Which of the following compounds would you expect to be an electrolyte?

N2
CH4
H2O
O2
КСІ

Answers

Answer 1

Answer:

N2 but i really didn't know

Answer 2

The compound that would be expected to be an electrolyte is : ( A ) N₂

What is an electrolyte

An electrolyte is any subsatnce which conducts electircity when dissolved in a solvent such as water. From the question the compound that can conduct electricty when dissolved in water is N₂

Hence we can conclude that The compound that would be expected to be an electrolyte is : ( A ) N₂

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Related Questions

2) If a brick has a length of 13.77 cm, a width of 8.50 cm, and a height of 5.12 cm:
a) What is the volume of the brick?
b) If the brick has a mass of 895.3 g, what is its density?

Answers

Answer:

a. 599 cm³

b. 1.49 g/cm³

Explanation:

A. Volume

Volume is the amount of space an object occupies. Since this is a brick, the object is a rectangular prism. The formula for the volume of a rectangular prism is the product of length, width, and height.

[tex]V= l *w*h[/tex]

The brick's length (l) is 13.77 centimeters, the width (w) is 8.50 centimeters, and the height (h) is 5.12 centimeters. Substitute these values into the formula.

[tex]V= 13.77 \ cm * 8.50 \ cm * 5.12 \ cm[/tex]

Multiply the numbers together.

[tex]V= 117.045 \ cm^ 2* 5.12 \ cm[/tex]

[tex]V= 599.2704 \ cm^3[/tex]

The original measurements have at least 3 significant figures, so our answer must have 3. For the number we calculated, that is the ones place. The 2 in the tenths place tells us to leave the 9 in the ones place.

[tex]V \approx 599 \ cm^3[/tex]

[tex]\bold {The \ volume \ of \ the \ brick \ is \ approximately \ 599 \ cubic \ centimeters}}[/tex]

2. Density

Density is the amount of matter in a specified space. The formula for density is mass over volume.

[tex]d= \frac{m}{v}[/tex]

The mass of the brick is 895.3 grams and we just found the volume to be 599.2704 cubic centimeters. Substitute the values into the formula.

[tex]d= \frac{895.3 \ g}{599 \ cm^3}[/tex]

Divide.

[tex]d= 1.494657763 \ g/cm^3[/tex]

Round to three significant figures. For the number we calculated, that is the hundredth place. The 4 in the thousandth place tells us to leave the 9 in the hundredth place.

[tex]d \approx 1.49 \ g/cm^3[/tex]

[tex]\bold {The \ density\ of \ the \ brick \ is \ approximately \ 1.49 \ grams /cubic \ centimeters}}[/tex]

Leaming Task 1:
Distinguish the process as spontaneous or non-spontaneous process. Write S it spontaneous and NSi non-spontaneous
on the bionk.
1. Melling ofice
2 Ruisting of ton
3. Marble going down the spiral.
4. Going up the
& Keeping the food fresh from spolage​

Answers

Solution :

Spontaneous Process

A spontaneous process is defined as the process that occurs without the help of any external aid or inputs. A spontaneous process is a natural process which occurs naturally in the environment.

Non Spontaneous process

A non spontaneous process is a process which does not occur naturally. Some inputs are provided for the process to occur. Energy from external source is applied into the process to start the process.

The following processes are :

1. Melling of ice   ---- Spontaneous

2 Rusting of iron  --- Spontaneous

3. Marble going down the spiral.   --- spontaneous

4. Going up the hill  ---- Non spontaneous

5. Keeping the food fresh from spoilage​  --- Non spontaneous

An HCl solution has a concentration of 0.09714 M. Then 10.00 mL of this solution was then diluted to 250.00 mL in a volumetric flask. The diluted solution was then used to titrate 250.0 mL of a saturated AgOH solution using methyl orange indicator to reach the endpoint.

Required:
a. What is the concentration of the diluted HCl solution?
b. If 7.93 mL of the diluted HCl solution was required to reach the endpoint, what is the concentration of OH- in solution?

Answers

Answer:

A. Concentration of diluted acid = 0.00389 M

B. Concentration of OH- in AgOH solution = 0.00012 M

Explanation:

A. Using the dilution formula: C1V1 = C2V2

Where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.

