In a typical situation, emissions from which category of activities comprise the largest amount of CO2 release into the atmosphere?

Answer:

0.76 M

30 g/L

Explanation:

Step 1: Given data

Step 2: Calculate the molarity of the base

We will use the following expression.

[tex]Ma \times Va = Mb \times Vb\\Mb = \frac{Ma \times Va}{Vb} = \frac{0.175M \times 375mL}{86mL} = 0.76 M[/tex]

Step 3: Calculate the concentration of the base in g/L

The molar mass of NaOH is 40.00 g/mol.

[tex]\frac{0.76mol}{L} \times \frac{40.00g}{mol} = 30 g/L[/tex]

Answer:

The correct answer is - 0.570 grams

Explanation:

The balanced chemical reaction is given by

Cu(NO3)2(aq)     + 2NaOH(aq)    -------->    Cu(OH)2(s)      + 2NaNO3(aq)

        1.0 mole            2.0 mole                 1.0 mole          2.0 mole

number of mol of Cu(OH)2,

n = Molarity * Volume

= [tex]35.0*0.167 = 5.845[/tex] millimoles

As clear in the equation, 1 mole of Cu(NO3)2 gives 1 mole of Cu(OH)2 , So, 5.845 millimoles of Cu(NO3)2 will produce 5.845 millimoles of Cu(OH)2

Mass of Cu(OH)2 = number of mol * molar mass

= [tex]97.5*5.845*10^-3[/tex]

= 0.570 grams

Thus, the correct answer is - 0.570 grams

Answer: 0.150 m [tex]Na_2SO_4(aq)[/tex] will have highest boiling point.

Explanation:

Formula used for Elevation in boiling point :

[tex]\Delta T_b=i\times k_b\times m[/tex]

where

[tex]\Delta T_b=T_b-T^o_b[/tex]= elevation in boiling point

[tex[k_b[/tex]  = boiling point constant  

m = molality

i = Van't Hoff factor

A) 0.100 m [tex]NiBr_2[/tex]

i = 3 as [tex]NiBr_2\rightarrrow Ni^{2+}+2Br^-[/tex]

concentration will be [tex]3\times 0.100=0.300[/tex]

B) 0.250 m [tex]CH_3OH[/tex]

i = 1 as [tex]CH_3OH[/tex] is a non electrolyte

concentration will be [tex]1\times 0.250=0.250[/tex]

C) 0.100 molal [tex]MgSO_4(aq)[/tex]

i = 2 as [tex]MgSO_4\rightarrrow Mg^{2+}+SO_4^{2-}[/tex]

concentration will be [tex]2\times 0.100=0.200[/tex]

D. 0.150 molal [tex]Na_2SO_4(aq)[/tex]

i = 3 as [tex]Na_2SO_4\rightarrrow 2Na^{+}+SO_4^{2-}[/tex]

concentration will be [tex]3\times 0.150=0.450[/tex]

E. 0.150 molal [tex]NH_4NO_3(aq)[/tex]

i = 2 as [tex]NH_4NO_3\rightarrrow NH_4^{+}+NO_3^{-}[/tex]

concentration will be [tex]2\times 0.150=0.300[/tex]

The solution having the highest concentration of ions will have the highest boiling point and thus 0.150 m [tex]Na_2SO_4(aq)[/tex] will have highest boiling point.

The aqueous solution that would have the highest temperature at boiling point would be:

D). 0.150 molal Na2SO4(aq)

The boiling point is described as the temperature at which the solution starts boiling or the vapor pressure becomes equivalent to the provided external/outer pressure.

To determine the elevation in boiling point, we will use:

Δ[tex]T_{b}[/tex] [tex]= i[/tex] × [tex]k_{b}[/tex] × [tex]m[/tex]

with

[tex]T_{b}[/tex] [tex]= T_{b} - T^{0}_{b}[/tex]

[tex]k_b[/tex] [tex]=[/tex] constant of boiling point

Using this formula,

0.150 molal Na2SO4(aq)

Given,

[tex]i = 3[/tex]

[tex]Na2So4[/tex] will have

[tex]2Na^{+}[/tex] [tex]+[/tex] [tex]SO^{2-}_{4}[/tex]

So,

Concentration [tex]= 3[/tex] × [tex]0.15[/tex][tex]0[/tex]

[tex]= 0.45[/tex][tex]0[/tex]

∵ 0.150 molal [tex]Na2SO4[/tex]Na2SO4(aq) has the maximum concentration.

