mol

Molar mass of compound = M'

M' = 1 × (atomic mass of M)+ 3 × (atomic mass of X)

195.52 g/mol = atomic mass of M + 3 × (35.5 g/mol)

Atomic mass of M = 89.02 g/mol

Atomic mass of 89.02 g/mol is of Yttrium.

Ytterium(III) chloride is the correct name for [tex]YCl_3[/tex].

[tex]2Y+3Cl_2\rightarrow 2YCl_3[/tex]

Moles of Yttrium = [tex]\frac{1g }{89.02 g/mol}=0.01123 mol[/tex]

Moles of chlorine gas= [tex]\frac{1 g}{71 g/mol}=0.01408 mol[/tex]

According to reaction, 3 moles of chlorine reacts with 2 moles of Y.

Then 0.01408 moles of chlorine gas will :

[tex]\frac{2}{3}\times 0.01408 mol=0.009387 mol[/tex] of Y.

This means that chlorine is in limiting amount., So, amount of yttrium (III) chloride will depend upon amount of chlorine gas.

According to reaction , 3 moles of chlorine gives 2 moles of [tex]YCl_3[/tex]

Then 0.01408 moles of chlorine will give :

[tex]\frac{2}{3}\times 0.01408 mol=0.009387 mol[/tex] of [tex]YCl_3[/tex]

Mass of 0.009387 moles of [tex]YCl_3[/tex]:

0.009387 mol × 195.52 g/mol = 1.835 g

1.835 grams of [tex]YCl_3[/tex]can be prepared.

To identify the elements in MX3, the calculations show that M is Iron (Fe) and X is Chlorine (Cl), forming Iron(III) Chloride (FeCl3). Given 1.00 g of each reactant, approximately 2.90 g of FeCl3 can be prepared. The limiting reagent in this process is Iron (Fe).

To determine the identities of M and X in the compound MX3 and the mass of MX3 that can be prepared, follow these steps:

Determine the molar mass of X2: The given data states that 0.105 g of X2 contains 8.92 × 1020 molecules. Using Avogadro's number (6.022 × 1023 molecules/mol), we can find the molar mass (MX2) of X2:

Calculate the atomic mass of X: Since X2 is diatomic, we divide the molar mass by 2:

Identify the element X - The atomic mass of 35.45 g/mol suggests that X is Chlorine (Cl).

Determine the molar mass and identity of M - Given that MX3 consists of 54.47% X by mass:

Solve for MM:

Identify the element M - The atomic mass of 55.85 g/mol suggests that M is Iron (Fe).

Name of MX3: Since M is Iron (Fe) and X is Chlorine (Cl), MX3 is Iron(III) Chloride (FeCl3).

Final Mass Calculation

Mole calculation for 1.00 g of M:

Limiting reagent: Each mole of X2 provides 2 moles of Cl, so: 0.0141 mol Cl2 × 2 = 0.0282 mol Cl (excess)

Iron (Fe) is the limiting reagent with 0.0179 mol producing 0.0179 mol of FeCl3

Calculate the mass of FeCl3:

Molar mass of FeCl3 = 55.85 g/mol + 3 × 35.45 g/mol ≈ 162.20 g/mol

Mass of FeCl3 = 0.0179 mol × 162.20 g/mol ≈ 2.90 g

Answer:

The reaction rate is increased.

Explanation:

The pressure is the force the gas molecules do when hitting each other and the walls of the container they're. The partial pressure is the pressure of a substance in a mixture would have if it was alone at the same conditions.

Thus, when the partial pressure increases, it means that the molecules collide more often. The reaction happens when the molecules collide in the right way, so, if the collisions are happening more often, the rate must be higher.

Answer:

C

Explanation:

Pb + H2O + O2 => Pb(OH)2

Number of oxygen atom on left = 3

Number of oxygen atom on the right = 2

To balance number of oxygen

Change the coefficient of Pb and H2O to two each

2Pb + 2H2O + O2 => 2Pb(OH)2

The coefficient of Pb and H2O is 2,2

The correct coefficients for Pb and H₂O when balancing the equation Pb + H₂O + O₂ ⇒ Pb(OH)₂ are 1 and 2, respectively; therefore, the answer is A. 1,2. This ensures that the chemical equation is fully balanced with the correct number of atoms for each element on both sides.

