Within the parietal cells of the stomach, CO2 and H2O are combined by ___________ to produce H2CO3, which quickly dissociates into H+ and HCO3-. H+ ions are pumped into the lumen of the stomach via a _____________, while HCO3- is pumped into the blood in exchange for Cl-, which ultimately enters into the lumen of the stomach as well.

Answer:

The maximum mass of carbon dioxide that could be produced by the chemical reaction is 5.96 grams

Explanation:

The balanced reaction between gaseous butane and oxygen occurs as follows:

[tex]2 C_{4} H_{10} + 13 O_{2}[/tex]⇒[tex]8 CO_{2} +10 H_{2} O[/tex]

In order for the equation to be balanced, it was taken into account that the law of conservation of matter states that no atom can be created or destroyed in a chemical reaction, so the number of atoms that are present in the reagents has to be equal to the number of atoms present in the products.

Knowing the reaction that occurs between both reagents, it is possible to know the stoichiometry of the reaction (that is, the quantities of reagents necessary for a certain amount of products to be produced). And assuming that 3.49 g of butane are mixed with 7.0 g of oxygen it is possible to determine the limiting reagent, that is to say the reagent that is consumed first, by determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

In order to determine the limiting reagent, you must first determine the reacting mass of each reagent. Then you must first know the molar mass of butane and oxygen, taking into account the atomic mass of each element that composes it and the amount present:

Atomic masses:

Molecular Mass:

Therefore, observing the reaction, 2 moles of butane and 13 moles of oxygen react. With the previously calculated molar masses it is possible to determine the mass that reacts by stoichiometry of each reagent.

Reactive mass of each reagent:

Assuming that 3.49 g of butane react, and taking into account stoichiometry, it is possible to make a rule of three to determine the limiting reagent: if for 116 grams of butane to react, 416 grams of oxygen are needed, how many moles of oxygen are needed to 3.49 grams react of butane?

[tex]grams of oxygen=\frac{3.49 grams of butane* 416 grams of oxygen}{116 grams of butane}[/tex]

grams of oxygen= 12.52

Then the limiting reagent will be oxygen because a smaller amount of reagent (7 grams) is available.  Then the following calculations are made from the available 7 grams of oxygen.

First, the amount of product that is produced by stoichiometry is determined, as was previously done with the reagents:

To determine the maximum mass of carbon dioxide that could cause the chemical reaction, a rule of three is applied taking into account the limiting reagent and stoichiometry: if by stoichiometry 416 grams of oxygen produce 354 grams of carbon dioxide, how many grams of product produce 7 grams of oxygen?

[tex]mass of carbon dioxide=\frac{7 grams of oxygen* 354 grams of carbon dioxide}{416 grams of oxygen}[/tex]

mass of carbon dioxide= 5.96 grams

Finally, the maximum mass of carbon dioxide that could be produced by the chemical reaction is 5.96 grams

Answer:

see explanation below

Explanation:

First, you are not providing any data of the bromide solution to calculate the mass. So, in order to help you, I will take some random values from a similar exercise, so you can solve this later with your data.

Let's suppose you add 360 mL of a 1.45 mol/L of a calcium bromide solution into the flask. To calculate the mass it was added, you need to calculate first the moles added. This can be done with the following expression:

M = n/V

Where:

M: molarity of solution

n: moles of solution

V: volume (in liters) of solution

here, you have to solve for n, so:

n = M*V

replacing the above data you have:

n = 1.45 * (0.360) = 0.522 moles

Now that we have the moles, you can calculate the mass by the following expression:

m = n * MM

Where MM it's the molar mass of calcium bromide. The reported MM of calcium bromide is 199.89 g/mol, so replacing:

m = 199.89 * 0.522

m = 104.34 g

And this is the mass that was added of the solution. As I stated before, use your data in this procedure, and you should get an accurate result.

Answer:

Pressure is inversely proportional to the volume of gas.

Explanation:

According to Boyle's law,

The volume of given amount of gas is inversely proportional to the pressure applied on gas at constant volume and number of moles of gas.

Mathematical expression:

P ∝ 1/ V

P = K/V

PV = K

when volume is changed from V1 to V2 and pressure from P1 to P2 then expression will be.

P1V1 = K         P2V2 = K

P1V1 = P2V2

Final answer:

In gases, pressure and volume are inversely related at constant temperature. Furthermore, pressure is directly proportional to temperature and volume is directly proportional to absolute temperature.

Explanation:

At constant temperature, the volume of a fixed number of moles of gas is inversely proportional to the pressure. This means that when the pressure doubles, the volume will halve.

Moreover, pressure is directly proportional to temperature when volume is constant, and the volume of a gas sample is directly proportional to its absolute temperature at constant pressure.

Answer:

9.82% of iron (II) will be sequestered by cyanide

Explanation:

We should first consider that Iron (II) and cyanide react to form the following structure:

[Fe(CN)₆]⁻⁴

Having considered this:

5.60 Lt Fe(II) 3.00x10⁻⁵ M ,this is, we have 5.60x3x10⁻⁵ =  1.68x10⁻⁴ moles of Fe⁺² (in 5.60 Lt)

Then , we have 9 ml NaCN 11.0 mM:

9 ml = 0.009 Lt

11.0 mM (milimolar) = 0.011 M (mol/lt)

So: 0.009x0.011 = 9.9x10⁻⁵ moles of CN⁻ ingested

As we now that the complex structure is formed by 1 Fe⁺² : 6 CN⁻ :

9.9x10⁻⁵ moles of CN⁻ will use 1.65x10⁻⁵ moles of Fe⁺² (this is, this amount of iron (II) will be sequestered

[(1.65x10⁻⁵ sequestred Fe⁺²)/(1.68x10⁻⁴ total available Fe⁺²)x100

% sequestered iron (II) = 9.82%

Answer: The client is experiencing Allostatic overload.