Thus, option D is the correct answer.

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Answer:

Formula for the salt: MCl₃

Explanation:

MClₓ → M⁺  +  xCl⁻

We apply the colligative property of boiliing point elevation.

We convert the boiling T° to °C

375.93 K - 273K = 102.93°C

ΔT = Kb . m . i

where ΔT means the difference of temperature, Keb, the ebulloscopic constant for water, m the molality of solution (mol of solute/kg of solvent) and i, the Van't Hoff factor (numbers of ions dissolved)

ΔT = 102.93°C - 100°C = 2.93°C

Kb = 0.512 °C/m

We replace data: 2.93°C = 0.512 °C/m . m . i

i = x + 1 (according to the equation)

22.9 g / (56g/m + 35.45x) = moles of salt / 0.1kg = molality

We have calculated the moles of salt in order to determine the molar mass, cause we do not have the data. We replace

2.93°C = 0.512 °C/m . [22.9 g / (56g/m + 35.45x)] / 0.1kg . (x+1)

2.93°C / 0.512 m/°C = [22.9 g / (56g/m + 35.45x)] / 0.1kg . (x+1)

5.72 m = [22.9 g / (56g/m + 35.45x)]/ 0.1 (x+1)

5.72 . 0.1 / [22.9 g / (56g/m + 35.45x)] = x+1

0.572 / (22.9 g / (56g/m + 35.45x) = x+1

0.572 (56 + 35.45x) / 22.9 = x+1

0.572 (56 + 35.45x) = 22.9x + 22.9

32.03 + 20.27x = 22.9x + 22.9

9.13 = 2.62x

x = 3.48 ≅ 3

Hey there!:

2 Al(NO)₃+ 3 H₂SO₄ → 1 Al₂O₁₂S₃+ 6 HNO₃

Reagents : Al(NO₃)₃ and H₂SO₄

Products : Al₂O₁₂S₃ and HNO₃

Coefficients  : 2 , 3 , 1  and 6

Hope this helps!

Answer:

A. Partial pressure of krypton, Kr is 346.97 mmHg

B. Partial pressure of nitrogen, N2 is 364.03 mmHg.

Explanation:

Step 1:

Data obtained from the question. This include the following:

Total pressure (Pt) = 711 mmHg

Mass of Kr = 11.7 g

Mass of N2 = 4.10 g

Partial pressure of Kr =..?

Partial pressure of N2 =...?

Step 2:

Determination of the number of mole of krypton, Kr and nitrogen, N2. This is illustrated below:

Molar mass of Kr = 84g/mol

Mass of Kr = 11.7g

Mole of Kr =?

Mole = mass /Molar mass

Mole of Kr = 11.7/84 = 0.139 mole

Molar mass of N2 = 2x14 = 28g/mol

Mass of N2 = 4.10g

Mole of N2 =?

Mole = mass /Molar mass

Mole of N2 = 4.1/28 = 0.146 mole

Step 3:

Determination of the mole fraction for each gas. This is illustrated below:

Mole of Kr = 0.139 mole

Mole of N2 = 0.146 mole

Total mole = 0.139 + 0.146 = 0.285 mole

Mole fraction of Kr = mol of Kr/total mol

Mole fraction of Kr = 0.139/0.285

Mole fraction of Kr = 0.488

Mole fraction of N2 = mol of N2/total mol

Mole fraction of N2 = 0.146/0.285

Mole fraction of N2 = 0.512

A. Determination of the partial pressure of krypton, Kr.

This is illustrated below:

Total pressure (Pt) = 711 mmHg

Mole fraction of Kr = 0.488

Partial pressure of Kr =..?

Partial pressure = mole fraction x total pressure

Partial pressure of Kr = 0.488 x 711

Partial pressure of Kr = 346.97 mmHg

B. Determination of the partial pressure of nitrogen, N2

This is illustrated below:

Total pressure (Pt) = 711 mmHg

Mole fraction of N2 = 0.512

Partial pressure of N2 =?