When balancing the chemical equation Pb + H₂O + O₂ ⇒ Pb(OH)₂, the correct coefficients for Pb and H₂O, respectively, are 1 and 2. Balancing chemical equations requires that the same number of atoms for each element be present on both sides of the equation. Starting with lead (Pb), we see that one Pb atom is needed on both sides, so we keep the coefficient for Pb as 1.

Next, we balance the hydrogen atoms. There are two hydrogens in each hydroxide ion, and since there are two hydroxide ions in Pb(OH)₂, this means there are a total of four hydrogen atoms in the product. To balance this, we need 2 molecules of H₂O (since each molecule of water contains two hydrogen atoms, giving us the four we need), making the coefficient for H₂O equal to 2.

Lastly, the oxygen atoms will balance themselves once H and Pb are balanced. With 2 H₂O molecules, we have 2 oxygen atoms, and the Pb(OH)₂ molecule has 2 oxygen atoms as well, which means we don't need any additional O₂ for balance. The balanced equation is 1 Pb + 2 H₂O + O₂ ⇒ 1 Pb(OH)₂. Therefore, the correct answer is A. 1,2.

Answer: No crystals of potassium sulfate will be seen at 0°C for the given amount.

Explanation:

We are given:

Mass of potassium nitrate = 47.6 g

Mass of potassium sulfate = 8.4 g

Mass of water = 130. g

Solubility of potassium sulfate in water at 0°C = 7.4 g/100 g

This means that 7.4 grams of potassium sulfate is soluble in 100 grams of water

Applying unitary method:

In 100 grams of water, the amount of potassium sulfate dissolved is 7.4 grams

So, in 130 grams of water, the amount of potassium sulfate dissolved will be [tex]\frac{7.4}{100}\times 130=9.62g[/tex]

As, the soluble amount is greater than the given amount of potassium sulfate

This means that, all of potassium sulfate will be dissolved.

Hence, no crystals of potassium sulfate will be seen at 0°C for the given amount.

Answer:

A) 0.20 cm³

B) 49.7 m²

C) 99.99%

D) 17.7 mg

Explanation:

A) The density of a material represents the mass that it occupies in a "piece" of volume. Thus, the density (d) is the mass (m) divided by the volume (v):

d =m/v

If the mass is 40.0 mg = 0.04 g, and the density is 0.20 g/cm³, the volume is:

0.20 = 0.04/v

v = 0.04/0.20

v = 0.20 cm³

B) The surface area (S) is the are that is presented in each gram of the material, so, it's the area (a) divided by the mass (m):

S = a/m

If the mass is 40.0 mg = 0.04 g, and the surface area is 1242 m²/g, so:

1242 = a/0.04

a = 49.7 m²

C) The percent of mercury removed is the mass removed divided by the initial mass, this multiplied by 100%. The mass removed is the initial mass (m0) less the final mass (m), so:

%removed = [(7.748 - 0.001)/7.748] *00%

%removed = 99.99%

D) The final mass of the spongy material is it mass (10 mg) plus the mass removed of the mercury (7.748 - 0.001 = 7.747 mg), so:

m = 10 + 7.747

m = 17.747 mg

m = 17.7 mg

Final answer:

This detailed answer explains how to calculate the volume, surface area, percentage of mercury removed, and final mass of a new material used to remove mercury from water.

Explanation:

A) To calculate the volume of a 40.0-mg sample of the material, you can use the formula: Volume = Mass / Density. Therefore, Volume = 40.0 mg / 0.20 g/cm³ = 200 cm³.

B) The surface area for a 40.0-mg sample can be calculated by multiplying the specific surface area by the mass: 1242 m²/g × 40.0 mg = 49680 m².

C) The percentage of mercury removed from the water is: ((Initial mercury mass - Final mercury mass) / Initial mercury mass) × 100 = ((7.748 mg - 0.001 mg) / 7.748 mg) × 100 = 99.9874%.

D) The final mass of the spongy material after exposure to mercury is the initial mass minus the mass used: 40.0 mg - 10.0 mg = 30.0 mg.

To calculate the pH of a solution containing a weak base and its conjugate weak acid, use the Henderson-Hasselbalch equation.

To calculate the pH of a solution that contains a weak base and its conjugate acid, we need to apply the Henderson-Hasselbalch equation.