From the data provided, C1 = 0.09714 M, V1 = 10.0 mL, V2 = 250.0 mL and C2 = ?

Making C2 subject of the formula above; C2 = C1V1/V2

C2 = 0.09714 M × 10 / 250 = 0.00389 M

B. Equation of the neutralization reaction is given below:

HCl + AgOH ---> AgCl + H₂O

From the equation, 1 mole of acid neutralizes 1 mole of base

Using the titration formula; CaVa/CbVb = na/nb

Where Ca is the concentration of the acid HCl = 0.00389 M

Va is the volume of acid = 7.93 mL

Cb is the concentration of base, AgOH = ?

Vb is volume of base = 250.0 mL

na/nb = mole ratio of acid and base = 1

Making Cb subject of the formula in the equation above; Cb = CaVa/Vb

Cb = 0.00389 M × 7.93 / 250

Cb = 0.00012 M

If a 520 mg sample of technetium-99 is used for diagnostic procedure, how much of Tc-99 remains after 30.0h? Half life of Tc-99 is 6.0 hours.

Answers

16.25 mg
I have attached an image of the half life formula.
In this problem, N0 would be 520, since that is the initial amount.
t = 30.0 Hours
t 1/2 = 6.0 hours
Now plug everything into the equation. You get 16.25 mg. No

The reversible reaction: 2SO2(g) O2(g) darrw-tn.gif 2SO3(g) has come to equilibrium in a vessel of specific volume at a given temperature. Before the reaction began, the concentrations of the reactants were 0.060 mol/L of SO2 and 0.050 mol/L of O2. After equilibrium is reached, the concentration of SO3 is 0.040 mol/L. What is the equilibrium concentration of O2

Answers

Answer:

0.030 M

Explanation:

Step 1: Make an ICE chart

        2 SO₂(g) + O₂(g) ⇄ 2 SO₃(g)

I        0.060      0.050          0

C         -2x            -x            +2x

E     0.060-2x  0.050-x       2x

Step 2: Find the value of x

The concentration of SO₃ at equilibrium is 0.040 mol/L. Then,

2x = 0.040

x = 0.020

Step 3: Calculate the concentration at equilibrium of O₂

[O₂] = 0.050 - x = 0.050 - 0.020 = 0.030 M

The equilibrium concentration of oxygen is 0.030 M.

A reversible reaction is a reaction that can move either in the forward or in the reverse direction. We have the reaction; 2SO2(g) + O2(g) ⇄ 2SO3(g). We can now set up the ICE table as shown below;

  2 SO₂(g) + O₂(g) ⇄ 2 SO₃(g)

I        0.060      0.050          0

C         -2x            -x            +2x

E     0.060-2x  0.050-x       2x

At equilibrium;

2x = 0.040 M

x = 0.040 M/2 =  0.020 M

For oxygen;

0.050-x  

0.050 M - 0.020 M = 0.030 M

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Carbonic anhydrase is strongly inhibited by the drug acetazolamide, which is used as a diuretic (i.e., to increase the production of urine) and to lower excessively high pressure in the eye (due to accumulation of intraocular fluid) in glaucoma.

a. True
b. False

Answers

Answer:

a. True

Explanation:

There is strong inhibition of Carbon Anhydrase by Aceta-zolamide Carbonic Anhydrase. The drug acetazolamide is used as diuretic which increase the urine production in human body. It lowers pressure in eye in glaucoma.

What is the molarity of a solution containing 150 g of zinc sulfate (ZnSO4) per liter?

Answers

Answer:

0.93 M

Step-by-step Explanation:

First, we have to calculate the molar mass (MM) of ZnSO₄ by using the molar mass of each chemical element:

MM(ZnSO₄) = 65.4 g/mol Zn + 32 g/mol S + (16 g/mol x 4) = 161.4 g/mol

Then, we divide the mass of ZnSO₄ into its molar mass to obtain the number of moles:

moles ZnSO₄ = mass/MM = 150 g/(161.4 g/mol)= 0.93 mol

Since the molarity of a solution expresses the number of moles of solute per liter of solution, we calculate the molarity (M) as follows:

M = moles ZnSO₄/1 L =  0.93 mol/1 L = 0.93 M

What kind of element is Phosphorus is

Answers

phosphorus is and non metal

Answer:

NON-METAL

Explanation:

Phosphorus is a non-metal that sits just below nitrogen in group 15 of the periodic table. This element exists in several forms, of which white and red are the best known.

