What is the molarity of a NaOH solution if 32.47 mL is required to titrate 0.6013 g of potassium hydrogen phthalate (C8H5O4K)

Answer:

Water and Salts

Explanation:

Hope it helps u....

Answer:

6.34917360^25g

Explanation:

It's been a while since I've done this type of problem so I'm not making any promises that its right hahaha, but I hope it helps anyway. Please let me know whether I'm right or not!

Answer:

Explanation:

Warm lake vapor pressure is higher than cold lake vapor pressure because it evaporates more quickly. In other words, as the temperature of warm water increases, the kinetic energy of the warm water molecule also increases. As the kinetic energy of the warm water molecule increases, the number of molecules conversion into a vapor also increases, thereby increasing the vapor pressure of warm water.

Answer:

B.

Warm water evaporates more quickly.

Explanation:

i got it right on edmentum!!

When ethanoic acid react with methanol in the presence of H⁺ or heat it gives methyl ethanoate and water.

The balanced chemical equation is the equation in which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation.

Now write the chemical equation

CH₃COOH + CH₃OH ⇄ CH₃COOCH₃ + H₂O

Here

CH₃COOH is Ethanoic acid (Acetic acid)

CH₃OH is Methanol (Methyl Alcohol)

CH₃COOCH₃ is Methyl ethanoate (Methyl Acetate)

H₂O is water

Thus from the above conclusion we can say that When ethanoic acid react with methanol in the presence of H⁺ or heat it gives methyl ethanoate and water.

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Because carbon dioxide dissolves in water and is transported by rain to the surface where it combines with rocks to generate carbonates, the Earth's atmosphere contains significantly less carbon dioxide than Venus's atmosphere.

Since Venus' atmosphere is around 100 times thicker than the earth's and extremely dense. Venus must be sufficiently close to the Sun for the little carbon dioxide in its early atmosphere to have heated the surface and released additional carbon dioxide into the atmosphere. Nitrogen makes up 79% of the atmosphere on Earth, along with oxygen (20%) and a trace amount of other gases including carbon dioxide and water vapor. On Venus and Mars, however, carbon dioxide makes up the majority of the atmosphere.

In Venus's original ocean drained, water vapor molecules were dispersed by ultraviolet light, and hydrogen was released into space. There was no longer any surface water, and the amount of carbon dioxide increased.

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Answer:

-----> 2 CH₄ + 2 NH₃ + 3 O₂ -----> 2 HCN + 6 H₂O

-----> CH₄ = Whichever reactant produces the smaller amount of product is the limiting reagent. This is because the limiting reagent runs out before all of the other reactant is completely used. Because CH₄ produces the smaller amount of product, it is the limiting reagent.

-----> 13.5 grams HCN

Explanation:

Part 1:

An equation is balanced when there is an equal amount of each reactant on both sides of the reaction. These amounts can be modified by adding coefficients in front of the molecules.

The unbalanced equation:

CH₄ + NH₃ + O₂ -----> HCN + H₂O

Reactants: 1 carbon, 7 hydrogen, 1 nitrogen, 2 oxygen

Products: 1 carbon, 3 hydrogen, 1 nitrogen, 1 oxygen

The balanced equation:

2 CH₄ + 2 NH₃ + 3 O₂ -----> 2 HCN + 6 H₂O

Reactants: 2 carbon, 14 hydrogen, 2 nitrogen, 6 oxygen

Products: 2 carbon, 14 hydrogen, 2 nitrogen, 6 oxygen

Part 2:

You can determine the limiting reactant by converting each reactant mass (besides O₂) to a product mass. Whichever reactant produces the smaller amount of product is the limiting reagent. This is because the limiting reagent runs out before all of the other reactant is completely used.

Let's convert to HCN because Part 3 also wants to know how much HCN is produced. To find this amount, you need to (1) convert grams reactant to moles reactant (via their molar masses), then (2) convert moles reactant to moles HCN (via the mole-to-mole ratio from equation coefficients), and then (3) convert moles HCN to grams HCN (via its molar mass).

