What volume of 6.9 M NaOH is needed to completely titrate 0.42 L of 2.39 M phosphoric acid according to
the equation:

H3PO4(aq) + 3NaOH(aq) + Na3PO4(aq) + 3H2O(aq)

A) O 0.05 L
B) O6.93 L
C) O0.44 L
D) 03.01 L
E) 436.43 L

What Volume Of 6.9 M NaOH Is Needed To Completely Titrate 0.42 L Of 2.39 M Phosphoric Acid According

Answers

Answer 1

Taking into account the definition of molarity and the stoichiometry of the reaction, the correct option is option C) 0.44 L of 6.9 M NaOH is needed to completely titrate 0.42 L of 2.39 M phosphoric acid.

The balanced reaction is:

H₃PO₄ (aq) + 3 NaOH (aq) → Na₃PO₄ (aq) + 3 H₂O(aq)

Then, by stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

H₃PO₄: 1 mole NaOH: 3 moles Na₃PO₄: 1 mole H₂O: 3 moles

Molarity is the number of moles of solute that are dissolved in a given volume.

Molarity is determined by:

[tex]Molarity=\frac{number of moles of solute}{volume}[/tex]

 Molarity is expressed in units [tex]\frac{moles}{liter}[/tex].

In this case, 0.42 L of 2.39 M phosphoric acid reacts. So, by definition of molarity, the number of moles that participate in the reaction is calculated as:

[tex]2.39 \frac{moles}{liter}=\frac{number of moles of phosphiric acid}{0.42 liters}[/tex]

Solving:

number of moles of phosphiric acid= 2.39 [tex]\frac{moles}{liter}[/tex]* 0.42 liters

number of moles of phosphiric acid= 1.0038 moles ≅ 1 mole

Approaching 1 mole of the amount of phosphoric acid required, then by stoichiometry of the reaction, 3 moles of NaOH are necessary to react with 1 mole of the acid.

Then by definition of molarity and knowing that 6.9 M NaOH is needed, you can calculate the necessary volume amount of NaOH by:

[tex]6.9 \frac{moles}{liter} =\frac{3 moles}{volume}[/tex]

Solving:

6.9 [tex]\frac{moles}{liter}[/tex]* volume= 3 moles

[tex]volume=\frac{3 moles}{6.9\frac{moles}{liter} }[/tex]

volume= 0.44 L

The correct option is option C) 0.44 L of 6.9 M NaOH is needed to completely titrate 0.42 L of 2.39 M phosphoric acid.

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Related Questions

g Which ONE of the following pairs of organic compounds are NOT pairs of isomers? A) butanol ( CH3-CH2-CH2-CH2-OH ) and diethyl ether ( CH3–CH2–O–CH2–CH3 ) B) isopentane ( (CH3)2-CH-CH2-CH3 ) and neopentane ( (CH3)4C ) C) ethanolamine ( H2N-CH2-CH2-OH ) and acetamide ( CH3-CO-NH2 ) D) acrylic acid ( CH2=CH-COOH ) and propanedial ( OHC–CH2–CHO ) E) trimethylamine ( (CH3)3N ) and propylamine ( CH3-CH2-CH2-NH2 )

Answers

Answer:

ethanolamine ( H2N-CH2-CH2-OH ) and acetamide ( CH3-CO-NH2 )

Explanation:

Isomers are compounds that have the same molecular formula but different structural formulas. Hence any pair of compounds that can be represented by exactly the same molecular formula are isomers of each other.

If we look at the pair of compounds; ethanolamine ( H2N-CH2-CH2-OH ) and acetamide ( CH3-CO-NH2 ), one compound has molecular formula, C2H7ON while the other has a molecular formula, C2H5ON, hence they are not isomers of each other.

What is the mass of 7.2 moles of titanium to the nearest tenth?

Answers

Answer:

7.2 moles of titanium weigh 344.7 grams.