Partial pressure = mole fraction x total pressure

Partial pressure of N2 = 0.512 x 711

Partial pressure of N2 = 364.03 mmHg

Answer:

4.16M/s

Explanation:

Based on the reaction:

CH₄ + N₂Cl₄ ⇄ CCl₄ + N₂ + 2H₂

1 mole of methane, CH₄, produce 2 moles of H₂.

That means whereas 1 mole of methane is consumed, 2 moles of H₂ are formed

Having this in mind, if you are consuming methane at a rate of 2.08M/s, the rate of formation of hydrogen must be twice this rate, because there are produced twice moles of H₂.

Thus, rate of formation of H₂ is:

2.08M/s ₓ 2 =

4.16M/s

The rate of formation of H2 is 4.16M/s

Based on the reaction:

CH₄ + N₂Cl₄ ⇄ CCl₄ + N₂ + 2H₂

here

1 mole of methane, CH₄, produce 2 moles of H₂.

In the case when you are consuming methane at a rate of 2.08M/s, the rate of formation of hydrogen must be twice this rate, because there are produced twice moles of H₂.

Thus, rate of formation of H₂ is:

2.08M/s ( 2) = 4.16M/s

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Answer:

The unknown compound is a weak acid.

Explanation:

Given that :

a 25 mL sample of a solution of an unknown compound is titrated with a 0.115 M NaOH solution.

A buffer region was found around a pH of 3.5. We know that a pH of 3.5 is a weak acid. So, it is likely to be an organic acid

Let assume the solution of the unknown sample to be CH₃COOH

Now :

25 mL of CH₃COOH reacted with 0.115 M of NaOH

The equation for the reaction will be :

CH₃COOH    +    NaOH -----> CH₃COONa + H₂O

at x mole of      0.115y M of

CH₃COOH       NaOH is present

If NaOH was added in excess;

CH₃COOH    +    NaOH -----> CH₃COONa , NaOH will be lost then CH₃COOH    and CH₃COONa will be present

Therefore;

At equilibrium : Only CH₃COONa will be present but if it is above equilibrium NaOH will be present   because the pH will increase due to the presence of the strong base

Answer:

Oxygen present in food items makes then rancid due to the presence of oils and fats. If the food is flushed with nitrogen, it prevents it from being oxidised (the nitrogen acts as an antioxidant).

Hope it helps ! :)

Answer:

E 1: cyclohexene

Explanation:

This reaction is an example of the dehydration of cyclic alcohols. The reaction proceeds in the following steps;

1) The first step of the process is the protonation of the cyclohexanol by the acid. This now yields H2O^+ attached to the cyclohexane ring.

2) the water molecule, which a good leaving group now leaves yielding a carbocation. This now leaves a cyclohexane carbocation which is highly reactive.

3) A water molecule now abstracts a proton from the carbon adjacent to the carbocation leading to the formation of cyclohexene and the regeneration of the acid catalyst. This is an E1 mechanism because it proceeds via a carbocation intermediate and not a concerted transition state, hence the answer.

Answer:

Option A. 1191.49 K

Explanation:

Data obtained from the question include:

The equation for the reaction is given below:

4HCl + O2 —> 2Cl2 + 2H2O

Enthalpy (H) = +280 KJ/mol = +280000 J/mol

Entropy (S) = +235 J/Kmol

Temperature (T) =..?

The temperature at which the reaction will be feasible can be obtained as follow:

Change in entropy (ΔS) = change in enthalphy (ΔH)/T

(ΔS) = (ΔH)/T

235 = 280000/T

Cross multiply

235 x T = 280000

Divide both side by 235

T = 280000/235

T = 1191.49 K

Therefore, the temperature at which the reaction will be feasible is 1191.49 K

Answer:

2.28 M

Explanation:

Step 1: Given data

Percent by mass (%m/m): 20.0 %

Density (ρ): 1.114 g/mL

Step 2: Calculate the percent by volume (%m/v)

We will use the following expression.

