You want to quickly set up a temporary water bath in your lab with a volume of 10.0 L and a temperature of 37.0°C. You only have hot water from your hot water faucet (temperature = 61.0°C) and cold water from your cold water faucet (temperature = 22.0°C). What volume of hot water (in liters) must you mix with cold water to get 10.0 L of 37.0°C water? Assume the specific heat of the water is 4.184 J/g・K and that the water has a density of 1.00 g/mL.

Answers

Answer 1

Answer:

Volume of hot water required = 3.85L

Explanation:

Suppose volume of hot Then volume of water required cold water = = x L (10.0-x) L

Heat given by hot water (Q₁)

= mass of hot water x heat capacity of water X AT

= x L * 4.184 * J / g. к x(61.0-37.0) °℃.

And Heat absorbed by cold water (Q₂) = (10.0-x) L x 4.184 J/g*k x(37+0 -220) C

Since energy is consumed, Q₁ = Q2.

i.e. X*l *4.184*J/g*k*24C = (10.0-x)L x 184 5

24 x 15 (10.0-x) = 150. - 15x

x = 150. (24+15) = 3.846

So, volume of hot water required. = 3.85 L

Answer 2

When the temperature of the water increases the water becomes hot.

According to the question the volume of hot water required = 3.85L.

Suppose volume of hot Then the volume of water required cold water  is [tex]x L (10.0-x) L[/tex]

All the data are given in the question, which is as follows:-

Heat has given by hot water (Q₁)

The formula we are going to use is as follows:-

= mass of hot water x heat capacity of water X AT

= [tex]x L * 4.184 *(61.0-37.0) ^oC[/tex]

The heat absorbed by cold water (Q₂) = [tex](10.0-x) L *4.184 *(37+0 -220) ^oC[/tex]

Since energy is consumed, Q₁ = Q2.

[tex]X*l *4.18424C = (10.0-x)L * 184 524 * 15 (10.0-x) = 150. - 15xx = 150. (24+15) = 3.846[/tex]

Hence, the volume of hot water required is = 3.85 L

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Related Questions

What are the uses of Sulphuric acid?

Answers

Answer:

The major use of sulfuric acid is in the production of fertilizers, e.g., superphosphate of lime and ammonium sulfate. It is widely used in the manufacture of chemicals, e.g., in making hydrochloric acid, nitric acid, sulfate salts, synthetic detergents, dyes and pigments, explosives, and drugs.

The major use of sulfuric acid is in the production of fertilizers, e.g., superphosphate of lime and ammonium sulfate. It is widely used in the manufacture of chemicals, e.g., in making hydrochloric acid, nitric acid, sulfate salts, synthetic detergents, dyes and pigments, explosives, and drugs.


A scientific hypothesis is
ANSWER:
predictive.
testable.
explanatory.
all of the above.

Answers

Answer:

All of the above.

Explanation:

For a scientific hypothesis to be considered a hypothesis, it has to be testable. When conducting a lab experiment, it also allows the tester to predict what might occur during and after the experimentation. They are also explanatory. For example, theories are hypotheses that have been verified and can explain why something in nature takes place.

what is the machine used to check melting point called?​

Answers

Answer:

Melting-point apparatus

A sample of oxygen gas is compressed from 30.6 L to 1.8 L at constant temperature pressure of 1.8 atm. Calculate the amount of energy in joules when the system releases 1.5 KJ of heat?