A hydrocarbon contains only the elements____?

Answers

Explanation:

elements are carbons and hydrogen

Answer:

Carbon and Hydrogen.

Explanation:

It’s in the name Hydro (H) Carbon (C)

from kinatic point of view explain the change from solid to liquied based on the effect of change of tempreture.​

Answers

Answer:

Temperature affects the kinetic energy in a gas the most, followed by a comparable liquid, and then a comparable solid. The higher the temperature, the higher the average kinetic energy, but the magnitude of this difference depends on the amount of motion intrinsically present within these phases.

Explanation:

Liquids have more kinetic energy than solids. When a substance increases in temperature, heat is being added, and its particles are gaining kinetic energy. Because of their close proximity to one another, liquid and solid particles experience intermolecular forces. These forces keep particles close together.

What is the molarity of a solution prepared by dissolving 12.0 g of potassium permanganate, KMnO4, in water to make 250.0 mL of solution

Answers

Answer:

Explanation:

Molecular weight of potassium permanganate is 158 g .

12 g = 12 g / 158 g moles

= .076 moles .

250 mL = .25 L .

Molarity of solution = moles dissolved / volume of solution

= .076 / .25 L

= .304 M .

Molarity of solution is .304 M.  

The molarity of a solution prepared by dissolving 12.0 g of potassium permanganate, KMnO4, in water to make 250.0 mL of solution will be 0.304 M. It can be calculated by using mole formula.

What is molarity ?

The amount of material in a given volume of the given solution has been measured in molarity (M). The moles of a particular solute per liter of a solution can be known as molarity. The molar solution of known concentration is indeed known as molarity.

What is mole?

A mole is just the quantity of a material that includes exactly 6.022 × [tex]10^{23}[/tex] of the particular chemical elementary entities.

Molecular weight potassium potassium permanganate = 12 g

Volume of the solution = 250mL = 25 L.

Calculation of mole:

Mole can be calculated by using the formula.

Mole = mass / molar mass

Mole = 12 g / 158 g

Mole = 0.076 mole.

Calculation of molarity:

Molarity can be calculated by using the molarity formula:

Molarity = moles of dissolved / volume of solution

Molarity = 0.076 / 0.25

Molarity = 0.304 M

Therefore, the molarity of the solution will be 0.304 M.

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A mixture of hydrogen and argon gases, at a total pressure of 980 mm Hg, contains 0.291 grams of hydrogen and 5.62 grams of
argon. What is the partial pressure of each gas in the mixture?
PH2
Par
mm Hg
mm Hg

Answers

Answer:

Partial pressure of H₂ = 499 mmHg

Partial pressure of Ar = 481 mmHg

Explanation:

We'll begin by calculating the number of mole of each gas. This can be obtained as follow:

For Hydrogen:

Molar mass of H₂ = 2 × 1 = 2 g/mol

Mass of H₂ = 0.291 g

Mole of H₂ =?

Mole = mass /molar mass

Mole of H₂ = 0.291/ 2

Mole of H₂ = 0.1455 mole

For Argon:

Molar mass of Ar = 40 g/mol

Mass of Ar = 5.62 g

Mole of Ar =?

Mole = mass /molar mass

Mole of Ar = 5.62 / 40

Mole of Ar = 0.1405 mole

Next, we shall determine the mole fraction of each gas. This can be obtained as follow:

Mole of H₂ = 0.1455 mole

Mole of Ar = 0.1405 mole

Total mole = 0.1455 + 0.1405

Total mole = 0.286 mole

Mole fraction of H₂ (nₕ₂) = mole of H₂ / total mole

Mole fraction of H₂ (nₕ₂) = 0.1455/0.286

Mole fraction of H₂ (nₕ₂) = 0.509

Mole fraction of Ar (nₐᵣ) = mole of Ar / total mole

Mole fraction of Ar (nₐᵣ) = 0.1405/0.286

Mole fraction of Ar (nₐᵣ) = 0.491

Finally, we shall determine the partial pressure of each gas. This can be obtained as follow:

For Hydrogen:

Mole fraction of H₂ (nₕ₂) = 0.509

Total pressure (Pₜ) = 980 mmHg

Partial pressure of H₂ (Pₕ₂) =?