Explanation:

Allostatic load is known as the wear and tear on the body which accumulates as an individual is exposed to repeated or chronic stress. "Chronic Stress" here, is caused by the difficulty of the client getting a new job coupled with the client caring for their ailing mother.

To determine if the reactions are exothermic or endothermic, we look at the energy required or released when bonds are formed or broken. The formation of NH₃, SO₂, and F₂ is exothermic, whereas the decomposition of H₂O to H₂ and O₂ is endothermic.

To predict whether the following reactions will be exothermic or endothermic, we can consider the energy changes involved in the process of breaking and forming chemical bonds.

A. N₂(g) + 3H₂(g) ------> 2NH₃(g)

This reaction is known to be exothermic. When nitrogen gas reacts with hydrogen gas to form ammonia, energy is released in the process.

B. S(g) + O₂(g) ------> SO₂(g)

The formation of sulfur dioxide from sulfur and oxygen is typically an exothermic reaction because energy is released when the SO₂ molecules are formed.

C. 2H₂O(g) ------> 2H₂(g) + O₂(g)

This reaction involves the decomposition of water vapor into hydrogen and oxygen gas, which requires energy to break the bonds of H₂O molecules. Therefore, it is endothermic.

D. 2F(g) ------> F₂(g)

The formation of fluorine molecules from individual fluorine atoms is an exothermic process. Energy is released when the F₂ molecule is formed due to the formation of a strong bond between the two fluorine atoms.

Answer:

Decay series

Explanation:

A succession of radioactive decay is termed decay series. The radios decay of an unstable nuclei usually continues until a stable isotope is reached. This continuous decay of radioactive isotopes is also known as a radioactive cascade.

It is important to note that most radioisotopes do not decay directly to form a stable nuclei. Instead, they undergo a series of decay until a stable isotope is formed. An example of a decay series can be seen in the decay of uranium-238 to uranium-234.

U-238 is more radioactive than U-234. U-238 first undergoes an alpha particle decay to form thorium 234. This is known as the daughter nuclei. Afterwards, thorium 234 undergoes decay to give protactinium 234. This then undergoes a beta negative decay to form the uranium 234 nuclei.

Answer:

Chemoselectivity

Explanation:

Chemoselectivity is a term that refers to the preferred result of a chemical reaction between two different functional groups.

In this case, the chemical reaction would be the bromination, which is preferred (or chemoselective) towards non-aromatic alkene groups, compared to aromatic alkene groups.

Answer:

Final temperature of calorimeter is 25.36^{0}\textrm{C}

Explanation:

Molar mass of anethole = 148.2 g/mol

So, 0.840 g of anethole = [tex]\frac{0.840}{148.2}moles[/tex] of anethole = 0.00567 moles of anethole

1 mol of anethole releases 5539 kJ of heat upon combustion

So, 0.00567 moles of anethole release [tex](5539\times 0.00567)kJ[/tex] of heat or 31.41 kJ of heat

6.60 kJ of heat increases [tex]1^{0}\textrm{C}[/tex] temperature of calorimeter.

So, 31.41 kJ of heat increases [tex](\frac{1}{6.60}\times 31.41)^{0}\textrm{C}[/tex] or [tex]4.76^{0}\textrm{C}[/tex] temperature of calorimeter

So, the final temperature of calorimeter = [tex](20.6+4.76)^{0}\textrm{C}=25.36^{0}\textrm{C}[/tex]

To determine the final temperature of the calorimeter, we need to calculate the heat released by the combustion of anethole using its enthalpy change. Then, we can use the equation q = mcΔT to find the change in temperature of the calorimeter. The final temperature is 16.65 °C.

To determine the final temperature of the calorimeter, we can make use of the equation q = mcΔT, where q is the heat absorbed or released, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. In this case, we need to consider the heat capacity of the calorimeter as well. We can start by calculating the heat released by the combustion of anethole using the given enthalpy change of -5539 kJ/mol. Then, we can use the equation to find the change in temperature of the calorimeter.

First, we need to calculate the moles of anethole. Using the molar mass of anethole, which is 178.26 g/mol, we can find:

Next, we can calculate the heat released by combustion:

Now, we can consider the calorimeter's heat capacity:

Since the initial temperature of the calorimeter is 20.6 °C, the final temperature will be:

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Answer:

The answer to your question is 1.1 moles of water

Explanation:

                     2Al(OH)₃  +   3H₂SO₄   ⇒   Al₂(SO₄)₃  +   6H₂O

                       0.45 mol      0.55 mol                                ?