Explanation:

Molar mass of titanium = 47.88 g

[tex]1 \: mole \: of \: titanium = 47.88 \: grams \\ 7.2 \: moles \: of \: titanium = g[/tex]

solve for g

[tex]g = ( \frac{7.2 \times 47.88}{1} ) \\ g = 344.7 \: grams[/tex]

Answer:

[tex]\boxed {\boxed {\sf 344.7 \ g \ Ti}}[/tex]

Explanation:

We are asked to find the mass of 7.2 moles of titanium.

To convert from moles to grams, we use the molar mass. This value is the mass of 1 mole of a substance. They are equivalent to the atomic masses found on the Periodic Table, but the units are grams per moles instead of atomic mass units.

Look up titanium's molar mass on the Periodic Table.

Ti: 47.87 g/mol

Convert moles to grams using dimensional analysis. First, set up a ratio using the molar mass.

[tex]\frac {47.87 \ g \ Ti}{1 \ mol \ Ti}[/tex]

We are converting 7.2 moles to grams, so we multiply the ratio by this value.

[tex]7.2 \ mol \ Ti *\frac {47.87 \ g \ Ti}{1 \ mol \ Ti}[/tex]

The units of moles of titanium cancel.

[tex]7.2 *\frac {47.87 \ g \ Ti}{1 }[/tex]

The denominator of 1 can be ignored (a number over 1 is just the number).

[tex]7.2 *47.87 \ g \ Ti}[/tex]

[tex]344.664 \ g \ Ti[/tex]

We are asked to round to the nearest tenth. The 6 in the hundredth place tells us to round the 6 in the tenth place up to a 7.

[tex]344.7 \ g \ Ti[/tex]

There are approximately 344.7 grams of titanium in 7.2 moles of titanium.

Draw the major condensation product obtained by treatment of ethyl 3-methylbutanoate with sodium ethoxide in ethanol.

Answers

Answer:

ethyl 3-ethoxy-3-hydroxy-2-isopropyl-5-methyl hexanoate

Explanation:

In this case, we have a very strong base (sodium ethoxide). Therefore, this compound will remove a hydrogen from ethyl 3-methyl butanoate generating a carbanion.

This carbanion, can attack another ethyl 3-methyl butanoate molecule on the carbonyl group generating a new C-C bond and producing a negative charge in the oxygen.

Then the ethanol can protonate the molecule generating an "OH" group and the ethoxide.

See figure 1

I hope it helps!

How is the mass percent of elements in a compound different for a 1.0-g sample versus a 100.-g sample versus a 1-mole sample of the compound?

Answers

Answer:

since mass percent is based on the proportions of elements in a compound, the mass of a sample does not matter

The mass percent or percentage by mass of a chemical compound is a constant.

What is mass percent?

Mass percent can be defined as percentage by mass and it is the concentration of a given amount (quantity) of element in a chemical compound or a component of a mixture.

The mass percent or percentage by mass of a chemical compound is a constant, regardless of the amount of substance that is present. Thus, all chemical compound always have a constant composition because mass percent is typically based on the proportions of the elements contained in a chemical compound.

Read more on mass here: brainly.com/question/3173452

Rank the following amine derivatives from highest acidity (lowest pKa value) to lowest acidity (highest pKa value).
Highest acidity
anilinium ion
aniline
ammonium ion
secondary amine
amide
Lowest acidity

Answers

Answer:

anilinium ion > ammonium ion > amide > aniline > secondary amine

Explanation:

Acidity of amine derivatives can derived from their pKa values.

The rule of thumb for acidity with relation to pKa values is that:

As the pKa decreases the acid strength increases and the conjugate base decreases. Similarly, as the pKa increases, the acid strength decreases and the conjugate base increase.

Hence the stronger the acid , the  lower pKa value  and the weaker the acid , the stronger the pKa value.

So the pKa value for anilinium ion = 4.6

ammonium ion = 9.4

Amide = 15

Similarly, for aniline and secondary amine, in order to determine the derivative with the higher acidity, we will consider the electron withdrawing substituent group.

The more difficult the electron are being withdraw from the electron withdrawing substituent , the more acidic the compound.

