0.415 g of an unknown triprotic acid are used to make a 100.00 mL solution. Then 25.00 mL of this solution is transferred to an Erlenmeyer flask and is titrated with 0.1029 M NaOH. The initial burette reading is 1.81 mL; when the titration endpoint is reached, the final burette reading is 39.70 mL. How many moles of triprotic acid are neutralized during the titration? Provide your answer in decimal form (e.g. 0.123) to the correct number of significant figures.

Answer:

Explanation:

29.4 mL of nitric acid having molarity of .187M reacts with 22.7 mL of potassium hydroxide .

moles contained in nitric acid solution

= .0294 x .187 moles

= 5.5 x 10⁻³ moles

This must have been reacted with same amount of moles of base

moles of sodium hydroxide in 22.7 mL of solution = 5.5 x 10⁻³ moles

molarity of base solution = 5.5 x 10⁻³ / 22.7 x 10⁻³

= .24 M .

Answer:

I think that gravity can be considered a force.

Explanation:

As the object falls, it moves faster and faster. Gravity is considered a universal force because it acts between any two masses anywhere in the universe. For example, there is a gravitational pull between the Sun and the Moon. Even small masses attract one another.

Hope this helped! :)

Answer:

maybe try to do something with electricity?

Explanation:

give me an update on when u find one :) cant wait to see it

Answer:

5.7 moles of O2

Explanation:

We'll begin by writing the balanced decomposition equation for the reaction. This is illustrated below:

2KClO3 —> 2KCl + 3O2

From the balanced equation above,

2 moles of KClO3 decomposed to produce 3 moles of O2.

Next, we shall determine the number of mole of O2 produced by the reaction of 3.8 moles of KClO3.

Since 100% yield of O2 is obtained, it means that both the actual yield and theoretical yield of O2 are the same. Thus, we can obtain the number of mole of O2 produced as follow:

From the balanced equation above,

2 moles of KClO3 decomposed to produce 3 moles of O2.

Therefore, 3.8 moles of KClO3 will decompose to produce = (3.8 × 3)/2 = 5.7 moles of O2.

Thus, 5.7 moles of O2 were obtained from the reaction.

Answer:

Gas D

Explanation:

liters at 30 °C

Answer:

so here is the answer

Explanation:

Kinetic is the answer.

Answer:

Metal, Copper, and Alluminum

Explanation: It is what it is

Answer:

λ = 5.68×10⁻⁷ m

Explanation:

Given data:

Energy of photon = 3.50 ×10⁻¹⁹ J

Wavelength of photon = ?

Solution:

E = hc/λ

h = planck's constant = 6.63×10⁻³⁴ Js

c = 3×10⁸ m/s

Now we will put the values in formula.

3.50 ×10⁻¹⁹ J =  6.63×10⁻³⁴ Js × 3×10⁸ m/s/ λ

λ =  6.63×10⁻³⁴ Js × 3×10⁸ m/s / 3.50 ×10⁻¹⁹ J

λ =  19.89×10⁻²⁶ J.m / 3.50 ×10⁻¹⁹ J

λ = 5.68×10⁻⁷ m

Answer:

a. the lithosphere

Explanation:

google

Answer:

A

Explanation:

as the temperature and speed is higher pressure will be higher

as the temperature is higher particles will be more active

and as speed is they will collide with more force

so the pressure will be higher in container B

Answer:

[Ne] 3s² 3p¹

Explanation:

Answer:

Explanation:

Chemicals to produce NH₃ ( ammonia )

1 ) Ammonium chloride ( NH₄Cl )

2 ) Calcium hydroxide  Ca(OH)₂

2NH₄Cl + Ca(OH)₂ = CaCl₂ + 2NH₃ + 2H₂O.

Net Ionic equation

2NH₄⁺ ( aq ) + 2OH⁻ (aq )  = 2NH₃(g)  + 2H₂O .

The pascal is sorthand for N/m^2

Answer:

C

Explanation:

CO2 is a molecule because it is compose of atoms which made it a molecule, it can also be call a compound.

Answer:

I believe it is c. Molecule, Pure compound

Explanation:

this is because it is composed of atoms which create molecule.

Answer:

1.1 × 10⁻⁶ M

Explanation:

Step 1: Given and required data

Step 2: Write the balanced equation for the solution reaction of calcium oxalate

CaC₂O₄(s) ⇄ Ca²⁺(aq) + C₂O₄²⁻(aq)

Step 3: Calculate the concentration of the oxalate ion to begin the precipitation of calcium oxalate

We will use the Ksp of calcium oxalate.

Ksp = 2.7 × 10⁻⁹ = [Ca²⁺].[C₂O₄²⁻]

[C₂O₄²⁻] = Ksp / [Ca²⁺]

[C₂O₄²⁻] = 2.7 × 10⁻⁹ / 2.5 × 10⁻³

[C₂O₄²⁻] = 1.1 × 10⁻⁶ M

Kidney stones are the calcareous stones formed in the kidney of the excretory system. The initial concentration of oxalate ion was [tex]1.1 \times 10^{-6} \;\rm M.[/tex]

Concentration is the amount of the substance present in the sample and can be calculated by the solubility constant of the products.