Molar Mass (CH₄): 12.011 g/mol + 4(1.008 g/mol)

Molar Mass (CH₄): 16.043 g/mol

Molar Mass (NH₃): 14.009 g/mol + 3(1.008 g/mol)

Molar Mass (NH₃): 17.033 g/mol

Molar Mass (HCN): 1.008 g/mol + 12.011 g/mol + 14.009 g/mol

Molar Mass (HCN): 27.028 g/mol

8 g CH₄           1 mole            2 moles HCN         27.028 g
--------------  x  ----------------  x  ----------------------  x  ----------------  =  13.5 g HCN
                       16.034 g          2 moles CH₄            1 mole

10 g NH₃           1 mole            2 moles HCN          27.028 g
---------------  x  ----------------  x  -----------------------  x  ----------------  =  15.9 g HCN
                         17.033 g          2 moles NH₃            1 mole

Because CH₄ produces the smaller amount of product, it is the limiting reagent. Therefore, the actual amount of HCN produced is 13.5 g.

The Henderson-hasselbalch equation for a propanoic acid solution is expressed as [A-]/[HA] = 2.7.

It is expressed as:

pH = pKa + log([A-]/[HA].

where,

[A-] is molar concentration of conjugate base and

[HA] is molar concentration of weak acid.

Given,

pH = 5.30

Substitute the value in the Henderson Hasselbalch equation

5.30 = 4.874 + log[A-]/ [HA]

log[A-]/[HA] = 5.30 - 4.874

[A-]/[HA] = 10^ 0.426

[A-]/[HA] = 2.7

Thus from the conclusion we can say that The henderson-hasselbalch equation for a propanoic acid solution is expressed as [A-]/[HA] = 2.7.

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The 4 major types of inhibition are  Competitive Inhibition, Non-competitive Reversible Inhibition,  Non-competitive Irreversible Inhibition, and  Uncompetitive Inhibition.

1. Competitive Inhibition: Competing inhibitors are substances that resemble an enzyme's natural substrate so closely that they vie for the active site. In order to stop additional reactions, the inhibitor binds to the active site and stays there.

The inhibitor and substrate fight for the enzyme's active site because the enzyme may react with the inhibitor and release the products as it normally would with its substrate

2. Non-competitive Inhibition:

And in the Michaelis equation, this kind of inhibition lowers V max but has no impact on Km.The rate at which the enzyme is fully saturated is given by V = V max [S] / Km + [S] (V max).

(Km): The substrate is the Michaelis constant, at which the reaction rate is halved, or at V maximum.

       Additionally, this kind of inhibition lowers Vmax but has no impact on Km.

3. Uncompetitive Inhibition: is the last type, where the inhibitor attaches to the [E-S] complex to create [E-S-I] even if it has no affinity for the enzyme.

However, this kind of inhibition lowers both Vmax and Km.

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Yes, the objects made of the same metal.

The volume of the first object, which is a cube, is equal to the edge length cubed.

This volume is converted to mL as follows:

V1 = [tex](3 cm)^{3}[/tex] = 27 [tex]cm^{3}[/tex] = 27 mL

Based on its given mass, the density will be:

d1 = m1/V1

   = 140.4g/ 27.0 mL

   = 5.20g/mL

The following thing is a sphere. We calculate the volume of a sphere using the following formula, with the result expressed in  mL:

V1 = 4/3 *[tex]\pi[/tex]* [tex](1.42)^{2}[/tex]

= 12 [tex]cm^{3}[/tex]

= 12 mL

The density based on the following mass will be 

d2= m2/V2

= 61.6 g/12.0 mL

= 5.14 g/mL

These two computed densities are separated by about 1%. We can determine that these objects are made of the same metal based on the precision of +/- 1% provided.

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The incorrect statement is e = 0 for a spontaneous reaction.

The cell potential is 0 at equilibrium.

For a spontaneous reaction the value cell potential E cell is positive i.e. >0

This statement is incorrect because the potential of a cell is positive if the reaction is spontaneous. Hence it cannot be zero.

For a spontaneous reaction gibbs free energy δ g value is negative i.e. < 0

At equilibrium the value of gibbs free energy is zero i.e. δg = 0

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Photosynthesis: water + carbon dioxide + sunlight → oxygen + glucose

Respiration: glucose + oxygen → carbon dioxide + water + energy

Combustion: methane + oxygen → water + carbon dioxide + heat

The process by which plants convert carbon dioxide, water, and sunshine into oxygen and sugar-based energy is known as photosynthesis.

water + carbon dioxide + sunlight → oxygen + glucose

The sugars created during photosynthesis are combined with oxygen during the process of respiration in plants to create energy for plant development.

glucose + oxygen → carbon dioxide + water + energy

Combustion is a chemical process between substances that often involves oxygen and produces heat and light in the form of flame.