Answers

Answer:

the change in the internal energy of the system is 3,752.67 J

Explanation:

Given;

initial volume of the gas, V₁ = 30.6 L

final volume of the gas, V₂ = 1.8 L

constant pressure of the gas, P = 1.8 atm

Energy released by the system, Q = 1.5 kJ = 1,500 J

Apply pressure-volume work equation, to determine the work done on the gas;

w = -PΔV

w = -P(V₂ - V₁)

w = - 1.8 atm(1.8 L - 30.6 L)

w = 51.84 L.atm

w = 51.84 L.atm x 101.325 J/L.atm

w = 5,252.67 J

The change in the internal energy of the system is calculated as;

ΔU = Q + w

Since the heat is given out, Q = - 1,500 J

ΔU = -1,500 J  +  5,252.67 J

ΔU = 3,752.67 J

Therefore, the change in the internal energy of the system is 3,752.67 J

Please please help help please

Answers

Acute toxin or D) 100% correct

1.rain pours from the sky
2.leaves of the plant dried
3.fluffy clouds form in the sky
4.bathing suit dries after swim
5.water puddles disappear

A.Evaporation
B.Condensation
C.Precipitation
D.Transpiration
Yan po pag pipilian

Answers

Answer:

1.Precipitation

2.Transpiration

3.Condensation

4.Evaporation

5.Evaporation

3.Condensation

Explanation:

Rain pours from the sky occurs due to the process of precipitation, leaves of the plant dried due to the process of transpiration in which the water is evaporated from the body of plant, fluffy clouds form in the sky occurs in the process of condensation, bathing suit dries after swim is due to evaporation in which water is removed and goes into the atmosphere and water puddles disappear due to the process of evaporation. Evaporation is the removal of water from the any surface whereas transpiration is the removal of water from plant body parts.

Linoleic acid is a polyunsaturated fatty acid found, in ester form, in many fats and oils. Its doubly allylic hydrogens are particularly susceptible to abstraction by radicals, a process that can lead to the oxidative degradation of the fat or oil.

a. True
b. Flase

Answers

Answer:

True.

Explanation:

The information presented in the question above regarding linoleic acid is true. Linoleic acid is, in fact, found in many oils and fats in the ester form. In addition, linoleic acid is considered a polyunsaturated fatty acid, due to the presence of two unsaturations in its composition. Its chemical formula is CH3-(CH2)4-CH=CH-CH2-CH=CH-(CH2)7COOH and it is an essential fatty acid for the human body, as it is essential in the composition of arachidonic acid that is responsible for building muscle, managing body fat thermogenesis, and regulating core protein synthesis.

If 0.250 L of a 5.90 M HNO₃ solution is diluted to 2.00 L, what is the molarity of the new solution?

Answers

Answer:

0.74 M

Explanation:

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

Molarity of stock solution (M₁) = 5.90 M

Volume of stock solution (V₁) = 0.250 L

Volume of diluted solution (V₂) = 2 L

Molarity of diluted solution (M₂) =?

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

M₁V₁ = M₂V₂

5.90 × 0.250 = M₂ × 2

1.475 = M₂ × 2

Divide both side by 2

M₂ = 1.475 / 2

M₂ = 0.74 M

Thus, the molarity of the diluted solution is 0.74 M

State two conditions necessary for an esterification reaction to take place​

Answers

Explanation:

Esterification occurs when a carboxylic acid reacts with an alcohol. This reaction can only occur in the presence of an acid catalyst and heat. It takes a lot of energy to remove the -OH from the carboxylic acid, so a catalyst and heat are needed to produce the necessary energy.

Answer:

The Esterification Process

The Esterification ProcessEsterification occurs when a carboxylic acid reacts with an alcohol. This reaction can only occur in the presence of an acid catalyst and heat. It takes a lot of energy to remove the -OH from the carboxylic acid, so a catalyst and heat are needed to produce the necessary energy.

The Esterification ProcessEsterification occurs when a carboxylic acid reacts with an alcohol. This reaction can only occur in the presence of an acid catalyst and heat. It takes a lot of energy to remove the -OH from the carboxylic acid, so a catalyst and heat are needed to produce the necessary energy.Once the -OH has been removed, the hydrogen on the alcohol can be removed and that oxygen can be connected to the carbon. Because the oxygen was already connected to a carbon, it is now connected to a carbon on both sides, and an ester is formed.