In aniline , the stabilized benzene ring attached to NH₂ makes it a less electron withdrawing group compared to the straight chains structure found in secondary amine where electron are easily withdraw by nucleophilic substitution reactions.

Thus, from highest acidity (lowest pKa value) to lowest acidity (highest pKa value).

the amine derivatives ranking is as follows:

anilinium ion > ammonium ion > amide > aniline > secondary amine

Which of the terms heat of vaporization and heat of fusion refers to condensation and which refers to melting?

Answers

Answer:

Heat of vaporization refers to condensation and heat of fusion refers to melting.

Explanation:

Heat of vaporization or heat of evaporation, is defined as the energy required to convert liquid substance into a gas which creates condensation. As a gas condenses to a liquid, heat is released.

Heat of fusion refers to melting because heat of fusion is defined as the energy required to change any amount of substance when it melts.

Hence, the correct answer is "Heat of vaporization refers to condensation and heat of fusion refers to melting.".

What type of bond would form between two atoms of phosphorus? A. Single covalent bond B. Single ionic bond C. Triple covalent bond D. Double covalent bond

Answers

Answer:

A double bond is formed when two pairs of electrons are shared between the two participating atoms. It is represented by two dashes (=). It is represented by two dashes (=). Double covalent bonds are much stronger than a single bond, but they are less stable

Explanation:

An ionic bond is a bond

Answers

Answer:

That involve the complete transfer of an electron from one atom of an element to another

A student sets up the following equation to convert a measurement. The (?) Stands for a number the student is going to calculate. Fill in the missing part of this equation. (0.030 cm^3) x ? =m^3

Answers

Answer:

\text{0.30 cm}^{3} \times \left (\dfrac{10^{-2}\text{ m}}{\text{1 cm}}\right )^{3} = 3.0 \times 10^{-7} \text{ m}^{3}  

Explanation:

0.030 cm³ × ? = x m³

You want to convert cubic centimetres to cubic metres, so you multiply the cubic centimetres by a conversion factor.

For example, you know that centi means "× 10⁻²", so  

1 cm = 10⁻² m

If we divide each side by 1 cm, we get 1 = (10⁻² m/1 cm).

If we divide each side by 10⁻² m, we get (1 cm/10⁻² m) = 1.

So, we can use either (10⁻² m/1 cm) or (1 cm/10⁻² m) as a conversion factor, because each fraction equals one.

We choose the former because it has the desired units on top.

The "cm" is cubed, so we must cube the conversion factor.

The calculation becomes

[tex]\text{0.30 cm}^{3} \times \left (\dfrac{10^{-2}\text{ m}}{\text{1 cm}}\right )^{3} = 0.30 \times 10^{-6}\text{ m}^{3} = \mathbf{3.0 \times 10^{-7}} \textbf{ m}^{\mathbf{3}}\\\\\textbf{0.30 cm}^{\mathbf{3}} \times \left (\dfrac{\mathbf{10^{-2}}\textbf{ m}}{\textbf{1 cm}}\right )^{\mathbf{3}} = \mathbf{3.0 \times 10^{-7}} \textbf{ m}^{\mathbf{3}}[/tex]

Assume that you are provided with the following materials:
• Strips of metallic zinc, metallic copper, metallic iron
• 1M aqueous solutions of ZnSO4, CuSO4, FeSO4, and aqueous iodine(I2)
• Other required materials to create Voltaic cells such as beakers, porous containers, graphite rods, a voltmeter, and a few wires with alligator clips.
In this modified version of the lab, after thoroughly studying the lab hand out and watching the videos,identify 4 different combinations of Voltaic cells that are possible to be created with the above materials.For each cell created, include the following details.
A) Which electrode was the anode,and which was the Cathode?
B) The anode and cathode half reactions.
C) Balanced equation for each cell you propose to construct.
D) Calculated Eocelle Short hand notation (line notation) for each cell (be sure to include the inactive electrode if needed).