Pₕ₂ = nₕ₂ × Pₜ

Pₕ₂ = 0.509 × 980

Partial pressure of H₂ (Pₕ₂) = 499 mmHg

For Argon:

Partial pressure of H₂ (Pₕ₂) = 499 mmHg

Total pressure (Pₜ) = 980 mmHg

Partial pressure of Ar (Pₐᵣ) =?

Pₜ = Pₕ₂ + Pₐᵣ

980 = 499 + Pₐᵣ

Collect like terms

980 – 499 = Pₐᵣ

481 = Pₐᵣ

Partial pressure of Ar (Pₐᵣ) = 481 mmHg

SUMMARY:

Partial pressure of H₂ (Pₕ₂) = 499 mmHg

Partial pressure of Ar (Pₐᵣ) = 481 mmHg

In aqueous solution the Ni2" ion forms a complex with four ammonia molecules. Write the formation constant expression for the equilibrium between the hydrated metal ion and the aqueous complex. Under that, write the balanced chemical equation for the first step in the formation of the complex K,=________.

Answers

Answer:

The correct equation is "[tex]\frac{[Ni(H_2O)_3 (NH_3)]^{2+}}{[Ni(H_2O)_4]^{2+} [NH_3]}[/tex]".

Explanation:

According to the question,

Throughout an aqueous solution, [tex]Ni^{2+}[/tex] exist as [tex][Ni(H_2O)_4]^{2+}[/tex]

So,

⇒ [tex][Ni(H_2O)_4]^{2+} + 4NH_3 \rightleftharpoons [Ni(NH_3)_4]^{2+} + H_2O[/tex]

⇒ [tex]K_f = \frac{[Ni(NH_3)_4]^{2+}}{[Ni(H_2O)_4^{2+}] [NH_3]^4}[/tex]

Here, we have excluded [tex][H_2O][/tex] as concentration of water will be const.

Now,

This formation of [tex][Ni(NH_3)_4]^{2+}[/tex] proceeds via several steps,

Step 1:

⇒ [tex][Ni(H_2O)_4]^{2+}+NH_3 \rightleftharpoons [Ni(H_2O)_3 (NH_3)]^{2+} + H_2O[/tex]

⇒ [tex]K_1 = \frac{[Ni(H_2O)_3 (NH_3)]^{2+}}{[Ni(H_2O)_4]^{2+} [NH_3]}[/tex]

Write a balanced half-reaction for the reduction of liquid water to gaseous hydrogen in basic aqueous solution. Be sure to add physical state symbols where appropriate.

Answers

Answer:

2 H₂O(l) + 2 e⁻ ⇒ H₂(g) + 2 OH⁻(aq)

Explanation:

Let's consider the unbalanced half-reaction for the reduction of liquid water to gaseous hydrogen in basic aqueous solution.

H₂O(l) ⇒ H₂(g)

First, we will perform the mass balance. We will balance oxygen atoms by multiplying H₂O by 2 and adding 2 OH⁻ to the right side.

2 H₂O(l) ⇒ H₂(g) + 2 OH⁻(aq)

Then, we perform the charge balance by adding 2 electrons to the left side.

2 H₂O(l) + 2 e⁻ ⇒ H₂(g) + 2 OH⁻(aq)

You prepare a solution by dissolving 25.3 g sucrose (C12H22O11) 705 mL of water. Calculate the molarity of the solution.