The Esterification ProcessEsterification occurs when a carboxylic acid reacts with an alcohol. This reaction can only occur in the presence of an acid catalyst and heat. It takes a lot of energy to remove the -OH from the carboxylic acid, so a catalyst and heat are needed to produce the necessary energy.Once the -OH has been removed, the hydrogen on the alcohol can be removed and that oxygen can be connected to the carbon. Because the oxygen was already connected to a carbon, it is now connected to a carbon on both sides, and an ester is formed.The methyl acetate that was formed is an ester. In this image, the green circle represents what was the carboxylic acid (in this case acetic acid), and the red circle represents what was the alcohol (in this case methanol):

This reaction lost an -OH from the carboxylic acid and a hydrogen from the alcohol. These two also combine to form water. So any esterification reaction will also form water as a side product.

Which equation obeys the law of conservation of
mass?

Answers

Answer:2C4H10+2C12+12O2 4CO2+CC14+H20

discuss the benefits of observing good safety measures in relation to increase in productivity within a pharmaceutical laboratory?

Answers

Answer:

Pharmaceutical laboratory helps in devloping and conducting research, vaccines. Various kinds of drugs and chemical substances used and are produced at a Pharmaceutical laboratory.

The pharmaceutical laboratories performs with various hazardous substances that results in exposure to various chemicals, biological substances and radiation. To avoid any injury or infection labs need to maintain all safety measures.

Spillage and relaseing chemical substances can be lethal during transportaions by safety measures for heling in for manufacturing of such therapeutic agents spillage and avoid wastage.

Maintaining good safety standards in the pharmaceuticals laboratory will help promote the health of technicians and workers which in turn will increase productivity and attain positive outcomes.

True or false: Boron contains 2s22p1 valence electrons, so only one p orbital is needed to form molecular orbitals.

Answers

Answer:

True

Explanation:

The valence orbitals of boron are 2s2 2p1. We have to recall that all the valence orbitals whether full or empty are involved in the formation of molecular orbitals.

The number of molecular orbitals formed is equal to the number of atomic orbitals that are combined.

Since there are two valence orbitals and there is only one p orbital among the valence orbitals, it is true that only one p orbital is needed to form molecular orbitals in boron.

What would be the specific mathematical effect on the reaction rate if you carried out the sodium iodide-in-acetone reactions on the alkyl halides using an iodide solution half as concentrated? ("Slower" or "faster" is not specific enough.)

Answers

Answer:

Slower

Explanation:

The reaction between alkyl halides and sodium iodide-in-acetone is an SN2 reaction. The rate of reaction depends on the concentration of the alkyl halide as well as the concentration of the sodium iodide. It is a bimolecular reaction.

This means that if the concentration of any of the reactants is halved, the rate of reaction decreases accordingly.

Therefore, if the iodide solution is half as concentrated, the reaction is observed to be slower in accordance with the rate law;

Rate = k[alkyl halide] [iodide]

Which of the following is a physical change?

Answers

the awnser i think is c

An atom has 20 electrons. Find out
i. It’s atomic numbers and total number of p-electrons
ii. The value of azimuthal quantum number (l) and magnetic quantum number (m) of the 19th electron of the atom.
iii. It’s group position in the periodic table.

Answers

Answer:

it's atomic number is 5 and total number is 10

The atom has an atomic number of 20 and has a total of 12 p electrons.

The azimuthal quantum number (l) of the 19th electron is 0 and the magnetic quantum number (m) of the 19th electron is 0.

It is an element of group 2

The number of electrons in the neutral atom is equal to the number of protons and is also the atomic number of an atom.

An atom is known to be electrically neutral. This is because the number of electrons in the atom is equal to the number of protons in the neutral atom.

The number of protons in the neutral atom is called the atomic number of the atom.

For an element that has 20 electrons, its electronic configuration is;

1s2 2s2 2p6 3s2 3p6 4s2.

The 19th electron is in the 4s orbital hence both the azimuthal and magnetic quantum numbers are zero.

The element has outermost electron configuration ns2 so it mus belong to group 2 of the periodic table.