methane + oxygen → water + carbon dioxide + heat

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Photosynthesis: water + carbon dioxide + sunlight → oxygen + glucose

Respiration: glucose + oxygen → carbon dioxide + water + energy

Combustion: methane + oxygen → water + carbon dioxide + heat

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Answer:

pH = 4

Explanation:

pH is a base 10 logarithmic scale

  1000 = 10^3

        7 -3  = pH = 4  

Answer:

1.16 moles CO₂

Explanation:

To find the moles of CO₂, you need to (1) convert grams C₈H₁₈ to moles (via the molar mass) and then (2) convert moles C₈H₁₈ to moles CO₂ (via the mole-to-mole ratio from equation coefficients). It is important that the conversions/ratios are arranged in a way that allows for the cancellation of units. The final answer should have 3 significant figures like the given value.

Molar Mass (C₈H₁₈): 8(12.011 g/mol) + 18(1.008 g/mol)

Molar Mass (C₈H₁₈): 114.232 g/mol

2 C₈H₁₈ + 25 O₂ -----> 16 CO₂ + 18 H₂O
^                                    ^

16.6 g C₈H₁₈            1 mole            16 moles CO₂          
--------------------  x  -----------------  x  -------------------------  = 1.16 moles CO₂  
                              114.232 g          2 moles C₈H₁₈          

If the solvent level in the developing jar is deeper than the origin (spotting line) of the TLC plate, the solvent will dissolve the compounds into the solvent reservoir instead of allowing them to move up the plate by capillary action. Thus, you will not see spots after the plate is developed.

What is Thin Layer Chromatography (TLC)?

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The weighted average of a penny is = 67.9g

The sample of penny for pre 1982 is a total of = 18 pennies.

The mass of each penny for pre 1982 = 3.1

Therefore the total weight for penny for pre 1982

= 18×3.1

= 55.8

The sample of penny for post 1982 is a total of = 32 pennies.

The mass of each penny for post 1982 = 2.5

Therefore the total weight for penny for post 1982

=32×2.5

= 80

The weighted average

= 55.8+80/2

= 135.8/2

= 67.9g

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We have five significant figures in the value 25 , 803 , 000 , 000 mg.

The term significant figures refers to the numbers of figures to which we can assign a mathematical values. The numbers that we refer to here must be non zero numbers.

The implies that we must look out for the values that are nonzero in the digits shown which is 25 , 803 , 000 , 000 mg. A close look at this values shows us that we have five figures that can be assigned values.

Thus we have five significant figures in the value 25 , 803 , 000 , 000 mg.

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Answer:

Alkanes

Explanation:

Alkanes are hydrocarbons containing only single bonds between the parent chain carbons.

Alkenes are hydrocarbons containing at least one double bond between the parent chain carbons.

Alkynes are hydrocarbons containing at least one triple bond between the parent chain carbons.

1.4406 × 10⁻⁶ is the Ka(dissociation constant) for the acid.

The equilibrium constant for the reaction of an acid with water is the acid dissociation constant, where the acid, HA separates into H⁺ and A⁻ ions.

The acid dissociation constant is represented by (Ka).

So let's first imagine that the given monoprotic acid is HA.

HA will dissociate into H⁺ and A⁻ ions.

HA ⇒ H⁺ and A⁻

The formula used for Ka is

Ka = [H⁺] [A⁻] / [HA]

Given

pH = 2.83

[HA] = 1.50 M

From the given pH, we can calculate [H⁺] and [A⁻]

[H⁺] = [A⁻] = 1 × [tex]10^{-2.83}[/tex] =  1.47 × 10⁻³ M

Ka = 1.47 × 10⁻³ × 1.47 × 10⁻³ / 1.50

Ka = 1.4406 × 10⁻⁶

Hence, 1.4406 × 10⁻⁶ is the Ka for the acid.

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According to the iupac convention, alkyl substituents on a hydrocarbon chain should be listed in alphabetical without taking prefixes into name.

A hydrocarbon would be an organic molecule in organic chemistry that is made completely of hydrogen as well as carbon. A good example of group 14 hydrides includes hydrocarbons. In general, hydrocarbons lack color and are hydrophobic. Typically, their faint smells are compared to that of gasoline and lighter fluid.

Therefore, according to the iupac convention, alkyl substituents on a hydrocarbon chain should be listed in alphabetical without taking prefixes into name.