Answers

Answer:

0.105 M

Explanation:

First we convert 25.3 grams of sucrose into moles, using sucrose's molar mass:

Molar Mass of C₁₂H₂₂O₁₁ = 342.3 g/mol25.3 g C₁₂H₂₂O₁₁ ÷ 342.3 g/mol = 0.0739 mol C₁₂H₂₂O₁₁

Now we calculate the molarity of the solution, using the given volume and the calculated number of moles:

Converting 705 mL ⇒ 705 mL / 1000 = 0.705 LMolarity = 0.0739 mol / 0.705 L = 0.105 M


0.28 M Ca(NO3)2
Express your answer using two significant figures.

Answers

Answer:

Mass=Moles × RFM

Mass= 0.28M× 164

Mass= 45.92 grammes


The absolute temperature of a gas is increased four times while maintaining a constant volume. What happens to the
pressure of the gas?
Olt decreases by a factor of four.
O It increases by a factor of four.
It decreases by a factor of eight
It increases by a factor of eight.

Answers

Answer:

The pressure increases by a factor of four.

Explanation:

Let's consider a gas at a given temperature and pressure (T₁, P₁). The absolute temperature of a gas is increased four times (T₂ = 4 T₁) while maintaining a constant volume. We can assess the effect on the pressure (P₂) by using Gay Lussac's law.

P₁/T₁ = P₂/T₂

P₂ = P₁ × T₂/T₁

P₂ = P₁ × 4 T₁/T₁

P₂ = 4 P₁

The pressure increases by a factor of four.

Calculate the boiling point of a 3.5 % solution (by weight) of sodium chloride in water.
Kb of H2O = 0.512 oC/M

Answers

Answer: The boiling point of the solution is [tex]101.02^oC[/tex]

Explanation:

We are given:

3.5 % (by weight) NaCl

This means that 3.5 g of NaCl is present in 100 g of solution

Mass of solvent = Mass of solution - Mass of solute

Mass of solvent (water) = (100 - 3.5) g = 96.5 g

Elevation in the boiling point is defined as the difference between the boiling point of the solution and the boiling point of the pure solvent.

The expression for the calculation of elevation in boiling point is:

[tex]\text{Boiling point of solution}-\text{boiling point of pure solvent}=i\times K_b\times m[/tex]

OR

[tex]\text{Boiling point of solution}-\text{Boiling point of pure solvent}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}[/tex] ......(1)

where,

Boiling point of pure solvent (water) = [tex]100^oC[/tex]

Boiling point of solution = ?

i = Vant Hoff factor = 2 (for NaCl)

[tex]K_b[/tex] = Boiling point elevation constant = [tex]0.512^oC/m[/tex]

[tex]m_{solute}[/tex] = Given mass of solute (NaCl) = 3.5 g

[tex]M_{solute}[/tex] = Molar mass of solute (NaCl) = 36.5 g/mol

[tex]w_{solvent}[/tex] = Mass of solvent (water) = 96.5 g

Putting values in equation 1, we get:

[tex]\text{Boiling point of solution}-(100)=2\times 0.512\times \frac{3.5\times 1000}{36.5\times 96.5}\\\\\text{Boiling point of solution}=(1.02+100)^oC\\\\\text{Boiling point of solution}=101.02^oC[/tex]

Hence, the boiling point of the solution is [tex]101.02^oC[/tex]

give one use of zinc​

Answers

Most Zinc are used to galvanise other metals such as iron which helps to prevent rusting
If you need any more let me know

How many atoms are in protons

Answers

Answer:

the number of protons in a atom is unique to each element

Explanation:

protons are about 99.86% as massive neutrons. The number of protons in a atom is unique to each element .For example carbon atoms have six protons in an atom is referred to as the atomic number of that element

To what volume should you dilute 55 mL of 12 M stock HNO3 solution to obtain a 0.145 HNO3 solution?

Answers

Answer:

4552 mL

Explanation:

From the question given above, the following data were obtained:

Volume of stock solution (V₁) = 55 mL

Molarity of stock solution (M₁) = 12 M

Molarity of diluted solution (M₂) = 0.145 M

Volume of diluted solution (V₂) =?