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Indicate how the concentration of each species in the chemical equation will change to reestablish equilibrium after reactant or product is added.

2CO(g) + O2(g) ⇌ 2CO2

Answers

Answer:

Indicate how the concentration of each species in the chemical equation will change to reestablish equilibrium after reactant or product is added.

[tex]2CO(g) + O2(g) <=> 2CO2[/tex]

Explanation:

When the reactants concentration increases, then the equilibrium will shift towards products and when the concentration of products increases, then equilibrium will shift towards reactants.

So, increases in concentration of carbon monoxide (CO) shifts the equilibrium to favor the formation of carbondioxide.

Similarly increase in concentration of oxygen also favor the formation of product carbon dioxide.

Increase in concentration of CO2 favors the formation of CO and O2.

Decrease in product concentration also favors the formation of product.

Decrease in reactant concentration favors the formation of reactants only.

If 12.3 g of Cu is deposited at the cathode of an electrolytic cell after 5.50 h, what was the current used?​

Answers

Answer:

1.88 A

Explanation:

Let's consider the reduction of copper in an electrolytic cell.

Cu²⁺ + 2 e⁻ ⇒ Cu

We can calculate the charge used to deposit 12.3 g of Cu using the following relations.

The molar mass of Cu is 63.55 g/mol.1 mole of Cu is deposited when 2 moles of electrons circulate.1 mole of electrons has a charge of 96486 C (Faraday's constant).

The charge used is:

[tex]12.3 g \times \frac{1 molCu}{63.55gCu} \times \frac{2molElectron}{1molCu} \times \frac{96486C}{1molElectron} = 3.73 \times 10^{4} C[/tex]

We can convert 5.50 h to seconds using the conversion factor 1 h = 3600 s.

5.50 h × 3600 s/1 h = 1.98 × 10⁴ s

The current used is:

I = q/t = 3.73 × 10⁴ C/1.98 × 10⁴ s = 1.88 A

Post-Lab Questions
1. A beverage company is having trouble with the production of the dye in their drinks. The color of their drink mix is supposed to be a pale green color, but they often get different results. For each unwanted result, choose the most plausible explanation to help the company improve the formula.
(1pts)
The color of the drink is too pale after adding the dye to the drink because
Choose...too much dye was added to the drink.the water in the drink is evaporating.not enough dye was added to the drink.the wrong dye was added to the drink.
(1pts)
The color of the dye is appearing as red, instead of green because
Choose...too much dye was added to the drink.the water in the drink is evaporating.not enough dye was added to the drink.the wrong dye was added to the drink.
(1pts)
The drink started out the correct color but it is getting darker over time, even though nothing has been added to the drink, because
Choose...too much dye was added to the drink.the water in the drink is evaporating.not enough dye was added to the drink.the wrong dye was added to the drink.
(1pts)
2. Beer's Law states that A=εbc, where A is the absorbance, ε is the molar absorptivity of the solute, b is the path length, and c is the concentration. Identify the experimental evidence from the activity that you have for the dependence of absorbance on each variable.
The evidence for the dependence of absorbance on the variable ε is
increasing the cuvette width increases the absorbance.
changing the compound changes the absorbance behavior.
adding more water decreases the absorbance.
Choose...ABC
(1pts)
The evidence for the dependence of absorbance on the variable b is
increasing the cuvette width increases the absorbance.
changing the compound changes the absorbance behavior.
adding more water decreases the absorbance.
Choose...ABC
(1pts)
The evidence for the dependence of absorbance on the variable c is
increasing the cuvette width increases the absorbance.
changing the compound changes the absorbance behavior.
adding more water decreases the absorbance.
Choose...ABC
(1pts)
3. Describe how you could use the Beer's Law simulation to experimentally determine the best wavelength at which to perform an experiment.
Measure the absorbance for solutions of multiple different solutes and find the minimum absorbance.
Measure the absorbance for solutions with different concentrations and find the slope of the trendline.
Measure the absorbance for the same solution at different wavelengths and find the maximum absorbance.
Measure the absorbance for the same solution in different cuvette sizes and find the y-intercept.