Given,

Concentration of calcium = 2.5mM

The solubility product constant = [tex]2.7 \times 10^{-9}[/tex]

The balanced chemical reaction for calcium oxalate can be written as:

[tex]\rm CaC_{2}O_{4}(s) \leftrightharpoons Ca^{2+}(aq) + C_{2}O_{4}^{2-}(aq)[/tex]

The concentration of oxalate ions by Ksp is calculated as:

[tex]\begin{aligned} [C_{2}O_{4}^{2-}] &= \rm \dfrac {Ksp}{[Ca^{2+}]}\\\\&= \dfrac{2.7 \times 10^{-9}}{2.5 \times 10^{-3}}\\\\&= 1.1 \times 10^{-6}\;\rm M\end{aligned}[/tex]

Therefore, [tex]1.1 \times 10^{-6} \;\rm M[/tex] is the concentration of oxalate ion.

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Answer:

4.2x10⁻⁴ M or 0.032 g/L.

Explanation:

Hello!

In this case, for solubility product problems, we apply the concepts of equilibrium because an insoluble salt is ionized until a certain point limited by the solubility product constant, Ksp. Thus, we first write the ionization reaction of aluminum hydroxide at equilibrium:

[tex]Al(OH)_3(s)\rightleftharpoons Al^{3+}(aq)+3OH^-(aq)[/tex]

Next, we write the corresponding equilibrium expression:

[tex]Ksp=[Al^{3+}][OH^-]^3[/tex]

Which in terms of [tex]x[/tex], the reaction extent, is:

[tex]Ksp=x*(3x)^3[/tex]

Because [tex]x[/tex] also represents the molar solubility of aluminum hydroxide at the considered temperature; now, we can write:

[tex]8.1x10^{-13}=x*(3x)^3[/tex]

Which can be solved for x as follows:

[tex]x=\sqrt[4]{\frac{8.1x10^{-13}}{27} } \\\\x=4.2x10^{-4}M[/tex]

Thus, the solubility is 4.2x10⁻⁴ M or mol/L and in g/L we have:

[tex]4.2x10^{-4}\frac{mol}{L}*\frac{78g}{1mol} =0.032\frac{g}{L}[/tex]

Best regards!

Answer:

I'm not an expert at this, but I assume its mercury.

Answer:

[tex][Cu^{2+}]=0.041 M[/tex]

Explanation:

Hello!

In this case, since the molarity of a solution is defined in terms of the moles of the solute and the volume of solution, given that the concentration of Cu(NH₃)₄²⁺ is 0.041 M, and there is only one copper atom per Cu(NH₃)₄²⁺ ion, we can compute the concentration of Cu²⁺ as shown below:

[tex][Cu^{2+}]=0.041\frac{molCu(NH_3)_4^{2+}}{L}*\frac{1molCu^{2+}}{1molCu(NH_3)_4^{2+}} =0.041 \frac{molCu(NH_3)_4^{2+}}{L}[/tex]

[tex][Cu^{2+}]=0.041 M[/tex]

Best regards!

Answer:

this is not chemistry

Explanation:

not chemistry

Answer: One of the foods was sour and one was sweet

Explanation:

The sweet one has changed because it has more sugar then the sour one, It either got moldy or melted.

Answer:

". It will mark the time since humans began altering Earth. Starting about 10,000 years ago, it is tentatively being called the Anthropocene."

i dont know if tht will help

100,000 years in the geologic history of Earth would be considered

as Anthropocene .

The first and most recent geological epoch in Earth's history when human life began to have a noticeable influence on the planet's climate, including ecosystems, and ecosystems, is referred to as the Anthropocene Epoch. This term refers to an undefined unit of geologic time.

It would be the first time since people started changing the planet. It must be tentatively referred to as the Anthropocene and began around 10,000 years ago.

To know more about Anthropocene

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#SPJ2

Answer:

urorrr

Explanation:

hahhahahahahhaahah oki nam omy mymy mymy

Answer:

-7.6755

Explanation:

Calculate

I calculate it

Answer:

8.279

Explanation:

The pH can be determined by hydrolysis of a conjugate base of weak acid at the equivalence point.

At the equivalence point, we have

[tex]$n_{NaOH}=n_{HNO_2}$[/tex]

           = 25.00 x 0.200

           = 5.00 m-mol

           = 0.005 mol

Volume of the base that is added to reach the equivalence point is

[tex]$\frac{0.005}{1.00} \times 1000= 5.00 \ mL$[/tex]

Number of moles of [tex]$NO^-_{2}=n_{HNO_2}$[/tex]

                                           = 0.005 mol

Volume at the equivalence point is 25 + 5 = 30.00 mL

Therefore, concentration of [tex]$NO^-_{2}= \frac{5}{30}$[/tex]

                                                        = 0.167 M

Now the ICE table :

            [tex]$NO^-_2 + H_2O \rightarrow HNO_3 + OH^-$[/tex]