Answers

Answer:

See explanation

Explanation:

First voltaic cell;

Zn(s)|Zn^2+(aq)||Cu^2+(aq)|Cu(s)

Anode;

Zinc

Cathode;

Copper

Oxidation half equation;

Zn(s)------> Zn^2+(aq) + 2e

Reduction half equation;

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

Overall; Zn(s) + Cu^2+(aq) -----> Zn^2+(aq) + Cu(s)

E°cell = 0.34 -(-0.76) =1.1 V

Second voltaic cell;

Zn(s)|Zn^2+(aq)||Fe^2+(aq)|Fe(s)

Anode;

Zinc

Cathode;

Iron

Oxidation half equation;

Zn(s)------> Zn^2+(aq) + 2e

Reduction half equation;

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

Overall; Zn(s) + Fe^2+(aq) -----> Zn^2+(aq) + Fe(s)

E°cell = (-0.44) -(-0.76) = 0.32 V

Third voltaic cell;

Fe(s)|Fe^2+(aq)||Cu^2+(aq)|Cu(s)

Anode;

Iron

Cathode;

Copper

Oxidation half equation;

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

Reduction half equation;

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

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

E°cell = 0.34 -(-0.44) = 0.78 V

Fourth voltaic cell

Cu(s)|Cu^2+(aq)||I2(aq)|C(s)|I^-(aq)

Anode;

Copper

Cathode;

Graphite rod

Oxidation half equation;

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

Reduction half equation;

I2(aq) +2e -----> 2I^-(aq)

Overall; Cu(s) + I2(aq) -----> Cu^2+(aq) + 2I^-(aq)

E°cell = 0.54 -0.34 = 0.20 V

Write the half-reactions as they occur at each electrode and the net cell reaction for this electrochemical cell containing indium and cadmium. In(s)|

Answers

Answer:

Oxidation half equation;

3Cd(s) -------> 3Cd^2+(aq) + 6e

Reduction half equation;

2In^3+(aq) + 6e -----> 2In(s)

Explanation:

Since the reduction potentials of Indium and Cadmium are -0.34 V and - 0.40 V respectively, we can see that cadmium will be oxidized while indium will the reduced.

We arrived at this conclusion by examining the reduction potential of both species. The specie with more negative reduction potential is oxidized in the process.

Oxidation half equation;

3Cd(s) -------> 3Cd^2+(aq) + 6e

Reduction half equation;

2In^3+(aq) + 6e -----> 2In(s)

A chemist fills a reaction vessel with 0.978 g aluminum hydroxide AlOH3 solid, 0.607 M aluminum Al+3 aqueous solution, and 0.396 M hydroxide OH− aqueous solution at a temperature of 25.0°C.
Under these conditions, calculate the reaction free energy ΔG for the following chemical reaction:
Al(OH)3(s) = A1+ (aq) +30H (aq)
Use the thermodynamic information in the ALEKS Data tab. Round your answer to the nearest kilojoule.
KJ

Answers

Answer: [tex]\Delta G^{0}[/tex] = 168.12 kJ

Explanation: Gibbs Free Energy, at any time, is defined as the enthalpy of the system minus product of temperature and entropy of the reaction, i.e.:

[tex]\Delta G^{0} = \Delta H^{0} - T.\Delta S^{0}[/tex]

Enthalpy is defined as internal heat existent in the system. It is calculated as:

[tex]\Delta H^{0} = \Sigma H^{0}_{product} - \Sigma H^{0}_{reagent}[/tex]

Using Enthalpy Formation Table:

[tex]\Delta H^{0} = [3*(-299.9)+(-524.7)] - (-1277)[/tex]

[tex]\Delta H^{0} = 62,6 kJ[/tex]

Entropy is the degree of disorder in the system. It is found by:

[tex]\Delta S^{0} = \Sigma S^{0}_{products} - \Sigma S^{0}_{reagents}[/tex]

Calculating:

[tex]\Delta S^{0} = (-321.7) + 3(-10.8) - 0[/tex]

[tex]\Delta S^{0} = -354.1J[/tex]

And so, Gibbs Free energy will be:

[tex]\Delta G^{0} = \Delta H^{0} - T.\Delta S^{0}[/tex]

[tex]\Delta G^{0} = 62600 - [298.(-354.1)][/tex]

[tex]\Delta G^{0} = 168121.8 J[/tex]

Rounding to the nearest kJ:

[tex]\Delta G^{0}[/tex] = 168.12 kJ

You have to prepare a pH 3.65 buffer, and you have the following 0.10M solutions available: HCOOH, CH3COOH, H3PO4, HCOONa, CH3COONa, and NaH2PO4. How many mL of HCOOH and HCOONa would you use to make approximately a liter of the buffer?

Answers

Answer:

550mL of HCOOH 0.1M and 450mL of HCOONa 0.1M

Explanation:

It is possible to find the pH of a buffer by using H-H equation:

pH = pKa + log [A⁻]/[HA]

For the formic buffer (HCOOH/HCOONa):

pH = 3.74 + log [HCOONa]/[HCOOH]

As you need a buffer of pH 3.65:

pH = 3.74 + log [HCOONa]/[HCOOH]

3.65 = 3.74 + log [HCOONa]/[HCOOH]

0.81283 = [HCOONa]/[HCOOH] (1)

Where  [HCOONa]/[HCOOH] can be taken as the moles of each specie.

As molarity of both solutions is 0.10M (0.10mol / L) and you need 1L of solution, total moles of the buffer are:

0.10 moles = [HCOONa] + [HCOOH] (2)

Replacing (2) in (1):

0.81283 = 0.10 - [HCOOH]  /[HCOOH]

0.81283[HCOOH] = 0.10 - [HCOOH]

1.81283[HCOOH] = 0.10

[HCOOH] = 0.055 moles

And moles of HCOONa are:

[HCOONa] = 0.1 mol - 0.055mol =

[HCOONa] = 0.045 moles

As concentration of the solutions is 0.1M, the volume you need to add of both solutions is:

HCOOH = 0.055 mol ₓ (1L / 0.1mol) = 0.55L = 550mL of HCOOH 0.1M

HCOONa = 0.045 mol ₓ (1L / 0.1mol) = 0.45L = 450mL of HCOONa 0.1M

The number should be considered like 550mL of HCOOH 0.1M and 450mL of HCOONa 0.1M.

Calculation of mL:

Here we used the H-H equation:

pH = pKa + log [A⁻]/[HA]

Now

For the formic buffer (HCOOH/HCOONa):

So,

pH = 3.74 + log [HCOONa]/[HCOOH]

Now

need a buffer of pH 3.65:

So,

pH = 3.74 + log [HCOONa]/[HCOOH]

3.65 = 3.74 + log [HCOONa]/[HCOOH]

0.81283 = [HCOONa]/[HCOOH] (1)

here  [HCOONa]/[HCOOH] can be considered as the moles of each specie.

Now the total moles should be

0.10 moles = [HCOONa] + [HCOOH] (2)

Now

0.81283 = 0.10 - [HCOOH]  /[HCOOH]

0.81283[HCOOH] = 0.10 - [HCOOH]

1.81283[HCOOH] = 0.10

[HCOOH] = 0.055 moles

And moles of HCOONa should be

[HCOONa] = 0.1 mol - 0.055mol =

[HCOONa] = 0.045 moles

Now

HCOOH = 0.055 mol ₓ (1L / 0.1mol) = 0.55L = 550mL of HCOOH 0.1M

HCOONa = 0.045 mol ₓ (1L / 0.1mol) = 0.45L = 450mL of HCOONa 0.1M

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Select the true statement concerning voltaic and electrolytic cells. Select one: a. Voltaic cells involve oxidation-reduction reactions while electrolytic cells involve decomposition reactions. b. Voltaic cells require applied electrical current while electrolytic cells do not. . c. all electrochemical cells, voltaic and electrolytic, must have spontaneous reactions. d. Electrical current drives nonspontaneous reactions in electrolytic cells.