Answers

Answer:

1. not enough dye was added to the drink.

The wrong dye was added to the drink

the water in the drink is evaporating

2. Changing the compound changes the absorbance behavior.

3. Measure the absorbance for the same solution in different cuvette sizes and find the y-intercept.

Explanation:

When the beverage company adds dye to the drink, there should be standard quantity added to the drink so that the color of the drink remains constant. When too much dye is added to the drink, the color will get dark brown or black. When the color of drink get lighter than green this means dye is not added in required quantity.

Given 0.60 mol CO2, 0.30 mol CO, and 0.10 mol H20, what is the partial pressure of the CO if the total pressure of the mixture was 0.80 atm?

Answers

Answer:

Explanation:

/ means divided by

* means multiply

1. formula is

partial pressure = no of moles(gas 1)/ no of moles(total)

0.30 mol CO/0.60 mol CO2 + 0.30 mol CO + 0.10 mol H20 ->

.3/(.6+.3+.1) =

.3/1 =

.3 =

partial pressure of CO

2.

.3 * .8 atm = .24

khanacademy

quizlet

The partial pressure of the CO is 0.24 atm if the total pressure of the mixture was 0.80 atm.

Dalton's Law of Partial pressure

Dalton's Law of partial pressure states that the total pressure exerted by non reacting gaseous mixture at a constant temperature and given volume is equal to the sum of partial pressure of all gases.

Dalton's Law of partial pressure using mole fraction of gas

Partial pressure of carbon monoxide (CO) = Mole fraction of carbon monoxide (CO) × Total pressure

Now, we have to find the first mole fraction of CO

Mole fraction of carbon monoxide (CO) = [tex]\frac{\text{moles of solute}}{\text{total moles of solute}}[/tex]

                                                                  = [tex]\frac{\text{moles of CO}}{\text{moles of CO}_2 + \text{moles of CO} + \text{moles of H}_{2}O}[/tex]

                                                                  = [tex]\frac{0.30}{0.60 + 0.30 + 0.10}[/tex]

                                                                  = [tex]\frac{0.30}{1}[/tex]

                                                                  = 0.3

Now, put the value in above equation, we get that

Partial pressure of carbon monoxide (CO)

= Mole fraction of carbon monoxide (CO) × Total pressure

= 0.3 × 0.8

= 0.24 atm

Thus, the partial pressure of the CO is 0.24 atm is the total pressure of the mixture was 0.80 atm.

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2- A 0.60 sample an unknown organic acid found in muscle cells is burned in air and found to contain 0.24 grams of carbon, 0.040 grams of hydrogen, with the rest being oxygen. If the molecular weight of the substance is 90 grams/n, what is the molecular formula

Answers

Answer:

C₃H₆O₃

Explanation:

To solve this question we need to find, as first, the moles of each atom in order to find empirical formula (Simplest whole-number ratio of atoms present in a molecule).

With the molar mass of the substance and the empirical formula we can find the molecular formula as follows:

Moles C -Molar mass:12.0g/mol-

0.24g * (1mol/12.0g) = 0.020 moles C

Moles H = Mass H because molar mass = 1g/mol:

0.040 moles H

Moles O -Molar mass: 16g/mol-

Mass O: 0.60g - 0.24g - 0.040g = 0.32g O

0.32g O * (1mol/16g) = 0.020 moles O

Ratio of atoms (Dividing in moles of C: Lower number of moles):

C = 0.020 moles C / 0.020 moles C = 1

H = 0.040 moles H / 0.020 moles C = 2

O = 0.020 moles O / 0.020 moles C = 1

Empirical formula:

CH₂O.