I (M)       0.167                   0            0

C (M)         -x                      +x          +x

E (M)      0.167-x                  x           x

Now, the value of the base dissociation constant is ,

[tex]$K_w=K_a \times K_b$[/tex]            [tex]$(K_w \text{ is the ionic product of water })$[/tex]

[tex]$K_b =\frac{K_w}{K_a}$[/tex]

[tex]$K_b =\frac{1 \times 10^{-14}}{4.6 \times 10^{-4}}$[/tex]

    = [tex]$2.174 \times 10^{-11}$[/tex]

Base ionization constant, [tex]$K_b = \frac{\left[HNO_2\right] \left[OH^- \right]}{\left[NO^-_2 \right]}$[/tex]

[tex]$2.174 \times 10^{-11}=\frac{x^2}{0.167 -x}$[/tex]

[tex]$x= 1.9054 \times 10^{-6}$[/tex]

So, [tex]$[OH^-]=1.9054 \times 10^{-6 } \ M$[/tex]

pOH =- [tex]$\log[OH^-]$[/tex]

       = [tex]$- \log(1.9054 \times 10^{-6} \ M)$[/tex]

        =5.72

Now, since pH + pOH = 14

           pH = 14.00 - 5.72

                = 8.279

Therefore the ph is 8.279 at the end of the titration.

Titration is a chemical process used to determine the concentration of an unknown sample from the known sample. At the equivalence point, the pH is 8.279.

In titration, the point at which the titrant neutralizes the unknown analyte solution is called the equivalence point.

The moles of equivalence point of sodium hydroxide and nitrous acid are:

[tex]\begin{aligned}\rm n_{NaOH} &= \rm n _{HNO_{2}}\\\\&= 25.00 \times 0.200\\\\&= 0.005 \;\rm mol\end{aligned}[/tex]

The volume of the added base is 5.00 mL.

The moles of nitrogen dioxide are 0.005 moles and the volume at the equivalence point is 30 mL.

The molar concentration of nitrogen dioxide will be, 0.167 M.

From the ICE table, the base dissociation constant is calculated as:

[tex]\begin{aligned}\rm K_{w} &= \rm K_{a} \times K_{b}\\\\\rm K_{b} &= \dfrac{1 \times 10^{-14}}{4.6 \times 10^{-14}}\\\\&= 2.174 \times 10^{-11}\end{aligned}[/tex]

The ionization constant of the base is given as,

[tex]\begin{aligned}\rm K_{b} &= \rm \dfrac{[HNO_{2}][OH^{-}]}{[NO_{2}^{-}]}\\\\2.174 \times 10^{-11} &= \rm \dfrac{x^{2}}{0.167- x}\\\\&= 1.905 \times 10^{-6} \end{aligned}[/tex]

Now, pOH is calculated as:

[tex]\begin{aligned} \rm pOH &=\rm - log [OH^{-}]\\\\&= \rm -log (1.905 \times 10^{-6})\\\\&= 5.72\end{aligned}[/tex]

The pH of the titration is calculated as:

[tex]\begin{aligned} \rm pH + pOH &= 14\\\\\rm pH &= 14.00 - 5.72\\\\&= 8.279\end{aligned}[/tex]

Therefore the pH is 8.27.

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Answer:

B

Explanation:

Answer:

a. 100%

b. 133%

c. 300%

Explanation:

To find yield first we need to determine theoretical yield converting each reactant to moles and find limitng reactant for each reaction:

Moles A:

4.5g * (1mol / 150g) = 0.03 moles

Moles B:

4.0g * (1mol / 100g) = 0.04 moles

a. For a complete reaction of 0.03 moles of A are needed:

0.03 moles A * (1 mole B / 3 moles A) = 0.01 moles of B

As there are 0.04 moles of B, A is limiting reactant.

Theoretical moles and mass of C are:

0.03 moles A * (1 mole C / 3 moles A) = 0.01 moles of C.

0.01 moles of C * (200g / mol) = 2g are produced.

Yield is:

2g / 2g * 100 = 100%

b. For a complete reaction of 0.03 moles of A are needed:

0.03 moles A * (3 mole B / 2 moles A) = 0.045 moles of B

As there are 0.04 moles of B, B is limiting reactant.

Theoretical moles and mass of C are:

0.04 moles B * (1 mole C / 3 moles B) = 0.0133 moles of C.

0.0133 moles of C * (200g / mol) = 2.67g are produced.

Yield is:

2.67g / 2g * 100 = 133%

c. For a complete reaction of 0.03 moles of A are needed:

0.03 moles A * (1 mole B / 1 moles A) = 0.03 moles of B

As there are 0.04 moles of B, A is limiting reactant.

Theoretical moles and mass of C are:

0.03 moles A * (1 mole C / 1 moles A) = 0.03 moles of C.

0.03 moles of C * (200g / mol) = 6g are produced.

Yield is:

6g / 2g * 100 = 300%

Answer:

Propionic acid reacts with NaOH(aq) to form sodium propionate and water.

Explanation:

.

Answer:

Oxygen and simple sugars.

Explanation:

Water, sunlight and carbon dioxide is taken in to photosynthesise and the product would be oxygen that is breathed out through the stomata and simple sugars that are used as fuel for the plant.