The volume of the diluted solution can be obtained by using the dilution formula as illustrated below:

M₁V₁ = M₂V₂

12 × 55 = 0.145 × V₂

660 = 0.145 × V₂

Divide both side by 0.145

V₂ = 660 / 0.145

V₂ ≈ 4552 mL

Thus, the volume of the diluted solution is 4552 mL

At 35 C, a sample of gas has a volume of 256 ml and a pressure of 720.torr. What would the volume
be if the temperature were changed to 22 C and the pressure to 1.25 atmospheres?

Answers

Answer: Volume would be 196.15 mL if the temperature were changed to [tex]22^{o}C[/tex] and the pressure to 1.25 atmospheres.

Explanation:

Given: [tex]T_{1} = 35^{o}C = (35 + 273) K = 308 K[/tex],     [tex]V_{1}[/tex] = 256 mL,    

[tex]P_{1}[/tex] = 720 torr (1 torr = 0.00131579 atm) = 0.947368 atm

[tex]T_{1} = 22^{o}C = (22 + 273) K = 295 K[/tex],       [tex]P_{2} = 1.25 atm[/tex]  

Formula used to calculate volume is as follows.

[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}[/tex]

Substitute the values into above formula as follows.

[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{1 atm \times 256 mL}{308 K} = \frac{1.25 atm \times V_{2}}{295 K}\\V_{2} = 196.15 mL[/tex]

Thus, we can conclude that the volume would be 196.15 mL if the temperature were changed to [tex]22^{o}C[/tex] and the pressure to 1.25 atmospheres.

Sobre ações relacionadas ao aquecimento global, assinale somente as alternativas corretas:
a) ( x) As ações humanas não influenciam no aumento da temperatura do planeta.
b) ( ) As mudanças climáticas são intensificadas pela emissão de gases das atividades humanas.
c) ( ) A queima de combustíveis fósseis e de florestas são as principais ações humanas que liberam gases que intensificam o efeito estufa.
d) ( ) O efeito estufa é um fenômeno natural.
e) ( ) Se as águas dos oceanos ficarem mais quentes, os furacões não terão tanta força.

Answers

Answer:

123456788012346778901234567890

When heated, carboxylic salts in which there is a good leaving group on the carbon beta to the carbonyl group undergo decarboxylation/elimination to give an alkene. Draw the structures of the products expected when this compound is heated.

Answers

Answer:

i dont know mate

Explanation:

A 2.584 g sample of a compound containing only carbon, hydrogen, and oxygen is burned in an excess of dioxygen, producing 5.874 g CO2 and 2.404 g H2O. What mass of oxygen is contained in the original sample?a. 0.7119 g.b. 0.8463 g.c. 0.29168 g.d. 0.1793 g.e. 0.6230 g.

Answers

Answer:

a. 0.7119 g

Explanation:

To solve this question we need to know that all carbon of the compound will react producing CO2 and all Hydrogen producing H2O.

Thus, we can find the mass of C and the mass of H and by difference regard to the 2.584g of the compound we can find the mass of oxygen as follows:

Moles CO2 = Moles C -Molar mass: 44.01g/mol-

5.874g CO2 * (1mol/44.01g) = 0.1335 moles CO2 = 0.1335 moles C

Mass C -Molar mass: 12.01g/mol-:

0.1335 moles C * (12.01g /mol) = 1.6030g C

Moles H2O -Molar mass: 18.01g/mol-

2.404gH2O * (1mol / 18.01g) = 0.1335 moles H2O * (2mol H / 1mol H2O) = 0.267 moles H

Mass H -Molar mass: 1g/mol-

0.267 moles H * (1g/mol) = 0.2670g H

Mass Oxygen =

Mass O = 2.584g compound - 1.6030g C - 0.2670g H

Mass O = 0.714g O ≈

a. 0.7119 g

Ethylene glycol (C2H6O2) is mixed with water to make auto engine coolants. How many grams of C2H6O2 are in 5.00 L of a 6.00 M aqueous solution

Answers

Answer:

1860g.

Explanation:

It is known that the molar mass of C2H6O2 is 62.08 g/mol.,

Now to  solve for the number of moles of solute, one must multiply both

sides by the volume:

moles of solute = (6.00 M)(5.00 L) = 30.0 mol

Notice since the definition of molarity is mol/L, the

product M × L gives mol, a unit of amount.