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Answer:

156.16 gm of ice

Explanation:

I will assume the ice starts at  0 C

As the ice melts, it produces water at  0 C  so we do not need to calculate THAT water being cooled....only the 356 ml being cooled to 0 C

356  gm *  4.186 j/gm-c  * (35 c)   = 52157.56 j

  this is the heat absorbed by the melting ice

x gm * 334 j / gm   = 52157.56 j           (334 j/gm is heat of fusion of water)

x = 156.16 gm of ice required

The molar mass of a substance can be calculated by dividing the mass of the substance by its number of moles.

However, the number of moles can be calculated using the following formula:

PV = nRT

Where;

0.92 × 1.6 = n × 0.0821 × 287

1.47 = 23.56n

n = 1.47/23.56

n = 0.624moles

Molar mass = 0.314g ÷ 0.624mol

Molar mass = 0.503g/mol

Therefore, the molar mass of a 0.314-gram sample of gas having a volume of 1.6L at 287K and 0.92 atm is 0.503g/mol.

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The 2.5 x 10²¹ molecules of CH₃OH have 0.00415 moles.

The number of moles is given by

[tex]n=\frac{m}{M}[/tex]

When given mass = molar mass of a substance, then no of moles is equal to 1 which is equivalent to 6.022 x 10²³.

We also know,

1 mole has 6.022 x 10²³ molecules of CH₃OH

According to the mole concept,

6.022 x 10²³ molecules of CH₃OH is equal to 1 mole

2.5 x 10²¹molecules of CH₃OH is equal to [tex]\frac{2.5\times10^2^1}{6.022 \times 10^2^3}[/tex] moles

Thus, 2.5 x 10²¹molecules of CH₃OH are equal to 4.15 x 10⁻³ moles or 0.00415 moles

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The vapor pressure of a warm pale higher than the vapor pressure of a cold lake because Warm water evaporates more quickly.

Hence, Option B is correct answer.

The process in which a liquid or solid is converted into vapour is called Evaporation. When water gains the heat energy then it changes in to gas. When the temperature increases it speeds up the rate of evaporation.

The temperature of warm water increases, the kinetic energy of the warm water molecule also increases.

Thus from the above conclusion we can say that The vapour pressure of a warm lake is higher than the vapor pressure of a cold lake because Warm water evaporates more quickly.

Hence, Option B is correct answer.

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Question: Why is the vapor pressure of a warm lake higher than the vapor pressure of a cold lake?

A. Warm water has a greater heat of vaporization.

B. Warm water evaporates more quickly.

C. Cool water evaporates more quickly.

D. Cool water has a greater heat of vaporization.

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Answer:

This substance would cause cause an extensive cellular damage to Anil.

The carbon nucleophile which attacks molecular bromine in the acid-catalyzed α-bromination of a ketone an acetylide. The Organometallic reagents like those used in the Grignard, Blaise, Reformatsky, and Barbier reactions as well as reactions involving the organolithium reagents and acetylides are frequently used as carbon nucleophiles.

These chemicals are frequently employed in nucleophilic additions. Compounds or intermediates that contain an electron-rich carbon atom are known as carbon-centered nucleophiles because they have the ability to donate an electron pair from that carbon atom to an electrophile. When writing resonance structures, the strong negative nature of a metal-bonded the carbon is apparent.

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1 to 2 feet (cubed) per person per year can be accumulated in lagoon systems.

Lagoons are basin-like to collect, retain, and treat wastewater for a predefined amount of time. They are pond-like bodies of water.

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The volume of the 18M HCl needed to make the solution will be 2.5 mL.

According to the dilution principle, the number of moles of solutes in a solution before and after dilution must remain the same.

Since, mole = molarity x volume

Thus, molarity x volume before dilution = molarity x volume after dilution.

Mathematically, the equation is written as: m1v1 = m2v2

In this case, m1 = 18 M, m2 = 1.5 M, and v2 = 30 mL.

What we are looking for is v1, the amount of the stock HCl needed for dilution.

v1 = m2v2/m1 = 1.5 x 30/18 = 2.5 mL.

Thus, 2.5 mL of the stock HCl would be needed.

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The percent abundance of the post-1982 pennies is 56%.

Percent abundance is the percentage amount of all naturally occurring isotopes of an element.

The percent abundance of the post-1982 pennies is calculated as follows;

percent abundance = number of  post-1982 pennies / total number of pennies x 100%

percent abundance =  (28) / (50) x 100%

percent abundance =  56%

Thus, the percent abundance of the post-1982 pennies is 56%.

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Answer:

FOR PRE 44

Explanation:

22/50X100=44