PLEASE HELP FAST!!

Which of the following ions is formed when an acid is dissolved in a solution?
H+
O−
OH−
SO42+

Answers

Answer 1

Answer:

H+

Explanation:

acid produce H+ when it is dissolved in a solution


Related Questions

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.

Para formar bronce, se mezclan 150g de cobre a 1100°C y 35g de estaño a 560°C. Determine la temperatura final del sistema.
Dato: Ce Cu: 0,093 cal/gºC; Ce Sn: 0,060 cal/gºC


URGENTEEEE

Answers

Answer:

La temperatura final del sistema es 1029,346 °C.

Explanation:

Asumamos que el sistema conformado por el cobre y el estaño no tiene interacciones con sus alrededores. Por la Primera Ley de la Termodinámica, el cobre cede calor al estaño con tal de alcanzar el equilibrio térmico. El cobre se encuentra inicialmente en su punto de fusión, mientras que el estaño está por encima de ese punto, de modo que la transferencia de calor es esencialmente sensible:

[tex]m_{Cu}\cdot c_{Cu}\cdot (T-T_{Cu}) = m_{Sn}\cdot c_{Sn}\cdot (T_{Sn}-T)[/tex]

[tex](m_{Cu}\cdot c_{Cu} + m_{Sn}\cdot c_{Sn})\cdot T = m_{Sn}\cdot c_{Sn}\cdot T_{Sn} + m_{Cu}\cdot c_{Cu}\cdot T_{Cu}[/tex]

[tex]T = \frac{m_{Sn}\cdot c_{Sn}\cdot T_{Sn}+m_{Cu}\cdot c_{Cu}\cdot T_{Cu}}{m_{Cu}\cdot c_{Cu}+m_{Sn}\cdot c_{Sn}}[/tex] (1)

Donde:

[tex]m_{Sn}[/tex] - Masa del estaño, en gramos.

[tex]m_{Cu}[/tex] - Masa del cobre, en gramos.

[tex]c_{Sn}[/tex] - Calor específico del estaño, en calorías por gramo-grados Celsius.

[tex]c_{Cu}[/tex] - Calor específico del cobre, en calorías por gramo-grados Celsius.

[tex]T_{Sn}[/tex] - Temperatura inicial del estaño, en grados Celsius.

[tex]T_{Cu}[/tex] - Temperatura inicial del cobre, en grados Celsius.

Si sabemos que [tex]m_{Cu} = 150\,g[/tex], [tex]m_{Sn} = 35\,g[/tex], [tex]c_{Cu} = 0,093\,\frac{cal}{g\cdot ^{\circ}C}[/tex], [tex]c_{Sn} = 0,060\,\frac{cal}{g\cdot ^{\circ}C}[/tex], [tex]T_{Sn} = 560\,^{\circ}C[/tex] y [tex]T_{Cu} = 1100\,^{\circ}C[/tex], entonces la temperatura final del sistema es:

[tex]T = \frac{(35\,g)\cdot \left(0,060\,\frac{cal}{g\cdot ^{\circ}C} \right)\cdot (560\,^{\circ}C)+(150\,g)\cdot \left(0,093\,\frac{cal}{g\cdot ^{\circ}C} \right)\cdot (1100\,^{\circ}C)}{(35\,g)\cdot \left(0,060\,\frac{cal}{g\cdot ^{\circ}C} \right)+(150\,g)\cdot \left(0,093\,\frac{cal}{g\cdot ^{\circ}C} \right)}[/tex]

[tex]T = 1029,346\,^{\circ}C[/tex]

La temperatura final del sistema es 1029,346 °C.

A chemical reaction occurring in a cylinder equipped with a moveable piston produces 0.601 mol of a gaseous product. If the cylinder contained 0.300 mol of gas before the reaction and had an initial volume of 2.16 L, what was its volume after the reaction

Answers

Answer:

4.33 L

Explanation:

Assuming ideal behaviour and that all 0.300 moles of gas reacted, we can solve this problem using Avogadro's law, which states that at constant temperature and pressure:

V₁n₂ = V₂n₁

Where in this case:

V₁ = 2.16 Ln₂ = 0.601 molV₂ = ?n₁ = 0.300 mol

We input the given data:

2.16 L * 0.601 mol = V₂ * 0.300 mol

And solve for V₂:

V₂ = 4.33 L

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

4.005 X 74 X 0.007 = 2.10049

Answers

Answer:

2.07459

Explanation:

this is the correct answer.

bio-chemisty of protain​

Answers

Answer:

Protein biochemistry is the study of proteins. Protein biochemistry is a scientific field dedicated to the study of proteins, complex chains of amino acids which make up the building blocks of all living organisms.