The Henderson-Hasselbalch equation is given by: pH = pKa + log([A-]/[HA]), where pKa is the negative logarithm of the acid dissociation constant (Ka), [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.

In this case, the weak base has a concentration of 0.047 M and the conjugate weak acid has a concentration of 0.057 M. The pKa value of the acid is 7.2 x 10^-8. Plugging these values into the Henderson-Hasselbalch equation, we can calculate the pH of the solution.

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The pH of a 75.0 mL solution that is 0.047 M in weak base and 0.057 M in the conjugate weak acid ( K a = 7.2 × 10⁻⁸) will be approximately 7.06.

To determine the pH of a solution that contains both a weak base and its conjugate acid, we can use the Henderson-Hasselbalch equation. The equation is:

pH = pKa + log([A⁻]/[HA])

Where:

Given that the Ka for the weak acid is 7.2 x 10⁻⁸, the pKa is:

pKa = -log(Ka) = -log(7.2 x 10⁻⁸)

The concentration of the weak base ([A⁻]) is given as 0.047 M, and the concentration of the conjugate weak acid ([HA]) is 0.057 M.

Substituting into the Henderson-Hasselbalch equation:

pH = pKa + log(0.047/0.057)

Performing the calculations:

pH = 7.14 + log(0.825)

pH 7.14 - 0.084 = 7.06

Therefore, the pH of the solution is approximately 7.06.

Answer:

The unknown gas is H2

Explanation:

Step 1: Data given

A mixture contains:

5.4 grams of Ar

5.4 grams of Ne

5.4 grams of X2

Molar mass of Ar = 39.95 g/mol

Molar mass of Ne = 20.18 g/mol

Step 2: Calculate moles of gases

.Moles of Ne = 4.5grams /20.18 g/mol = 0.223 moles

Moles of Ar= 4.5 grams /39.95 g/mol = 0.113 moles

Step 3: Calculate volume of gases

Volume of Ne =22.4 * 0.223 = 5

Volume of Ar =22.4 * 0.1525 = 2.53 L

Volume of unknown gas = 69.03 - 5 - 2.53 = 61.5 L

Step 4: Calculate moles of unknown gas

Moles of unkown gas =61.5/22.4 = 2.75 moles

Step 5: Calculate molar mass of unknown gas

Molar mass = mass / moles

Molar mass = 5.4 grams /2.75 moles ≈ 2 g/mol

The unknown gas is H2

The question is incomplete, the complete question is:

For a certain chemical reaction, the standard Gibbs free energy of reaction is 144. kJ. Calculate the temperature at which the equilibrium constant K = [tex]5.9\times 10^{-26}[/tex] .

Round your answer to the nearest degree.

Answer:

25°C is the temperature at which the equilibrium constant is [tex]5.9\times 10^{-26}[/tex].

Explanation:

[tex]\Delta G^o=-RT\ln K[/tex]

where,

[tex]\Delta G^o[/tex] = standard Gibbs free energy = 144.0 kJ=144,000 J  (Conversion factor: 1kJ = 1000J)

R = Gas constant = [tex]8.314 J/K mol[/tex]

T = temperature at which reaction is occurring = ?

K = Equilibrium constant of the reaction =[tex]5.9\times 10^{-26}[/tex]

Putting values in above equation, we get:

[tex]144,000 J/mol=-(8.3145J/Kmol)\times T\times \ln [5.9\times 10^{-26}][/tex]

[tex]T=\frac{144,000 J/mol}{-(8.314 J/Kmol)\times \ln [5.9\times 10^{-26}]}[/tex]

T = 298.15 K

T = 298.15 - 273 °C = 25°C

25°C is the temperature at which the equilibrium constant is [tex]5.9\times 10^{-26}[/tex].

Using the equation ΔG° = -RTlnK, which relates the Gibbs free energy change to the equilibrium constant, you can solve for temperature by rearranging the formula. Ensure that the units for Gibbs energy and the gas constant match. Plugging the given values into the rearranged formula will provide the temperature.

The student's question asked how to calculate the temperature at which the equilibrium constant K equals 5.9 x 10 with a standard Gibbs free energy of reaction of 144 kJ.