Use the molar  mass of C3H8O3, one can convert mol to g:

Mass m =30 mol × 62.08 g/mol

m = 1860g.

Hence, there are 1,860 g of C2H6O2 in the specified amount of

engine coolant.

A mixture of hydrocarbons contains 38.3% hexane, C6H14, 13.9% octane, C8H18, and 47.8% decane, C10H22. The mixture is combusted in an excess of oxygen. What mass of CO2 is generated from the combustion of 19.3 kg of the mixture

Answers

Answer:

52.206 kg

Explanation:

From the given information:

Mass of hexane C6H14 = [tex]19.3*10^3 \ g \times \dfrac{38.3}{100}[/tex]

= 7391.9 g

Mass of octane C8H18 = [tex]19.3*10^3 \ g \times \dfrac{13.9}{100}[/tex]

= 2682.7 g

Mass of decane C10H22 = [tex]19.3*10^3 \ g \times \dfrac{47.8}{100}[/tex]

= 9225.4 g

However, recall that:

number of moles of an atom = mass/molar mass

For hexane, no of moles = 7391.9 g/86.18 g/mol

= 85.77 moles

For octane, no of moles = 2682.7 g/114.23 g/mol

= 23.49 moles

For decane, no of moles = 9225.4 g/142.29 g/mol

= 64.84 moles

Therefore:

number of moles of CO2 produced = (6 × 85.77)+(23.49)+(10×64.84) moles

= 1186.51 moles

Finally, the mass of CO2 produced is:

= 1186.51 mol × 44 g/mol

= 52206.44  g

= 52.206 kg

Which technique is best suited to each application?

a. In the second week of a four week biochemistry experiement, you have 50 fractions collected from a gel filtration column to determine which fractions contain lactate dehydrogenase. You are given only 400 uL of 0.100 mg/mL lactate dehydrogenase to prepare your calibration curve. 96-well microplate
b. Your environmental lab has 2000 samples to be analyzed for trace ammonia by next week. discrete analyzer.
c. Twenty water samples must be analyzed for Cl-, NH3, PO3-, and So during each work shift. flow injection analysis colorimeter.
d. Your professor heard you will be hiking the Appalachian Trail next summer. She asks you to collect 100-mL water samples from the ten streams with the highest concentration of phosphate.

Answers

Answer:

a. discrete analyzer

b. 96 well microplate

c. flow injection analysis

d. colorimeter

Explanation:

96 well microplates are instruments designed for sample collection and throughput screening. If an environment lab has collected 2000 samples then 96 well microplate is best suited application. Discrete analyzer is automated chemical analyzer which performs test on samples kept in discrete cells. Flow injection analysis is approach used for chemical analysis. It injects a plug of sample into a flowing carrier stream. Colorimeter is a device which measures absorbance of wavelength of light by a specific solution.

what would happen if the number of protons and electrons in the Atom did change

Answers

When you change the number of protons in an atom, you will change the atom from one element to a different element. Sometimes, when you add a proton to an element, the element will become radioactive. If you change the number of electrons in an atom, you will get an ion of the element.

You need to produce a buffer solution that has a pH of 5.50. You already have a solution that contains 10 mmol (millimoles) of acetic acid. How many millimoles of acetate (the conjugate base of acetic acid) will you need to add to this solution?

Answers

Answer:

56.9 mmoles of acetate are required in this buffer

Explanation:

To solve this, we can think in the Henderson Hasselbach equation:

pH = pKa + log ([CH₃COO⁻] / [CH₃COOH])

To make the buffer we know:

CH₃COOH  +  H₂O  ⇄   CH₃COO⁻  +  H₃O⁺     Ka

We know that Ka from acetic acid is: 1.8×10⁻⁵

pKa = - log Ka

pKa = 4.74

We replace data:

5.5 = 4.74 + log ([acetate] / 10 mmol)

5.5 - 4.74 = log ([acetate] / 10 mmol)

0.755 = log ([acetate] / 10 mmol)

10⁰'⁷⁵⁵ = ([acetate] / 10 mmol)

5.69 = ([acetate] / 10 mmol)

5.69 . 10 = [acetate] → 56.9 mmoles

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