Explanation:

I hope that helped

Copy and Pasted!

Answer:

Listen to what guy said on top.

Explanation:

polypeptide structures consisting of one or more long chains of amino acids residue.....

or my answer

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.

What were the advantages for Mendel in using pea plants for his breeding experiments?

Answers

Answer:

He chose pea plants because they were easy to grow, could be bred rapidly, and had several observable characteristics, like petal color and pea color.

Explanation:

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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The gas law for an ideal gas at absolute temperature T (in kelvins), pressure P (in atmospheres), and volume V (in liters) is PV = nRT, where n is the number of moles of the gas and R = 0.0821 is the gas constant. Suppose that, at a certain instant, P = 8.0 atm and is increasing at a rate of 0.13 atm/min and V = 13 L and is decreasing at a rate of 0.17 L/min. Find the rate of change of T with respect to time (in K/min) at that instant if n = 10 mol.

Answers

Answer:

The rate of change of T with respect to time is 0.40 K/min

Explanation:

The gas law equation is:

[tex] PV = nRT [/tex]

We can find the rate of change of T with respect to time by solving the above equation for T and derivating with respect to time:

[tex] \frac{dT}{dt} = \frac{d}{dt}(\frac{PV}{nR}) [/tex]

[tex] \frac{dT}{dt} = \frac{1}{nR}(V\frac{dP}{dt} + P\frac{dV}{dt}) [/tex]

Where:

n: is the number of moles = 10 mol

R: is the gas constant = 0.0821

V: is the volume = 13 L

P: is the pressure = 8.0 atm

dP/dt: is the variation of the pressure with respect to time = 0.13 atm/min

dV/dt: is the variation of the volume with respect to time = -0.17 L/min

Hence, the rate of change of T is:

[tex] \frac{dT}{dt} = \frac{1}{10*0.0821}(13*0.13 - 8.0*0.17) = 0.40 K/min [/tex]    

Therefore, the rate of change of T with respect to time is 0.40 K/min

I hope it helps you!    

Which equation obeys the law of conservation of
mass?

Answers

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

When you hold an object with a positive electric charge near an object with a
negative electric charge, it causes the negatively charged object to move. How
could you model this interaction with magnets? Describe the types of forces
involved and the energy transformations that take place.

Answers

Answer:

A positively charged object will exert a repulsive force upon a second positively charged. This repulsive force will push the two objects apart while a negatively charged object will exert a repulsive force upon a second negatively charged object. Objects with like charge repel each other

The interaction between objects with positive and negative electric charges can be analogously modeled using magnets. The Types of Forces Involved are; Attractive Magnetic Force, Repulsive Magnetic Force and the Energy Transformations are; Potential Energy Transformation, and Kinetic Energy Transformation.

In this analogy, magnets can represent the charges, and magnetic forces can represent the electric forces.

Interaction Between Magnets

Imagine we have two magnets: one with a north pole (N) and the other with a south pole (S). When you bring the north pole of one magnet close to the south pole of the other magnet, they are attracted to each other. Conversely, if you bring the north pole of one magnet near the north pole of the other magnet, they repel each other.

Types of Forces Involved:

Attractive Magnetic Force (Analogous to Electric Attraction):

When the north pole of one magnet is brought close to the south pole of another magnet, they experience an attractive magnetic force. Similarly, when objects with opposite electric charges were brought close together, then they will experience an attractive electric force.

Repulsive Magnetic Force (Analogous to Electric Repulsion):

When two magnets with the same pole (both north or both south) are brought close to each other, they experience a repulsive magnetic force. This is analogous to the repulsion between objects with like electric charges (both positive or both negative).

Energy Transformations;

When you bring the magnets closer together or move them apart, energy transformations occur:

Potential Energy Transformation;

As the magnets are moved closer together, the potential energy of the magnetic interaction decreases. This is because the magnets' magnetic fields interact more strongly, and they tend to move toward each other due to the attractive or repulsive forces.

Kinetic Energy Transformation;

If you let the magnets go after bringing them close together, they will move towards each other (in the case of attraction) or move apart (in the case of repulsion). This movement involves a transformation of potential energy into kinetic energy. The kinetic energy increases as the magnets move, and it's at its maximum when the magnets are farthest apart (in the case of repulsion) or when they collide (in the case of attraction).