Answers

Answer:

Electrical current drives nonspontaneous reactions in electrolytic cells.

Explanation:

Electrochemical cells are cells that produce electrical energy from chemical energy.

There are two types of electrochemical cells; voltaic cells and electrolytic cells.

A voltaic cell is an electrochemical cell in which electrical energy is produced from spontaneous chemical process while an electrolytic cell is an electrochemical cell where electrical energy is produced from nonspontaneous chemical processes. Current is needed to drive these nonspontaneous chemical processes in an electrolytic cell.

Answer:

electrolytic cells generate electricity through a non-spontaneous reaction while voltaic cells absorb electricity to drive a spontaneous reaction.

Explanation:

Answer via Educere/ Founder's Education

28.What is the correct IUPAC name for the following compound?A)12-crown-5B)12-crown-4C)4-crown-12D)12-crown-12E)Cyclododecane tetraether

Answers

Answer:

12-crown-4

Explanation:

We must recall that any structural moiety in organic chemistry having the R-O-R unit is an ether. If the oxygen form a ring in which they are sandwiched in between carbon atoms, the compound is known as a crown ether. The name emanates from the close resemblance of the compound to an actual crown.

If we want to name the crown ether, we first count the number of carbon atoms present and the number of oxygen atoms present. The correct name is now, total number of carbon + oxygen atoms -crown- number of oxygen atoms, in this case; 12-crown-4, hence the answer.

Answer:

12-crown-4

Explanation:

A balloon is filled to a volume of 1.50 L with 3.00 moles of gas at 25 °C. With pressure and temperature held constant, what will be the volume (in L) of the balloon if 0.50 moles of gas are released?

Answers

Answer:

Volume : 1.25 L

Explanation:

We are given here that the volume ( V[tex]_1[/tex] ) = 1.50 Liters, the initial moles ( held at 25 °C ) = 3.00 mol, and the final moles ( n[tex]_2[/tex] ) = 3.00 - 0.5 = 2.5 mol. The final mol is calculated given that 0.50 mol of gas are released from the prior 3.00 moles of gas.

Volume ( V[tex]_1[/tex] ) = 1.50 L,

Initial moles ( n[tex]_1[/tex] ) = 3.00 mol,

Final Volume ( n[tex]_2[/tex] ) = 3.00 - 0.5 = 2.5 mol

Applying the combined gas law, we can calculate the final volume ( V[tex]_2[/tex] ).

P[tex]_1[/tex]V[tex]_1[/tex] / n[tex]_1[/tex]T[tex]_1[/tex] = P[tex]_2[/tex]V[tex]_2[/tex] / n[tex]_2[/tex]T[tex]_2[/tex] - we know that the pressure and temperature are constant, and therefore we can apply the following formula,

V[tex]_1[/tex] / n[tex]_1[/tex] = V[tex]_2[/tex] / n[tex]_2[/tex] - isolate V[tex]_2[/tex],

V[tex]_2[/tex] = V[tex]_1[/tex] n[tex]_2[/tex] / n[tex]_1[/tex] = 1.50 L [tex]*[/tex] 2.5 mol / 3.00 mol = ( 1.5 [tex]*[/tex] 2.5 / 3 ) L = 1.25 L

The volume of the balloon will be 1.25 L.

Given the following values of pKa, determine which is the weakest base of the answers listed. Acid pKa HClO2 1.95 HClO 7.54 HCOOH 3.74 HF 3.17 HNO2 3.15

Answers

Answer:

HClO 7.54

Explanation:

Hypochlorous acid (HClO) is a weakest acid because the pKa value of Hypochlorous acid is very high among the options given in the activity. pKa is a method which is used in order to identify the strength of an acid. The higher the value of pKa of a liquid, lower the strength of an acid while lower the value of pKa of chemical, higher the strength of an acid. In the options, HClO2  is a strong acid due to high lower pKa value.

Evaluate the exponential expression (−2)6.

Answers

A general exponential expression is something like:

A^n

This means that we need to multiply the number A by itself n times.