Process

1.- Calculate the limiting reactant

Theoretical proportion

       Al(OH)₃ / H₂SO₄ = 2/3 = 0.667

Experimental proportion

       Al(OH)₃ / H₂SO₄ = 0.45 / 0.55 = 0.81

From the proportions, we conclude that the limiting reactant is H₂SO₄

2.- Calculate the moles of H₂O

                        3 moles of H₂SO₄ ----------------  6 moles of water

                        0.55 moles of H₂SO₄ -----------    x

                        x = (0.55 x 6) / 3

                        x = 3.3 / 3

                       x = 1.1 moles of water

Answer:

Maximum amount of [tex]CO_{2}[/tex] can be produced is 37.5 g

Explanation:

Balanced equation: [tex]2C_{4}H_{10}+13O_{2}\rightarrow 8CO_{2}+10H_{2}O[/tex]

Molar mass of butane ([tex]C_{4}H_{10}[/tex])  = 58.12 g/mol

Molar mass of [tex]O_{2}[/tex] = 32 g/mol

Molar mass of [tex]CO_{2}[/tex] = 44.01 g/mol

So, 24 g of butane  = [tex]\frac{58.12}{24}mol[/tex] of butane = 2.422 mol of butane

Also, 44.3 g of [tex]O_{2}[/tex]  = [tex]\frac{44.3}{32}mol[/tex] of [tex]O_{2}[/tex] = 1.384 mol of [tex]O_{2}[/tex]

According to balanced equation-

2 moles of butane produce 8 mol of [tex]CO_{2}[/tex]

So, 2.422 moles of butane produce [tex](\frac{8}{2}\times 2.422)moles[/tex] of [tex]CO_{2}[/tex] = 9.688 moles of [tex]CO_{2}[/tex]

13 moles of [tex]O_{2}[/tex] produce 8 mol of [tex]CO_{2}[/tex]

So, 1.384 moles of [tex]O_{2}[/tex] produce [tex](\frac{8}{13}\times 1.384)moles[/tex] of [tex]CO_{2}[/tex] = 0.8517 moles of [tex]CO_{2}[/tex]

As least number of moles of [tex]CO_{2}[/tex] are produced from [tex]O_{2}[/tex] therefore [tex]O_{2}[/tex] is the limiting reagent.

So, maximum amount of [tex]CO_{2}[/tex] can be produced = 0.8517 moles = [tex](44.01\times 0.8517)g=37.5 g[/tex]

To calculate the maximum mass of carbon dioxide produced, use the balanced equation and determine the limiting reactant. Then, calculate the moles of carbon dioxide produced using the mole ratio from the balanced equation and convert it to grams using the molar mass of carbon dioxide.

To calculate the maximum mass of carbon dioxide that could be produced by the reaction between gaseous butane (C4H10) and gaseous oxygen (O2), we need to use the balanced equation for the reaction:

C4H10 + 13/2 O2 → 4 CO2 + 5 H2O

First, we need to determine the limiting reactant by comparing the moles of butane and oxygen. The molar mass of butane is 58.1 g/mol, so 24 g of butane is equal to 24/58.1 moles. The molar mass of oxygen is 32 g/mol, so 44.3 g of oxygen is equal to 44.3/32 moles.

Next, we calculate the moles of carbon dioxide produced using the mole ratio from the balanced equation. Since the mole ratio between butane and carbon dioxide is 1:4, the moles of carbon dioxide produced is 4 times the moles of butane. Finally, we convert the moles of carbon dioxide to grams by multiplying by the molar mass of carbon dioxide (44 g/mol).

Using this information, we can calculate the maximum mass of carbon dioxide that could be produced. The answer should be rounded to 3 significant digits to match the rounding specified in the question.

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Answer: Option (c) is the correct answer.

Explanation:

It is known that relation between charge, current and time is as follows.

   Total charge passed = current (A) x time (s)

                      = [tex]3.86 \times 16.2 \times 60[/tex]

                    = 3751.92 C

Moles of electrons passed = [tex]\frac{\text{total charge}}{F} [/tex]

                                             = [tex]\frac{3751.92}{96485}[/tex]

                                            = 0.03889 mol

As, the given metal salt is MCl. Therefore, the reduction reaction is as follows.

         [tex]M^{+} + e^{-} \rightarrow M [/tex]

Thus, moles of M = moles of electrons = 0.03889 mol

As we known that molar mass is calculated using the formula:

     Molar mass of M = [tex]\frac{mass}{moles}[/tex]

                                  = [tex]\frac{1.52}{0.03889}[/tex]

                                 = 39.1 g/mol

We know that potassium is the metal which has molar mass as 39 g/mol.

Thus, we can conclude that the metal is identified as K (potassium).

The current passed for cell in the unit time gives the charge passes to the cell. The metal deposited in the electrolysis is Potassium.

The electrolysis is given as the breaking of the salt for the formation of the ions under the influence of the electric current.

The charge transferred ([tex]Q[/tex]) to the cell in the given time is calculated as:

[tex]Q=\rm current\;\times\;time\\\\\textit Q=3.86\;amp\;\times\;16.2\;\times\;60\;sec\\\\\textit Q=3751.92\;C[/tex]

The moles of sample is given as:

[tex]\rm Moles=\dfrac{Charge}{Faraday} \\\\Moles=\dfrac{3751.92}{96,485} \\\\Moles=0.03889\;mol[/tex]

The mass of sample deposited is 1.5 grams, the molar mass of the sample is calculated as:

[tex]\rm Molar\;mass=\dfrac{mass}{moles} \\\\Molar\;mass=\dfrac{1.52}{0.03889}\\\\ Molar\;mass=39.1\;g/mol[/tex]

The molar mass of the compound is 39.1 g/mol. The element with molar mass 39.1 g/mol is Potassium.

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Answer:

I would interpret the statement by using a formula.

Explanation:

In order to be scientifical in a research, the scientific method must be used, this means that in this case, Lavoisier should´ve follow a Hypothesis, Objectives, a Methodlogy, Results and made a Discussion and Conclusion.