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Me please answer as follows

Answers

Answer:

no reaction occurs .that is no product

The action of some commercial drain cleaners is based on the following reaction:
2 NaOH(s) + 2 Al(s) + 6 H2O(l) â 2 NaAl(OH)4(s) + 3 H2(g)
What is the volume of H2 gas formed at STP when 4.32 g of Al reacts with excess NaOH?
A. 3.59 L
B. 2.39 L
C. 5.87 L
D. 5.38 L

Answers

Answer:

5.38 L

Option D.

Explanation:

2 NaOH(s) + 2 Al(s) + 6 H₂O(l)   →  2 NaAl(OH)₄(s)  + 3 H₂(g)

We convert mass of Al to moles:

4.32 g . 1 mol /26.98g = 0.160 moles

As NaOH is in excess, aluminum is the limiting reactant.

We see stoichiometry, were ratio is 2:3.

2 moles of Al can produce 3 moles of hydrogen

Our 0.160 moles may  produce (0.160 . 3)/2 = 0.240 moles of H₂.

We know that 1 mol of any gas at STP conditions is contained in 22.4L

So let's make the conversion factor:

0.240 mol . 22.4L / 1mol = 5.38 L

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%

The chemical formula for hydrogen chloride is HCl A chemist determined by measurements that moles of hydrogen chloride participate in a chemical reaction. Calculate the mass of hydrogen chloride that participates. Be sure your answer has the correct number of significant digits.

Answers

The question is incomplete, the complete question is;

. A chemist determined by measurements that 0.015 moles of hydrogen chloride participate in a chemical reaction. Calculate the mass of hydrogen chloride that participates Round your answer to 2 significant digits. x S. ?

Answer:

0.54 g

Explanation:

Recall that;

Number of moles = mass/molar mass

Molar mass of HCl =36.5 g/mol

Mass= number of moles × molar mass

Mass= 0.015 moles × 36 g/mol

Mass= 0.54 g

What is the name for CH 3 CH 2 COCHCH 3 CH(CH 3 ) 2 ?

Answers

Answer:

Butanoic acid

Explanation:

The IUPAC name of CH3CH2CH2COOH is:

The IUPAC name for a given compound is Butanoic acid.

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:

Explain carefully what happen to the propanol-water system if approximately
50% of propanol by mass is fractionally distilled. What will be the distillate and
the residue?

Answers

Answer:

In the case of mixtures of ethanol and water, this minimum occurs with 95.6% by mass of ethanol in the mixture. The boiling point of this mixture is 78.2°C, compared with the boiling point of pure ethanol at 78.5°C, and water at 100°C. You might think that this 0.3°C doesn't matter much, but it has huge implications for the separation of ethanol / water mixtures. The next diagram shows the boiling point / composition curve for ethanol / water mixtures. I've also included on the same diagram a vapor composition curve in exactly the same way as we looked at on the previous pages about phase diagrams for ideal mixtures.

he equation for the dissociation of pyridine is
C5H5N(aq) + H2O(l) ⇌ C 5H5NH+(aq) + OH-(aq) Kb = 1.9 × 10-9

Calculate the pH of a pyridine solution that has a concentration of 9.2 M. Round your answer to two decimal places.

Answers

Answer:

10.10

Explanation:

Step 1: Write the basic dissociation reaction for pyridine

C₅H₅N(aq) + H₂O(l) ⇌ C₅H₅NH⁺(aq) + OH⁻(aq)      Kb = 1.9 × 10⁻⁹

Step 2: Calculate [OH⁻]

For a weak base, we will use the following expression.

[OH⁻] = √(Cb × Kb) = √(9.2 × 1.9 × 10⁻⁹) = 1.3 × 10⁻⁴ M

Step 3: Calculate pOH

We will use the definition of pOH.

pOH = -log [OH⁻] = -log 1.3 × 10⁻⁴ = 3.9

Step 4: Calculate pH

We will use the following expression.

pH = 14 - pOH = 14 - 3.9 = 10.10

refer to pic plssss

Answers

Answer:

fgufyifyifyiyduhyufyiddjyfjyf86yif

o Calculate the pH of 500 ml of buffer solution containing .2M of ammonium sulphate and 0.3M ammonium hydroxide which 5 2.1% ionized in dilute solution. (kb of ammonium hydroxide is 1.8×10^-5)

Answers

Answer is 9

pKb=−logK

b=−log10^-5=5

A mixture of ammonium hydroxide and ammonium chloride forms a basic buffer solution.