Molar mass CH2O:

12g/mol + 2*1g/mol + 16g/mol = 30g/mol

As molecular formula has a molar mass 3 times higher than empirical formula, the molecular formula is 3 times empirical formula:

C₃H₆O₃

The molecular formula of the organic acid would be C3H6O3

Molecular formula

Molecular formula = [empirical formula]n

Where n = molar mass/mass of empirical formula

Empirical formula

C = 0.24/12 = 0.02

H = 0.040/1 = 0.04

O = 0.6 - (0.24+0.04) = 0.32/16 = 0.02

Divide by the smallest

C = 1

H = 2

O = 1

Empirical formula = CH2O

Empirical formula mass = 12 + 2 + 16 = 30

n = 90/30 = 3

Molecular formula = [CH2O]3

                               = C3H6O3

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En la fermentación del alcohol, la levadura convierte la glucosa en etanol y dióxido de carbono:
C6H12O6(s) → 2C2H5OH(l) + 2CO2(g)
Si reaccionan 5.97 g de glucosa y se recolectan 1.44 L de CO2 gaseoso, a 293 K y 0.984 atm, ¿cuál
es el rendimiento porcentual de la reacción

Answers

Answer:

88.9%

Explanation:

Primero convertimos 5.97 g de glucosa a moles, usando su masa molar:

5.97 g ÷ 180 g/mol = 0.0332 mol

Después calculamos la cantidad máxima de moles de CO₂ que se hubieran podido producir:

0.0332 mol C₆H₁₂O₆ * [tex]\frac{2molCO_2}{1molC_6H_{12}O_6}[/tex] = 0.0664 mol CO₂

Ahora calculamos los moles de CO₂ producidos, usando los datos de recolección dados y la ecuación PV=nRT:

0.984 atm * 1.44 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 293 Kn = 0.0590 mol

Finalmente calculamos el rendimiento porcentual:

0.0590 mol / 0.0664 mol * 100% = 88.9%

A quantity of 1.435 g of naphthalene , was burned in a constant-volume bomb calorimeter. Consequently, the temperature of the water rose from 20.28oC to 25.95oC If the heat capacity of the bomb plus water was , calculate the heat of combustion of naphthalene on a molar basis; that is, find the molar heat of combustion.

Answers

Answer:

molar heat of combustion = -5156 *10³ kJ/mol

Explanation:

A quantity of 1.435 g of naphthalene , was burned in a constant-volume bomb calorimeter. Consequently, the temperature of the water rose from 20.28oC to 25.95oC If the heat capacity of the bomb plus water was 10.17 kJ/°C, calculate the heat of combustion of naphthalene on a molar basis; that is, find the molar heat of combustion.

Step 1: Data given

Mass of naphthalene = 1.435 grams

Initial temperature of water = 20.28 °C

Final temperature of water = 25.95 °C

heat capacity of the bomb plus water was 10.17 kJ/°C

Molar mass naphtalene = 128.2 g/mol

Step 2:

Qcal = Ccal * ΔT

⇒with Qcal =the heat of combustion

⇒with Ccal = heat capacity of the bomb plus water = 10.17 kJ/°C

⇒with ΔT = the difference in temperature = T2 - T1 = 25.95 - 20.28 = 5.67°C

Qcal = 10.17 kJ/°C * 5.67 °C

Qcal = 57.7 kJ

Step 3: Calculate moles

Moles naphthalene = 1.435 grams / 128.2 g/mol

Moles naphthalene = 0.01119 moles

Step 4: Calculate the molar heat of combustion

molar heat of combustion = Qcal/ moles

molar heat of combustion = -57.7 kJ/ 0.01119 moles

molar heat of combustion = -5156 *10³ kJ/mol

#6 and #7. How many carbon atoms are in a mixture of 7.00 mol c2F2 and 0.400 mol carbon dioxide and also #7

Answers

Answer:

#6  8.67x10²⁴ atoms

#7  

1. Atom

2. Formula unit

3. Molecule

4. Ion

Explanation:

#6 First we calculate how many carbon moles are there in 7.00 moles of C₂F₂, keeping in mind that there are 2 C moles per C₂F₂ mol:

7.00 mol C₂F₂ * 2 = 14.00 mol C

As for carbon dioxide, there are 0.400 C moles in 0.400 moles of CO₂.