[tex]\% m/v = \% m/m \times \rho = 20.0 \% \times 1.114g/mL = 22.3 g \% mL[/tex]

Step 3: Calculate the moles of solute in 100 mL of solution

The molar mass of phosphoric acid is 97.99 g/mol. The moles corresponding to 22.3 g are:

[tex]22.3g \times \frac{1mol}{97.99g} = 0.228mol[/tex]

Step 4: Calculate the molarity of the solution

0.228 moles of solute are in 100 mL (0.100 L) of solution. The molarity of the solution is:

[tex]M = \frac{0.228mol}{0.100L} =2.28 M[/tex]

Answer:

c. 0.34 M hydrocyanic acid + 0.27 M sodium cyanide .

Explanation:

A buffer is defined as the aqueous mixture of a weak acid with its conjugate base or vice versa. Based on the systems:

a. 0.12 M calcium hydroxide + 0.29 M calcium bromide. IS NOT A GOOD BUFFER SYSTEM because Ca(OH)₂ is a strong base.

b. 0.25 M perchloric acid + 0.16 M sodium perchlorate. IS NOT A GOOD BUFFER SYSTEM because perchloric acid is a strong acid

c. 0.34 M hydrocyanic acid + 0.27 M sodium cyanide. IS A GOOD BUFFER SYSTEM because HCN is a weak acid, and its conjugate base, CN⁻, is obtained in the dissolution of NaCN as Na⁺ and CN⁻ ions.

Answer:

B

Explanation:

All the elements in a period have valence electrons in the same shell. The number of valence electrons increases from left to right in the period. When the shell is full, a new row is started and the process repeats.

A new period starts when a new neutron cycle starts. Hence, option B is correct.

A period in the periodic table is a row of chemical elements. All elements in a row have the same number of electron shells.

All the elements in a period have valence electrons in the same shell.

The number of valence electrons increases from left to right in the period.

When the shell is full, a new row is started and the process repeats.

Hence, option B is correct.

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Answer:

Explanation:

Following are a few consequences of fossil fuels

1. It causes air pollution.

2. When they are burned, they produce toxic substances which leads to global warming.

3. Waste products are hazardous to public health and environment.

4. They are non - renewable and unsustainable.

5. Drilling fossil fuels is a dangerous process

Hope this helps

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Answer:

HCl(aq) + KOH(aq) ⇒ KCl(aq) + H₂O(l)

Explanation:

Hydrochloric acid is an acid because it releases H⁺ in an aqueous solution.

Potassium hydroxide is a base because it releases OH⁻ in an aqueous solution.

When an acid reacts with a base they form a salt and water. This is a neutralization reaction. The neutralization reaction between hydrochloric acid and potassium hydroxide is:

HCl(aq) + KOH(aq) ⇒ KCl(aq) + H₂O(l)

Answer:

1. It dissolves much more ice faster than sodium chloride

2. Calcium chloride is more effective in melting ice at lower temperatures.

Explanation:

Salts are used to melt ice on roadways and sidewalks because they help to lower the freezing point of water.

Sodium chloride and calcium chloride are both salts used for this purpose but calcium chloride is usually preferred for the following two reasons:

1. It dissolves much more ice faster than sodium chloride: Calcium chloride dissolves much more ice faster than sodium chloride because when it dissociates, it produces three ions instead of the two produced when sodium chloride. Therefore, the heat of hydration of its ions is greater than that of sodium chloride.

2. Calcium chloride is more effective in melting ice at lower temperatures. It lowers the freezing point of water more than sodium chloride. Calcium chloride is able to lower the freezing point of water to about -52°C while sodium chloride only lowers it to about -6°C.

Answer:

351.1g/mol

Explanation:

you can find the answer using The ideal gas equation

n= PV/RT

n=(1.21*1.9/0.082*301)mol

n=0.092 mol

molar mass=Mass/mole

m=32.3g/0.092mol

m=351.1g/mol

Answer:

THE MASS OF 7.68 *10^24 MOLECULES OF PHOSPHORUS TRICHLORIDE IS 1746.25 g.

Explanation:

Molar mass of PCl3 = ( 31 + 35.5 *3) = 137.5 g/mol

At 7.68 * 10^24 molecules, how many number of mole is present?