Using that we will get (-2)^6 = 64

With that definition, we can rewrite:

(-2)^6 = (-2)*(-2)*(-2)*(-2)*(-2)*(-2)

So we just need to solve the above expression.

Also, remember the rule of signs:

(-)*(-) = (+)

We will get:

(-2)*(-2)*(-2)*(-2)*(-2)*(-2) =  [(-2)*(-2)]*[(-2)*(-2)]*[(-2)*(-2)]

                                        =  4*4*4 = 16*4 = 64

Then we got:

(-2)^6 = 64

If you want to learn more, you can read:

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The following two compounds each exhibit two heteroatoms (one nitrogen atom and one oxygen atom). In compound A, the lone pair on the nitrogen atom is more likely to function as a base. However, in compound B, the lone pair on the oxygen atom is more likely to function as a base. Explain this difference by selecting the correct reason(s)

Answers

Answer:

The answer is the photo attached

Explanation:

A solution of HCOOH has 0.16M HCOOH at equilibrium. The Ka for HCOOH is 1.8×10−4. What is the pH of this solution at equilibrium? Express the pH numerically.

Answers

Answer:

[tex]pH=2.28[/tex]

Explanation:

Hello,

In this case, for the acid dissociation of formic acid (HCOOH) we have:

[tex]HCOOH(aq)\rightarrow H^+(aq)+HCOO^-(aq)[/tex]

Whose equilibrium expression is:

[tex]Ka=\frac{[H^+][HCOO^-]}{[HCOOH]}[/tex]

That in terms of the reaction extent is:

[tex]1.8x10^{-4}=\frac{x*x}{0.16-x}[/tex]

Thus, solving for [tex]x[/tex] which is also equal to the concentration of hydrogen ions we obtain:

[tex]x=0.00528M[/tex]

[tex][H^+]=0.00528M[/tex]

Then, as the pH is computed as:

[tex]pH=-log([H^+])[/tex]

The pH turns out:

[tex]pH=-log(0.00528M)\\\\pH=2.28[/tex]

Regards.

s-Block compounds give a characteristic flame colour in the flame test. Based on this, can you give one use of s-block compounds?​

Answers

Answer:

Lithium is used in making electrochemical cells.

21. What are the two main ways of working with clay?

Answers

Answer:

Diferentes tipos de arcilla

ARCILLA DE LADRILLOS. Contiene muchas impurezas. ...

ARCILLA DE ALFARERO. Llamada también barro rojo y utilizada en alfarería y para modelar. ...

ARCILLA DE GRES. Es una arcilla con gran contenido de feldespato. ...

ARCILLAS “BALL CLAY” O DE BOLA. ...

CAOLIN. ...

ARCILLA REFRACTARIA. ...

BENTONITA.

Explanation:

Answer:

Coil method and the slab method.

Explanation:

The entropy of a substance above absolute zero will always be:

a. Negative
b. Positive
c. Neither Negative nor positive

Answers

i will be positive. just because it’s positive

How many grams of Al were reacted with excess HCl if 3.86 L of hydrogen gas were collected at STP in the following reaction?
2 Al (s) + 6 HCl (aq) → 2 AlCl₃ (aq) + 3 H₂ (g)

Answers

The amount, in grams, of Al that reacted with excess HCl if 3.86 L of hydrogen gas were collected at STP in the reaction would be 3.099 grams.

Stoichiometric problem

From the equation of the reaction, the mole ratio of Al that reacts with that of hydrogen gas that forms is 2:3.

At STP, 1 mole of any gas is equivalent to 22.4 Liters of the gas.

But only 3.86 L of hydrogen was formed in the reaction.

The equivalent mole of 3.86 L hydrogen at STP would be:

      3.86 x 1/22.4 = 0.1723 moles

From the mole ratio, the equivalent mole of Al that will produce 0.1723 moles of hydrogen gas would be:

          0.1723 x 2/3 = 0.1149 moles

Recall that: mass = mole x molar mass

Mass of 0.1149 moles Al = 0.1149 x 26.98

                                        = 3.099 grams

Thus, the amount of Al that reacted with excess HCl is 3.099 grams.