To prove that the mass gained by the pipe plus the mass of the collected gas "exactly" equaled the lost mass of water, Lavoisier should´ve used this statement as a formula where it proves with numbers that is correct.

For example:

mg= mass gained by the pipe

mc=mass of the collected gas

ml=lost mass of water

mg+mc=ml

Answer:

a. K₂CO₃(aq) + Pb(NO₃)₂(aq) → 2KNO₃(aq) + PbCO₃(s)

b. Li₂SO₄(aq) + Pb(C₂H₃O₂)₂(aq) → 2Li(C₂H₃O₂) + PbSO₄(s)

c. Cu(NO₃)₂(aq) + MgS(aq) → Mg(NO₃)₂(aq) + CuS(s)

d. NO REACTION

Explanation:

For the reactions, the cation and the anion of the compounds will be replaced. The reaction will occur if at least one of the products is insoluble and will form a precipitated.

a. Potassium carbonate = K₂CO₃

Lead(II) nitrate = Pb(NO₃)₂

Products = KNO₃ and PbCO₃.

According to the solubility rules, all K⁺ ions are soluble, with no exceptions, so KNO₃ is soluble. All CO₃⁻² ions are insoluble, and Pb⁺² is not an exception, so PbCO₃ will be insoluble and will form a precipitated, so the reaction happen:

K₂CO₃(aq) + Pb(NO₃)₂(aq) → 2KNO₃(aq) + PbCO₃(s)

b. Lithium sulfate = Li₂SO₄

Lead(II) acetate = Pb(C₂H₃O₂)₂

Products = Li(C₂H₃O₂) and PbSO₄

All Li⁺ are solubles, without exceptions, so Li(C₂H₃O₂) is soluble, and all SO₄⁻² are soluble, but Pb⁺² is an exception, so PbSO₄ is insoluble and will form a precipitated, then the reaction happens:

Li₂SO₄(aq) + Pb(C₂H₃O₂)₂(aq) → 2Li(C₂H₃O₂) + PbSO₄(s)

c. Copper(II) nitrate = Cu(NO₃)₂

Magnesium sulfide = MgS

Products = CuS and Mg(NO₃)₂

All NO₃⁻ are soluble, with no exceptions, so Mg(NO₃)₂ is soluble, and all S⁺² are insoluble, and Cu⁺² is not an exception, so CuS is insoluble, and will form a precipitated, then the reaction happens:

Cu(NO₃)₂(aq) + MgS(aq) → Mg(NO₃)₂(aq) + CuS(s)

d. Strontium nitrate = Sr(NO₃)₂

Potassium iodi = KI

Products = K(NO₃)₂ and SrI₂

All K⁺ are soluble, with no exceptions, so K(NO₃)₂ is soluble, and all I⁻ are soluble, and Sr⁺² are not an exception, then SrI₂ is soluble. Therefore, no precipitated is formed and the reaction doesn't happen.

Precipitation reactions occur for some pairs of aqueous solutions, while others do not.

a. The combination of potassium carbonate and lead(II) nitrate will result in the formation of a precipitate. The balanced molecular equation for this reaction is:

Pb(NO3)2 + K2CO3 → PbCO3 + 2 KNO3

b. The combination of lithium sulfate and lead(II) acetate will not result in a precipitation reaction.

c. The combination of copper(II) nitrate and magnesium sulfide will result in the formation of a precipitate. The balanced molecular equation for this reaction is:

Cu(NO3)2 + MgS → CuS + Mg(NO3)2

d. The combination of strontium nitrate and potassium iodide will not result in a precipitation reaction.

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Answer : The correct option is, (d) [tex]5.0\times 10^{-12}mole[/tex]

Explanation :

First we have to calculate the [tex]H^+[/tex] concentration.

[tex]pH=-\log [H^+][/tex]

[tex]11.5=-\log [H^+][/tex]

[tex][H^+]=3.16\times 10^{-12}M[/tex]

Now we have to calculate the [tex]OH^-[/tex] concentration.

[tex][H^+][OH^-]=K_w[/tex]

[tex]3.16\times 10^{-12}\times [OH^-]=1.0\times 10^{-14}[/tex]

[tex][OH^-]=3.16\times 10^{-3}M[/tex]

Now we have to calculate the molar solubility of [tex]Zn(OH)_2[/tex].

The balanced equilibrium reaction will be:

[tex]Zn(OH)_2\rightleftharpoons Zn^{2+}+2OH^-[/tex]

The expression for solubility constant for this reaction will be,

[tex]K_{sp}=[Zn^{2+}][OH^-]^2[/tex]

Now put all the given values in this expression, we get:

[tex]5.0\times 10^{-17}=[Zn^{2+}]\times (3.16\times 10^{-3})^2[/tex]

[tex][Zn^{2+}]=5.0\times 10^{-12}M[/tex]

Therefore, the molar solubility of [tex]Zn(OH)_2[/tex] is, [tex]5.0\times 10^{-12}mole[/tex]

Answer:

[tex]m_{Fe}=71.982 kg[/tex]

Explanation:

First of all, we need  to calculate the moles of H2. Assuming the H2 is an ideal gas:

[tex]n=\frac{P*V}{T*R}[/tex]

[tex]n=\frac{1atm*31100L}{294K*0.082 L*atm*mol^{-1}*K^{-1}}[/tex]

[tex]n=1290 mol[/tex]

Now, to produce 1 mol of H2 is required 1 mol of Fe:

[tex]m_{Fe}=n*M=1290mol*\{55.8 g}{mol}[/tex]

[tex]m_{Fe}=71982 g=71.982 kg[/tex]

The study of chemicals and bonds is called chemistry. There are two types of elements are there and these are metals and nonmetals.