6.03 * 10^23 molecules = 1 mole

7.68*10^24 molecules = x mole

x mole = 7.68 *10^24 molecules/ 6.03 *10^23

x mole = 1.27 *10 moles

x mole = 12.7 moles

Using mole = mass / molar mass

mass = mole * molar mass

mass = 12.7 moles * 137.5 g/mol

mass = 1746.25 g

Hence, the mass of 7.68 *10^24 molecules is 1746.25 g

Answer:

Partial pressure of hydrogen H₂ = 0.32 atm

Explanation:

Given:

Total pressure = 0.48 atm

Find:

Partial pressure of hydrogen

Computation:

Number of mole of H₂ = 1 / 2 = 0.5 moles

Number of mole of He  = 1 / 4 = 0.25 moles

Total moles = 0.5 + 0.25 = 0.75

Partial pressure of hydrogen H₂ = [moles / total moles] Total pressure

Partial pressure of hydrogen H₂ = [0.50 / 0.75]0.48 atm

Partial pressure of hydrogen H₂ = 0.32 atm

Answer: 0.32 atm

Explanation:

First convert the mass of H2 to moles using the molar mass.

(1.0 gram H2 ⋅ (1.0 mol H2 / 2.016 g H2)) ≈ 0.50 mol H2

Next, convert the mass of helium He to moles using the atomic mass.

(1.0 gram He ⋅ (1.0 mol He / 4.003 g He)) ≈ 0.25mol He

The total number of moles is about 0.75 moles . The partial pressure of a component of a gas mixture can be found by multiplying the mole fraction by the total pressure.

PH2 = XH2 × Ptotal

PH2 = (0.50 mol / 0.75 mol)(0.48 atm) = 0.32 atm

Answer:

THE AMOUNT OF HEAT REQUIRED TO CONVERT ICE FROM -20 C TO STEAM AT 120 C IS 30 946 J OR 30.946 KJ OF HEAT.

Explanation:

Mass = 10 g

To convert 10 g of ice at -20°C to steam at 120°C, the heat involved is:

1. Heat involved in converting the ice from -20 °c to ice at 0 °C:

Heat = mass * specific heat of water solid * change in temperature

heat = 10g * 2.09 J/g°C * ( 0- (-20))

Heat = 10 * 2.09 * 20

heat = 418 J

2. Heat required to convert the ice from 0°C to water at 0°C:

Heat = mass * specific heat of fusion of water solid

Heat = 10 * 333

Heat = 3330 J

3. Heat required to convert water at 0 C to water at 100 C:

Heat = mass * specific heat of water * change in teperature

Heat = 10 * 4.18 * (100 -0)

Heat = 4180 J

4. Heat required to convert water at 100 C to steam at 100 C:

Heat = mass * specific heat of vaporization

Heat = 10 * 2260

Heat = 22600 J

5. Heat required to convert steam from 100 C to steam at 120 C:

Heat = mass * specific heat of water * change in temperature

Heat = 10 * 2.09 * (120 -100)

Heat = 10 * 2.09 * 20

Heat = 418 J

T

he heat required to convert 10 g of ice at -20 C to steam at 120 C is therefore the total of the individual heat of reactions

Total amount of heat = ( 418 J + 3330 J + 4180 J + 22600 J + 418 J)

Total heat =  30946 J

Answer:

THE HEAT CAPACITY OF THE CALORIMETER IS 3666.67 J/C

Explanation:

Mass = 2 g

Temperature difference = 32 C - 29 C = 3 C

Heat of combustion = 11 kJ/g

Heat capacity of the calorimeter = unknown

It is important to note that the heat of combustion of the reaction is the heat absorbed by the calorimeter in raising the mixture by 3 C

So therefore,

Heat = heat capacity * temperature difference

Heat capacity = Heat / temperature difference

Heat capacoty = 11 000 J / 3 C

Heat capacity = 3666.67 J/ C

Answer:

#1: 0.00144 mmolHCl/mg Sample

#2: 0.00155 mmolHCl/mg Sample

#3: 0.00153 mmolHCl/mg Sample

Explanation:

A antiacid (weak base) will react with the HCl thus:

Antiacid + HCl → Water + Salt.