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Bomb calorimetry is a poor choice to determine the number of nutritional Calories in food; it consistently overestimates the Caloric content because options: A) dietary fiber isn't used by the body. B) carbohydrates don't burn to completion. C) proteins don't burn. D) water has Calories and isn't burnable.

Answers

Answer:

A) dietary fiber isn't used by the body.

Explanation:

The food we eat contains certain nutritional contents that provides energy, measured in calories (CAL) to the body. A procedure called BOMB CALORIMETRY can be used to determine the energy contents of these foods. The energy-supplying macromolecules contained in food substances we eat are carbohydrate, protein, fats etc.

Bomb calorimetry uses the method of burning the food substance in a device called bomb calorimeter, and measure the caloric content of the burnt food. Bomb calorimetry measures all the present calories in a food substance, which can include dietary fibers. Due to this reason, it is considered a poor choice in determining the number of nutritional calories in a food substance.

Dietary fibers are indigestible carbohydrates that cannot be broken down and used by the body. They pass along the alimentary canal until they are egested. Hence, they are no source of nutrients to the body. Since bomb calorimetry measures all calories including dietary fibers, it is said to overestimate the caloric content of food substances.

Chemistry
What is a chemical reaction

Answers

Answer:

A process that involves rearrangement

Explanation:

A chemical reaction is the process that involves rearrangement of the molecular or ironic structure of a substance, as a distinct from a change in physical form or a nuclear reaction.

Answer:

Explanation:

Chemistry

The chemical reaction H2(g) + ½ O2(g) → H2O(l) describes the formation of water from its elements.

The reaction between iron and sulfur to form iron(II) sulfide is another chemical reaction, represented by the chemical equation:

8 Fe + S8 → 8 FeS

2. The reaction of a triglyceride with methanol in the presence of a strong base to form
methyl esters and glycerol is called
O A. transesterification.
O B. saponification.
O C. ester formation.
O D. dehydration condensation.

Answers

Answer:

The answer it's A. transesterification

what is the valency of element sulphur ​

Answers

I am not to sure because I have not studied this

sorry not sure wish I can help u

In the following reaction, what element is gaining mass?

Mg(s) + CuSO4(aq) -> MgSO4(aq) + Cu(s)

A. Mg
B. None of these
C. Cu
D. O

Answers

Mg gained mass because it went from being a single element (on the reactant side) to being a molecule (on the product side).

Mg element is gaining mass. Hence, option A is correct.

What is a chemical equation?

A chemical reaction is a representation of symbols of the elements to indicate the number of substances and moles of reactant and product.

Mg gained mass because it went from being a single element (on the reactant side) to being a molecule (on the product side).

Hence, option A is correct.

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Different vinegars can be 5-20% acetic acid solutions and have been used for medicinal purposes for thousands of years. If a person takes 2.0 tablespoons of vinegar a day and the Molarity of the vinegar is .84 M, then how many grams of acetic acid (HC2H3O2) will be consumed? 1 Tablespoon is 15 mL.

.013 g
.026 g
.76 g
1.5 g

Answers

Answer:

1.5g

Explanation:

Remember that Molarity = (#moles of solute)/(#liters of solution)

This problem informs us that the Molarity of the vinegar is 0.84 and that the solution is 15mL.

First let's get your SI units to the correct ones.

15mL (1L/1000mL) = 0.015L

Molarity = (#moles of solute)/(#liters of solution) ~

(Molarity)(#liters of solution) = #moles of solute

(0.84M)(.015L) = 0.0126moles of acetic acid per tablespoon

2 tablespoons a day = 0.0126moles*2 =  0.0252 moles of acetic acid.

Now that we have the # of moles of acetic acid we need to get our answer into grams. The molecular weight of HC2H3O2 is 60g/mole.

0.0252mole HC2H3O2 (60g HC2H3O2/1mole HC2H3O2) = 1.512g ~ 1.5g HC2H3O2.

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