The correct answer is 71.982kg.

The formula used as:-

[tex]n=\frac{P*V}{T*R} \\\\n=\frac{1*31100}{294*0.082} \\\\\\n=1290moles[/tex]

The mass will be:-

[tex]M= n*M\\1290*55.8\\\\=71.982[/tex]

Hence, the correct answer is 71.982.

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Answer:

The electrons stripped from glucose in cellular respiration end up in compound water by the reduction of oxygen.

Explanation:

During electron transport chain electrons are donated by various reducing equivalents such as NADH,FADH2.The donated electrons then moves through various electron carriers .

       During electron transport chain oxygen(O2) act as terminal electron acceptor which accept the electron from complex 4 and thereby get reduced to form water.(H2O).

In cellular respiration, electrons stripped from glucose are eventually incorporated into water. They are carried through the electron transport chain via NADH and FADH2 and combine with oxygen to form water in the process of oxidative phosphorylation.

In the process of cellular respiration, the electrons that are stripped from glucose ultimately end up in water. The process starts with glucose undergoing glycolysis and the Krebs cycle, forming NADH and FADH2 compounds. These compounds then donate their electrons to the electron transport chain in a series of redox reactions.

During this process, free oxygen acts as the final electron acceptor in the chain. The electrons combine with hydrogen ions and the accepted oxygen to form water. This is a part of the process called oxidative phosphorylation. The other product of this process is ATP, which is the main energy currency in cells.

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Answer:

a) 1 litre of  10% solution and 7 litre of 20% solution

b) 1.67 litres of 50% solution and 8.33 litres of the 20% solution

c) 3.75 litres of 50% solution and 6.25 litres of 20% solution

Explanation:

Given:

chemist needs = 10 liters of a 25% acid solution

Concentration of three solutions that are to be mixed = 10%, 20% and 50%.

Solution:

A) Use 2 liters of the 50% solution

Let us mix this with 10% and 20% solution

They will have to equal 8 litres

Let x=20% solution

Then (8-x) =10%

So the equation becomes,

10%(8-x)+ 20%x+50%(2)=25(10)

(0.1)(8-x) +0.2x+0.50(2)= 0.25(10)

0.8-0.1x+0.2x+1.0=2.5

0.2x-0.1x=2.5-0.8-1.0

0.1x=0.7

[tex]x=\frac{0.7}{0.1}[/tex]

x= 7

so, 8-x = 8 -7= 1 litre of  10% solution and 7 litre of 20% solution

B)Use as little as possible of the 50% solution

Let x be the amount of 50% solution.

Then(10-x) be the 20% solution

Now the equation becomes,

50%(x)+20%(10-x)=25%(10)

0.50x+0.2(10-x)=0.25(10)

0.5x+2.0-0.2x=2.5

0.3x=2.5-2.0

0.3x=0.5

[tex]x=\frac{0.5}{0.3}[/tex]

x=1.67  

now (10-x)=(10-1.67)=8.33

so there will be 1.67 litres of 50% solution and 8.33 litres of the 20% solution

c) ) Use as much as possible of the 50% solution

Let x be the amount of 50% solution.

Then(10-x) be the 20% solution

Now the equation becomes,

50%(x)+10%(10-x)=25%(10)

0.50x+0.1(10-x)=0.25(10)

0.5x+1.0-0.1x=2.5

0.4x=2.5-1.0

0.4x=1.5

[tex]x=\frac{1.5}{0.3}[/tex]

x=3.75

Now, (10-x)=(10- 3.75)=6.26

So there will be 3.75 litres of 50% solution and 6.25 litres of 20% solution

The amount of liters of each solution to satisfy each given condition are;

A) 8.33 liters of 20% solution

B) 6.25 liters of 10% solution

C) 1 liter of the 10% solution

A) Use as little as possible the 50% solution.

Mix it with 20% solution only

Let x be the amount of 50% solution

Thus;

(10 - x) = 20% solution

equation:

0.50x + 0.20(10 - x) = 0.25(10)

0.5x + 2 - 0.2x = 2.5

0.3x = 2.5 - 2

0.3x  = 0.5

x = 0.5/0.3

x = 1.67 liters of 50% solution required

Thus; 10 - 1.67 = 8.33 liters of 20% solution

B) Use as little as possible of the 50% solution;

Mix it with the 10% solution only

Let x be amount of 50% solution

Thus;

(10 - x) = 10% solution

equation:

0.50x + 0.10(10 - x) = 0.25(10)

0.5x + 1 - 0.1x = 2.5

0.4x = 2.5 - 1

x = 1.5/0.4

x = 3.75 liters of 50% solution required. Thus;

10 - 3.75 = 6.25 liters of 10% solution

C) Use 2 liters of the 50% solution

Mix it with 10% and the 20% and they will have to equal 8 liters.