In the titration of antiacid, the strong acid (HCl)  is added in excess, and you're titrating with NaOH moles of HCl that doesn't react.

Moles that react are the difference between mmoles of HCl - mmoles NaOH added (mmoles are Molarity×mL added). Thus:

Trial 1: 0.391M×14.00mL - 0.0962M×34.26mL = 2.178 mmoles HCl

Trial 2: 0.391M×14.00mL - 0.0962M×33.48mL = 2.253 mmoles HCl

Trial 3: 0.391M×14.00mL - 0.0962M×33.84mL = 2.219 mmoles HCl

The mass of tablet in mg in the 3 experiments is 1515mg, 1452mg and 1443mg.

Thus, mmoles HCl /mg OF SAMPLE for each trial is:

#1: 2.178mmol / 1515mg

#2: 2.253mmol / 1452mg

#3: 2.219mmol / 1443mg

Answer:

The enthalpy change is −636.9798 kJ

Explanation:

The equation of the reaction between Carbon and Oxygen can be represented as :

[tex]C_{(s)} + O_{2(g)} \to CO_{2_(g)} \ \ \ \ \ \ \Delta H = -394 \ \ kJ/mol[/tex]

molar mass of Carbon = 12 mole

when mass of  19.399999999999999 g of carbon reacted with oxygen;

the number of moles = mass/molar mass

the number of moles = 19.399999999999999/12

the number of moles = 1.6167 mol

Thus the enthalpy change when  1 mole of carbon react with oxygen = -394 kJ/mol

Therefore when 1.6167 moles react with oxygen; we have:

= 1.6167 mol ×  -394 kJ/mol

= −636.9798 kJ

Answer:

Thorium-235

Explanation:

Half-life is defined as the time taken for a radioactive material to reduce to half of its original amount. If the original amount of a radioactive substance N is 100%, then;

1st half-life- 50% of N is left

2nd half life - 25% of N is left

3rd half-life- 12.5% of N is left

The half-life of Thorium-235 is 15 billion years, hence three half lives will take place in 45 billion years. Hence 12.5% of the original amount of Thorium-235 present will remain after about 42 billion years.

Answer:

See explanation.

Explanation:

Hello,

In this case, for the described chemical reaction:

2 HCl(aq) + Mg(OH)2(aq) → MgCl2(aq) + 2 H2O(l)

We can notice there is a 2:1 molar ratio between the moles of hydrochloric acid and magnesium hydroxide, therefore, at the equivalence point:

[tex]n_{HCl}=2*n_{Mg(OH)_2}[/tex]

And in terms of volumes and concentrations we verify:

[tex]V_{HCl}M_{HCl}=2*V_{Mg(OH)_2}M_{Mg(OH)_2}[/tex]

So we use the given data to proof it:

[tex]4.3mL*0.417M=2*11.9mL*0.151M\\1.793=3.594[/tex]

Therefore, we can conclude the data is wrong by means of the 2:1 mole ratio that for sure was not taken into account. This is also supported by the fact that normalities are actually the same, but the nomality of magnesium hydroxide is the half of the hydrochloric acid normality since the acid is monoprotic and the base has two hydroxyl ions.

Best regards.

Answer:

The substance with the highest heat gives heat to the lowest temperature, equating both temperatures,

In this situation there is talk of giving up heat but not matter, it is here that the law of conservation of energy comes into play.

Explanation:

The law of conservation of energy talks about that energy is transformed and never lost between two substances or two bodies that interact with each other, these interactions can be heat exchanges, as in this example.

Answer:

the same and 14

Explanation:

ed tell chem

Answer: the same 10

Explanation:

Answer:

23.36mL of the stock solution are required.

Explanation:

A dilution consist in the addition of solvent to decreases the concentration of a stock solution (The solution more concentrated).

As you want to prepare 250.0mL = 0.2500L of a 0.1342M NaOH, moles of NaOH you require to make this concentration in this volume are:

0.2500L × (0.1342mol / L) = 0.03355 moles of NaOH you require in the diluted solution.

These moles comes from the 1.436M stock solution. The volume of the stock solution you need to add is:

0.03355moles NaOH × (1L / 1.436mol) = 0.02336L of the 1.436M solution =