We calculate the total number of C moles:

14.00 mol + 0.400 mol = 14.4 mol C

Finally we calculate the number of atoms in 14.4 C moles, using Avogadro's number:

14.4 mol * 6.023x10²³ atoms/mol = 8.67x10²⁴ atoms

#7

1. Radon - Atom (Ra)2. Formula unit (It is a crystalline solid, BaBr₂)3. Molecule (NH₃)4. Ion (It has a formal charge, +2)

PLEASE HELP!!

How does temperature, agitation, and particle size affect solubility?

Answers

Answer:

At higher temperatures, particles move faster and collide more, increasing solubility rates.

Agitation increases solubility rates as well, by bringing fresh solvent into contact with the undissolved solute

The smaller the particle size, the higher (faster) solubility rate. Vice versa, the bigger the particle size, the lower (slower) solubility rate.

Explanation:

Liquid octane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 10.3 g of octane is mixed with 23. g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to significant digits.

Answers

Answer:

9.36 g

Explanation:

The equation of the reaction is;

C8H18(g) + 25/2 O2(g) ----> 8CO2(g) + 9H2O(g)

Number of moles of octane = 10.3g/ 114 g/mol = 0.09 moles

1 mole of octane yields 9 moles of water

0.09 moles of octane yields 0.09 × 9/1 = 0.81 moles of water

Number of moles of oxygen = 23g/32g/mol = 0.72 moles

12.5 moles of oxygen yields 9 moles of water

0.72 moles of oxygen yields 0.72 × 9/12.5 = 0.52 moles of water

Hence oxygen is the limiting reactant;

Maximum mass of water produced = 0.52 moles of water × 18 g/mol = 9.36 g

You are asked to prepare a buffer solution with a pH of 3.50. The following solutions, all 0.100 M, are available to you: HCOOH, CH3COOH, H3PO4 , NaCHOO, NaCH3COO, and NaH2PO4.  What would be the best combination to make the required buffer solution? Select one:
a. NaH2PO4 and NaCHOO  
b. H3PO4 and NaH2PO4
c. NaH2PO4 and HCOOH
d. CH3COOH and NaCH3COO e. HCOOH and NaCHOO
can someone helo me with this​

Answers

Answer:

e. HCOOH and NaCHOO

Explanation:

For a buffer solution, both an acid and its conjugate base are required.

With the information above in mind, we can discard options a) and c), as those combinations are not of an acid and its conjugate base.

Now it is a matter of comparing the pKa (found in literature tables) of the acids of the remaining three acids:

H₃PO₄ pKa = 2.12CH₃COOH pKa = 2.8HCOOH pKa = 3.74

The acid with the pKa closest to the desired pH is HCOOH, so the correct answer is e. HCOOH and NaCHOO

4.106
Calculate the moles and the mass of solute in each of the following solutions.
(a) 150.0 mL of 0.245 M CaCl2

Answers

Solution: (moles of solute)

molarity = moles of solute / volume of solution

moles of solute = molarity × volume of solution

moles of solute = 0.245 mol/L × 0.1500 L

moles of solute = 0.03675 mol

moles of solute = 0.0368 mol

-----------------------------------------------------------

Solution: (mass of solute)

Step 1: Calculate the molar mass of solute.

molar mass of solute = (40.08 g/mol × 1) + (35.45 g/mol × 2)

molar mass of solute = 110.98 g/mol

Step 2: Calculate the mass of solute.

mass of solute = moles of solute × molar mass of solute

mass of solute = 0.03675 mol × 110.98 g/mol

mass of solute = 4.08 g

Note: The volume of solution must be expressed in liters (L).

Answer:

[tex]\boxed {\sf \bold {0.0368 \ mol \ CaCl_2}}}}[/tex]

[tex]\boxed {\sf \bold {4.08 \ g \ CaCl_2}}}}}[/tex]

Explanation:

1. Moles of Solute

Molarity is a measure of concentration in moles per liter.