Let x be the amount of 20% solution

Thus;

8 - x = 10% solution

Equation:

0.20x + 0.10(8 - x) + 0.50(2) = 0.25(10)

0.20x + 0.8 - 0.10x + 1 = 2.5

0.2x - 0.1x + 1.8 = 2.5

0.1x = 2.5 - 1.8

0.1x = 0.7

x = 0.7/0..1

x = 7 liters of 20% solution

Thus; 8 - 7 = 1 liter of the 10% solution

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To find the total number of hours the battery can be used over its lifetime, we need to determine the number of times the battery can be recharged and calculate the cumulative usage hours. The battery loses 2% of its charge after each charging, retaining 98% of the previous charge. Using the formula for the sum of a geometric series, the total number of hours the battery can be used is calculated to be 1000 hours.

To find the total number of hours the battery can be used over its lifetime, we need to determine the number of times the battery can be recharged and calculate the cumulative usage hours. Since the battery loses 2% of its charge after each charging, it retains 98% of the previous charge. We can use this information to create a geometric sequence

First term (a) = 20 hours

Common ratio (r) = 98% or 0.98

Number of terms (n) = number of times the battery can be recharged

Using the formula for the sum of a geometric series, we can calculate the total number of hours the battery can be used:

Sum = a(1 - r^n) / (1 - r)

Substituting the given values:

Sum = 20(1 - 0.98^n) / (1 - 0.98)

This means that the battery can be used for a total of 1000 hours over its lifetime.

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Answer:

0.272g

Explanation:

To calculate the mass of oxygen collected, we can calculate the number of moles of oxygen collected and multiply this by the molar mass of oxygen.

To calculate the number of moles of oxygen collected, we can use the ideal gas equation I.e PV = nRT

Rearranging the equation, n =PV/RT

We now identify each of the terms below before substituting and calculating.

n = number of moles, which we are calculating.

R = molar gas constant = 62.64 L.Torr. K^-1. mol^-1

V = volume = 500ml : 1000ml = 1L, hence , 500ml = 500/1000 = 0.5L

T = temperature = 25 degrees Celsius = 273 + 25 = 298K

P = pressure. But since the gas was collected over water, we subtract the vapour pressure of water from the total pressure = 642- 23.8= 618.2torr

We substitute these values into the equation to yield the following:

n = (618.2 × 0.5) ÷ ( 62.64 × 298)

n = apprx 0.017moles

To calculate the mass of oxygen collected, we need the atomic mass of oxygen. = 16 amu

Thus the mass of oxygen collected = 0.017mole × 16g= 0.272g

To find the mass of oxygen collected over water, we can use the ideal gas law equation and the given values of total pressure, volume, temperature, and vapor pressure of water. By rearranging the equation and calculating the number of moles, we can then determine the mass of oxygen using its molar mass.

To find the mass of oxygen collected, we need to use the ideal gas law equation, PV = nRT, where P is the total pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

First, we need to convert the temperature from Celsius to Kelvin by adding 273.15: 25°C + 273.15 = 298.15K.

To calculate the number of moles, we need to subtract the vapor pressure of water (23.8 torr) from the total pressure (642 torr) to obtain the partial pressure of oxygen: 642 torr - 23.8 torr = 618.2 torr.

Now we can rearrange the ideal gas law equation to solve for n:

n = (PV) / (RT) = (618.2 torr * 0.5 L) / (0.0821 L·atm/mol·K * 298.15K)

Using this value of n, we can calculate the mass of oxygen using the molar mass of oxygen (32 g/mol):

mass = n * molar mass = 0.0126 mol * 32 g/mol = 0.4032 g.

Answer:

[tex]Molarity=1.22\ M[/tex]

Explanation:

Given:  

Pressure = 745 mm Hg

Also, P (mm Hg) = P (atm) / 760

Pressure = 745 / 760 = 0.9803 atm

Temperature = 19 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (19 + 273.15) K = 292.15 K  

Volume = 0.200 L

Using ideal gas equation as:

[tex]PV=nRT[/tex]

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.9803 atm × 0.200 L = n × 0.0821 L.atm/K.mol × 292.15 K  

⇒n = 0.008174 moles

From the reaction shown below:-

[tex]H_2SO_3+2NaOH\rightarrow Na_2SO_3+2H_2O[/tex]

1 mole of [tex]H_2SO_4[/tex] react with 2 moles of [tex]NaOH[/tex]

0.008174 mole of [tex]H_2SO_4[/tex] react with 2*0.008174 moles of [tex]NaOH[/tex]

Moles of [tex]NaOH[/tex] = 0.016348 moles

Volume = 13.4 mL = 0.0134 L ( 1 mL = 0.001 L)

So,

[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]

[tex]Molarity=\frac{0.016348}{0.0134}\ M[/tex]

[tex]Molarity=1.22\ M[/tex]

Answer: Glucose is an example of carbon-based macromolecule known as carbohydrates

Explanation:

carbon based macromolecule are important cellular components and they perform a variety of functions necessary for growth and development of living organisms. There are 4 major types of carbon based molecules and these includes;

Carbohydrate

Lipids

Proteins and

Nucleic acids.

Carbon is the primary components of these macromolecules. Carbohydrate macromolecules are made up of monosaccharide, disaccharide and polysaccharides. Glucose is an example of a monosaccharide and it has two important types of functional groups: a carbonyl group and a hydroxyl group. I hope this helps. Thanks

Glucose is a simple sugar and is an example of a carbohydrate, which is a type of carbon-based macromolecule. Glucose is a monosaccharide that can join together with other sugars to form complex carbohydrates or polysaccharides. It serves as a vital energy source for cells.