The relationship between the Gibbs free energy change and the equilibrium constant is defined by the equation ΔG° = -RTlnK, where R is the gas constant (8.314 J/K mol), T is the absolute temperature in Kelvin, and K is the equilibrium constant.

To solve for temperature, rearrange the equation as T = -ΔG / (R * lnK). But first, you must ensure that the units for Gibbs energy and the gas constant match. If ΔG is given in kJ, it should be converted to J (1 kJ = 1000 J).

So, T = -(144,000 J) / (8.314 J/K mol * ln(5.9 x 10)), which should give you the answer.

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Answer:

Basic

Explanation:

[tex]Sodium\ Fluoride\ +\ Water = Hydrogen\ Fluoride +\ Sodium\ Hydroxide[/tex]  

[tex]NaF + H_{2}O\rightarrow HF + NaOH[/tex]

Sodium fluoride, NaF, is a soluble salt that dissociates completely in aqueous solution to give sodium cations, Na+, and fluoride anions, F-

[tex]NaF\rightarrow Na^{+} +F^{-}[/tex]

and when it dissolve in water the pH of the solution becomes greater than seven thereby becoming basic.

Final answer:

After sodium fluoride is dissolved in pure water, the solution becomes slightly basic due to the hydrolysis of fluoride ions which generates hydroxide ions.

Explanation:

When sodium fluoride is added to pure water and stirred until it dissolves, fluoride ions (F⁻) are released into the solution. These ions are capable of reacting, to a small extent, with water in a process known as hydrolysis. During this reaction, fluoride ions accept a proton from the water molecules, resulting in the formation of hydrofluoric acid (HF) and hydroxide ions (OH⁻). Since hydroxide ions increase the pH level of the solution, the new solution becomes slightly basic compared to pure water. Therefore, the corrected statement is: Sodium fluoride is added to pure water and stirred to dissolve. Compared to pure water, the new solution is slightly basic.

Answer:

Hi

Williamson's ether reactions imply that an alkoxide reacts with a primary haloalkane. Alkoxides consisting of the conjugate base of an alcohol and are formed by a group R attached to an oxygen atom. They are often written as RO–, where R is the organic substituent (Step 1).

Sn2 reactions are characterized by the reversal of stereochemistry at the site of the leaving group. Williamson's synthesis is no exception and the reaction is initiated by the subsequent attack of the nucleophile. This requires that the nucleophile and electrophile be in anti-configuration (Step 2).

As an example (figure 3).

In the attached file are each of the steps of Williamson's synthesis.

Explanation:

The Williamson ether synthesis involves the reaction of an alkyl halide and an alcohol to form an ether. In the specific case of methyl t-butyl ether, methanol and t-butyl bromide are used. The reaction proceeds via the formation of an alkoxide ion which attacks the alkyl halide.

The Williamson ether synthesis is a method used to synthesize ethers. In this reaction, an alkyl halide (in this case, an alkyl bromide) reacts with an alcohol in the presence of a strong base to form an ether. The general mechanism involves the formation of an alkoxide ion, which then attacks the alkyl halide to form the desired ether.

In the specific case of methyl t-butyl ether, the alkyl bromide used is t-butyl bromide and the alcohol used is methanol. The reaction is as follows:

CH3OH + (CH3)3CBr → [(CH3)3CO]– + CH3Br
CH3Br + [(CH3)3CO]– → (CH3)3COCH3 + Br–

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Answer:

MnO- Manganese Oxide

Explanation:

Empirical formula: This is the formula that shows the ratio of elements

present in a  

compound.

How to determine Empirical formula

1. First arrange the symbols of the elements present in the compound

alphabetically to  determine the real empirical formula. Although, there

are exceptions to this rule, E.g H2So4

2. Divide the percentage composition by the mass number.

3. Then divide through by the smallest number.

4. The resulting answer is the ratio attached to the elements present in

a compound.

                                                                              Mn                         O    

% composition                                                      72.1                      27.9    

Divide by mass number                                       54.94                     16  

                                                                               1.31                      1.74    

Divide by the smallest number                         1.31                      1.31                          

                                                                               1                    1.3

The resulting ratio is 1:1

Hence the Empirical formula is MnO, Manganese oxide

Answer:

The absolute error = 3.1m

The relative error = 0.31

Explanation:Please see attachment for explanation