Henderson's equation for the basic buffer solution is as given below-

pOH=pKb+log[acid]*[salt]

Substitute values in the above expression.

pOH=5+log0.1*0.1=5

Hence, the pH of the solution is pH=14−pOH=14−5=9

What do you mean by pH ?potential of hydrogena measure of the acidity or alkalinity of a solution equal to the common logarithm of the reciprocal of the concentration of hydrogen ions in moles per cubic decimetre of solution. Pure water has a pH of 7, acid solutions have a pH of less than 7, and alkaline solution with a pH greater than 7.

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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]

A gas mixture is made by combining 8.7 g each of Ar, Ne, and an unknown diatomic gas. At STP, the mixture occupies a volume of 17.28 L. What is the molar mass of the unknown gas

Answers

Answer: Molar mass of the unknown gas is 73.153 g/mol.

Explanation:

Given: Mass of each gas = 8.7 g

Volume = 17.28 L

Let us assume that the molar mass of gas is m g/mol.

Molar mass of Ar is 40 g/mol and Ne is 20 g/mol.

Hence, total moles of each gas are as follows.

[tex](\frac{8.7}{40} + \frac{8.7}{20} + \frac{8.7}{m}) mol[/tex]

At STP, the total volume of these gases is as follows.

[tex](\frac{8.7}{40} + \frac{8.7}{20} + \frac{8.7}{m}) mol \times 22.4 L = 17.28 L\\(\frac{8.7}{40} + \frac{8.7}{20})22.4 L + \frac{8.7}{m} \times 22.4 L = 17.28 L\\14.616 + \frac{8.7}{m} \times 22.4 L = 17.28 L\\\frac{8.7}{m} \times 22.4 L = (17.28 L - 14.616)\\\frac{8.7}{m} \times 22.4 L = 2.664 \\m = 73.153 g/mol[/tex]

Thus, we can conclude that molar mass of the unknown gas is 73.153 g/mol.

when a polar bond is formed between 2 atoms which atom receives a partial positive charge

Answers

Answer:

The more electronegative atom in a covalent bond

What is the difference between conjugate acid-base pair?

a. a H atom. c. a mole water
b. a H+ ion d. a OH– ion​

Answers

Answer:

b. a H+ ion

Explanation:

The concept of conjugate acid-base pair is related to Bronsted-Lowry acid-base theory and according to this theory, acid is a proton acceptor.

In short,

conjugate base is formed when an acid donates a proton.

conjugate acid is formed when a base accepts a proton.

One of the purposes of this lab is to determine the order of the reaction with respect to the Allura Red dye by creating first and second-order graphs for all four trials. The correct order of the reaction is the one where the slopes of the graphs for the four trials are roughly the same. Why is this important when choosing the order of the reaction

Answers

Answer: Hello the options related to your question are attached below

The slope is related to the rate constant so all four trials should have the same slope since the reactions are all the same ( Option C )

Explanation:

It is important when choosing the order of the reaction because the concentration of the bleaches used in the four trials are in excess hence their slopes have to be roughly the same and also because the reactions are similar and they where done at the same temperature, hence the slope of the first and second-order graphs will be the same.

Write the number of sig. fig. in four numbers given in the sentence below. An (one) octopus has 8 legs. 13 octopi have 104 legs.
Give four answers.
A. Infinity, Infinity, Infinity, Infinity
B. 1, 1, 2, 3
C. Infinity, Infinity, 2, 3
D. No answer text provided.​

Answers

Answer:

1, 1, 2, 3

Explanation:

The numbers 1 and 8 both have 1 sig. fig.

The number 13 has 2 sig. figs.

The number 104 has 3 sig. figs.

100.0 mL of a 0.780 M solution of KBr is diluted to 500.0 mL. What is the new concentration of the solution?

Answers

5 times dilution
0.780M x 1/5 = 0.156M
Hope this help.

classify each of the following as a pure substance or a mixture.
a) baking soda
b) ice
c)blueberry muffin
d) zink

Answers

Baking soda- mixture (I think)
Ice- purse substance
Blueberry muffin- mixture
Zink- mixture

Answer:

c

Explanation:

its c because it has multiple mixture blueberries flower water and others thats why i says c

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