Glucose is a type of carbon-based macromolecule known as a carbohydrate. Carbohydrates are organic molecules composed of carbon, hydrogen, and oxygen atoms and they serve as a vital energy source for cells. The family of carbohydrates includes both simple and complex sugars. Glucose and fructose are examples of simple sugars, whereas starch, glycogen, and cellulose are examples of complex sugars, also called polysaccharides.

Monosaccharides like glucose are the simplest form of carbohydrates. These small molecules can combine to form larger, more complex structures. When many monosaccharides join together, they form polysaccharides such as starch and glycogen, which serve as energy storage in plants and animals respectively.

While glucose serves as the most common fuel for ATP production within cells, excess glucose is either stored for future energy needs in the liver and skeletal muscles as glycogen, or it's converted into fat in adipose cells.

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Answer:

The enthalpy of the reaction is -2855.622 kilo Joules.

Explanation:

[tex]2C_2H_6(g) + 7O_2(g)\rightarrow 4CO_2(g) + 6H_2O(g)[/tex]

We are given:

[tex]\Delta H^o_f_{(C_{2}H_6(g))}= -84.667 kJ/mol[/tex]

[tex]\Delta H^o_f_{O_2((g))}= 0 kJ/mol[/tex]

[tex]\Delta H^o_f_{CO_2((g))}= -393.5 kJ/mol[/tex]

[tex]\Delta H^o_f_{H_2O((g))}= -241.826 kJ/mol[/tex]

The equation used to calculate enthalpy of reaction :

[tex]\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)][/tex]

The equation for the enthalpy change of the above reaction is:

[tex]\Delta H^o_{rxn}=(4 mol\times \Delta H^o_f_{(CO_2(g))}+6 mol\times \Delta H^o_f_{(H_2O(g)))}-(2 mol\times \Delta G^o_f_{(C_{2}H_6(g))}+7 mol\times \Delta H^o_f_{(O_2(g)))[/tex]

Putting values in above equation, we get:

[tex]\Delta H^o_{rxn}=[4 mol\times (-393.5 kJ/mol)+6 mol\times (-241.826 kJ/mol)]-[2 mol\times (-84.667 kJ/mol)+7 mol\times 0 kJ/mol][/tex]

[tex]=-2855.622 kJ[/tex]

The enthalpy of the reaction is -2855.622 kilo Joules.

The enthalpy change for the reaction is   -2855.622  kJ/mol

Recall that enthalpy(ΔH) is a state function so;

ΔHreaction = ∑ΔHproducts - ΔHreactants

So;

The equation of the reaction is; (Recall that the question specified that we should use the smallest whole number coefficients)

2C2H6(g) + 7 O2(g) -----> 4CO2(g) + 6H2O(g)

The enthalpy of each of the reactants and products are given below;

[C2H6(g)] = −84.667 kJ/mol

O2 g = 0 KJ/mol ( O2 exists in its standard state)

[CO2(g)]  =  −393.5 kJ/mol

[H2O(g)] = −241.826 kJ/mol

Hence;

ΔHreaction = ∑[4 × (−393.5) + 6 × (−241.826)] - [2 × (−84.667) + (7 × 0)]

ΔHreaction = -2855.622  kJ/mol

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Answer:

0.1 moles

Explanation:

Okay, the explanation here is pretty long. Let’s go!

Firstly, we know that the compound contains only carbon, hydrogen, oxygen and nitrogen. So the molecular formula would look like this CxHyOzNp

Where x , y, z and p are number of atoms of each element present respectively. According to the law of constant composition, the ratio of the number of atoms are fixed irrespective of the source or method of preparation. From this, we now know that in both samples, we have the same number of atoms. What is proper to do is to calculate the numbers, we do that as follows.

Firstly, we will need to calculate these numbers using the masses given in the first sample. Over the calculations, we should note that the formula we would be using is the relation: mass = number of moles * atomic mass or molar mass. Rearranging the equation gives different variations of the formula.

Now let’s do some mathematics.

There is 3.94g of carbon iv oxide, we can calculate the number of moles of it present which eventually would yield the number of moles of carbon present.

The molar mass of carbon iv oxide is 44g/mol.

The number of moles of carbon iv oxide present is thus 3.94/44 = 0.0895 moles

Since there is just 1 atom of carbon present in carbon iv oxide, this means the number of moles of carbon present is also 0.0895 moles

The mass of carbon present is the number of moles of it present multiplied by the atomic mass unit of carbon which is 12. This mass is 0.0895 * 12 = 1.0745g

Next, we calculate the number of moles of hydrogen and consequently its mass present.

To get this, we can access it from the number of moles of water present.

We get this by dividing the mass of water present by the molar mass of water. This is equals 1.89/18 = 0.105 moles. Now we know that 1 mole of water has 2 atoms of hydrogen, hence 1 mole of water will yield 2 moles of hydrogen. The number of moles of hydrogen present is thus 0.105 * 2 = 0.21 moles.

The mass of hydrogen thus present is 0.21 * 1 = 0.21g

Now, we know the mass of hydrogen present and the mass of carbon present. But, we do not know the mass of oxygen and nitrogen present.

To get this, we subtract the mass of hydrogen and carbon present from the mass of the total= 2.18 - 1.0745 - 0.21 = 0.8955g

Now we know the mass of oxygen and nitrogen combined. We can access their number of moles using their respective atomic masses. The total number of moles present is equal 0.8955/30 = 0.02985 moles

Wondering where 30 came from? The atomic mass of nitrogen and oxygen are 14 and 16 respectively. We now get the number of moles of both present. This is equal to the atomic mass divided by the total mass multiplied by the number of moles.