[tex]molarity= \frac {moles \ of \ solute}{liters \ of \ solution}[/tex]

In this solution, there are 150.0 milliliters of solution and the molarity is 0.245 M CaCl₂ or 0.245 mol CaCl₂ per liter.

First, convert the milliliters to liters. There are 1000 milliliters in 1 liter.

[tex]{150 \ mL * \frac{1 \ L}{1000 \ mL}= \frac{150}{1000} \ L = 0.150 \ L[/tex]

Now, substitute the known values (molarity and liters of solution) into the formula. The moles of solution are unknown, so we can use x.

[tex]0.245 \ mol \ CaCl_2 /L= \frac{ x}{0.150 \ L}[/tex]

We are solving for x, so we must isolate this variable. It is being divided by 0.150 L. The inverse of divisions is multiplication, so we multiply both sides by 0.150 L.

[tex]0.150 \ L *0.245 \ mol \ CaCl_2 /L= \frac{ x}{0.150 \ L} * 0.150 L[/tex]

[tex]0.150 \ L *0.245 \ mol \ CaCl_2 /L=x[/tex]

The units of liters cancel.

[tex]0.150 *0.245 \ mol \ CaCl_2 =x[/tex]

[tex]0.03675 \ mol \ CaCl_2[/tex]

The original measurements have 3 significant figures, so our answer must have the same.

We should round to the ten thousandths place. The 5 to the right of this place tells us to round the 7 up to an 8.

[tex]\bold {0.0368 \ mol \ CaCl_2}[/tex]

2. Mass of the Solute

We can convert mass to moles using the molar mass. These values are found on the Periodic Table. They are the same as the atomic masses, but the units are grams per mole (g/mol) instead of atomic mass units.

The solute is calcium chloride: CaCl₂. Look up the molar masses of the individual elements.

Ca: 40.08 g/mol Cl:  35.45 g/mol

Notice that chlorine has a subscript of 2. We must multiply the molar mass by 2.

Cl₂: 35.45 *2= 70.9 g/mol

Add calcium's molar mass.

CaCl₂: 40.08 + 70.9 =110.98 g/mol

Use the molar mass as a ratio.

[tex]\frac {110.98 \ g\ CaCL_2}{ 1 \ mol \ CaCl_2}[/tex]

Multiply the moles of calcium chloride we calculated above.

[tex]0.0368 \ mol \ CaCl_2 *\frac {110.98 \ g\ CaCL_2}{ 1 \ mol \ CaCl_2}[/tex]

The units of moles of calcium chloride cancel.

[tex]0.0368 *\frac {110.98 \ g\ CaCL_2}{ 1 }[/tex]

[tex]4.084064 \ g\ CaCl_2[/tex]

Round to 3 significant figures again. For this number, it is the hundredths place. The 4 in the thousandths place tells us to leave the 8.

[tex]\bold {4.08 \ g \ CaCl_2}[/tex]

Suppose you ran this reaction without triethylamine and simply used an excess of reactant 1. At the end of the reaction, your methylene chloride solution would contain mostly reactant 1 and the product. What would you do to remove reactant 1 from the solution

Answers

ummm is that chemistry?

Answer:

is this chem

Explanation:

Based upon the intermolecular forces present, rank the following substances according to the expected boiling point for the substance.

a. HCl
b. NaCl
c. N2
d. H2O

Answers

It would be N2!!!!!!!!!!!!!!!!!!

Que es la actividad física y en qué mejora

Answers

La actividad física regular puede mejorar su fuerza muscular y aumentar su resistencia. El ejercicio proporciona oxígeno y nutrientes a sus tejidos y ayuda a que su sistema cardiovascular funcione de manera más eficiente. Y cuando la salud de su corazón y pulmones mejoran, tiene más energía para hacer frente a las tareas diarias. Encantado de ayudarle
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