For oxygen = 16/30 * 0.02985 = 0.01592

For nitrogen = 14/30 * 0.02985 = 0.01393

From this we can try and get an empirical formula for the compound. This helps us to know the ratios of the number of atoms. To get this , we divide the number of moles by the smallest number of moles. The smallest number of moles is unarguably that of nitrogen. The empirical formula is calculated as follows:

C = 0.0895/0.01393 = 6

H = 0.21/0.01393 = 15

O = 0.01592/0.01393 = 1

N = 0.01393/0.01393 = 1

Thus, the empirical formula looks like this :

C6H15NO

Now, we can move to the second sample.

We know that the sample contains 0.235g of nitrogen. We first need to get the number of moles of nitrogen present in the sample. To get this, we simply divide this mass by the atomic mass. That is: 0.235/14 = 0.0168 moles

Now since the question asks us to get the number of moles of carbon and we know that in any elemental analysis, the ratio of carbon to nitrogen is 6 to 1, we simply multiply the number of moles of nitrogen by 6.

Hence, this is 0.0168 * 6 = 0.1 moles

Final answer:

The moles of carbon in the sample can be calculated using the mass of CO2 produced during combustion. The calculated moles of carbon are 0.0895 mol, based on the provided mass of CO2 and its molar mass.

Explanation:

The question asks how to calculate the moles of carbon in a sample based on combustion analysis data. To find the moles of carbon, we use the mass of CO2 produced during combustion. Since each mole of CO2 contains one mole of carbon, the moles of CO2 will equal the moles of carbon. The mass of CO2 produced is 3.94 g. The molar mass of CO2 is 44.01 g/mol.

Using the formula moles = mass / molar mass, the moles of carbon can be calculated as:

Moles of C = 3.94 g CO2 / 44.01 g/mol = 0.0895 mol C

This calculation determines the amount of carbon present in the sample by analyzing the carbon dioxide produced during combustion.

Answer:

16

Explanation:

The number of orbitals can be calculated when energy level is given.

For example:

n = 4

So in 4th energy level number of orbitals are,

n² = 4² = 16

There are 16 orbitals when n=4

One is 4s three are 4P five is 4d and seven are 4f.

while the number of electrons in energy level can be calculated as

2n²

n is energy level.

For n=4

number of electrons are,

2(4)² = 32

Answer:

15.24 J/K

Explanation:

First, let's write the expression to calculate the change in entropy:

ΔS = n*C*ln(T2/T1)

C is the heat capacity. It will be Cp if the heating is isobaric, and Cv if it is isochoric. (In this case is Cv)

Now, in order to do this, we need to calculate first the pressure of the gas at first, and then, after the chance:

P = nRT/V

P1 = 4.67 *¨0.082 * 344 / 8.9 = 14.801 atm

P2 = 4.67 * 0.082 * 294 / 7.6 = 14.814 atm

Now, let's calculate the change in entropy:

ΔS = 4.67 * (12.47 + 8.314) ln(344/294)

ΔS  = 15.24 J/K

Answer:

Number of protons and electrons stay constant

Number of neutrons Differs

Explanation:

Isotopes are the different kinds of same element. Now, as we know quite well that an element can only have one atomic number, this means that the proton number is irrespective of the type of atom

Of the element. The proton number is the identity of the element.

As we know that the atom is electrically neutral, it means the number of electrons will always stay the same too.

Since isotopes are not alike in every respect, the number of neutrons differ. This means they have same atomic numbers but different mass numbers.

Answer:

Number of protons and electrons stay constant

Explanation:

Answer:

0.744 M

Explanation:

IO⁻⁴(aq) + 2H₂O(l) ⇌ H₄IO⁻⁶(aq)

Kc = 3.5×10⁻²= [H₄IO⁻⁶] / [IO⁻⁴]

First let's calculate the new concentration of IO⁻⁴ at equilibrium:

0.904 M * 26.0 mL / 500.0 mL = 0.047 M = [IO⁻⁴]

Now we can calculate [H₄IO⁻⁶] using the formula for Kc:

3.5×10⁻²= [H₄IO⁻⁶] / [IO⁻⁴]

3.5×10⁻²= [H₄IO⁻⁶] / 0.047 M

[H₄IO⁻⁶] = 0.744 M

Final answer:

The question asks for the equilibrium concentration of H4IO-6 after dilution and reaction has reached equilibrium, implying the use of equilibrium concepts and calculations surrounding concentration and reaction constants. However, without enough detail or context provided on changes in concentration or how the equilibrium constant is applied, an exact answer cannot be directly calculated from the given information.

Explanation:

The question involves calculating the equilibrium concentration of H4IO-6 after diluting NaIO4 and allowing the reaction IO-4(aq) + 2H2O(l) ⇌ H4IO-6(aq); Kc=3.5×10-2 to reach equilibrium. Firstly, calculate the initial concentration of IO-4 after dilution: C1V1 = C2V2, where C1 = 0.904 M and V1 = 26.0 mL, V2 = 500.0 mL. Solving gives C2, the concentration of NaIO4 after dilution. To find the concentration of H4IO-6 at equilibrium, you would typically use the equilibrium constant (Kc), but the question's information does not provide a direct route to calculate this without additional context regarding the relationship between concentrations of reactants and products at equilibrium. Normally, you would set up an ICE table and solve for the equilibrium concentrations using Kc, but without the concentration change (ΔC), this calculation cannot be directly completed.