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2 years ago

Acids increase the concentration of hydronium ions in solution by donating hydrogen ions to water molecules. True False

Chemistry

2 answers:

Wittaler [7]2 years ago

6 0

Answer:

its true just took it and got it right :)

Explanation:

Kitty [74]2 years ago

3 0

Answer:

Explanation:

General reaction of acid in water is as follows:

HCl + H2O = H3O+ + Cl-

Thus Acids increase the concentration of hydronium ions in solution by donating hydrogen ions to water molecules is true

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An unknown solid is entirely soluble in water. On addition of dilute HCl, a precipitate forms. After the precipitate is filtered

Karo-lina-s [1.5K]

Answer:

Pb(NO3)2

Cd(NO3)2

Na2SO4

Explanation:

In the first part, addition of HCl leads to the formation of PbCl2 which is poorly soluble in water. This is the first precipitate that is filtered off.

When the pH is adjusted to 1 and H2S is bubbled in, CdS is formed. This is the second precipitate that is filtered off.

After this precipitate has been filtered off and the pH is adjusted to 8, addition of H2S and (NH4)2HPO4 does not lead to the formation of any other precipitate.

The yellow flame colour indicates the presence of Na^+ which must come from the presence of Na2SO4.

5 0

1 year ago

A tank of 0.1m3 volume contains air at 25∘C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at

Dmitriy789 [7]

Answer:

Amount of Energy = 23,467.9278J

Explanation:

Given

Cv = 5/2R

Cp = 7/2R wjere R = Boltzmann constant = 8.314

The energy balance in the tank is given as

∆U = Q + W

According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

3 0

2 years ago

A solution is prepared by condensing 4.00 l of a gas, measured at 27°c and 748 mmhg pressure, into 58.0 g of benzene. calculate

fgiga [73]

First, we are using the ideal gas law to get n the number of moles:

PV = nRT

when P is the pressure = 748 mmHg/760 = 0.984 atm

V is the volume = 4 L

R is ideal gas constant = 0.0821

T is the temperature in Kelvin = 300 K

∴ n = 0.984atm*4L/0.0821*300

= 0.1598 moles

when the concentration = moles * (1000g / mass)

= 0.1598 * (1000g / 58 g )

= 2.755 M

when the freezing point = 5.5 °C

and Kf = - 5.12 °C/m

∴ the freezing point for the solution = 5.5 °C + (Kf*m)

= 5.5 °C - (5.12°C/m * 2.755m)

= -8.6 °C

8 0

2 years ago

Each day, the stomach produces 2.0 L of gastric juice that contains 0.10 M HCl. Phillips Milk of Magnesia is a white-colored, aq

trasher [3.6K]

Answer:

It would take 72.9 mL of milk of magnesia.

Explanation:

First of all we have to think how the compounds react with each other and what are the products formed. In this case, the hydrochloric acid reacts with magnesium hydroxide to generate magnesium chloride and water as a subproduct. Having said that, we have to state the balanced chemical reaction to know the associated stoichiometry:

2 HCl + Mg(OH)2 → MgCl2 + 2 H2O

According to the balanced equation we know that 2 mol of HCl reacts with 1 mol of Mg(OH)2.

Now we calculate the quantity of moles of HCl that we have present in 2.0 lts of 0.10 M solution:

0.1 M HCl = 0.1 moles HCl / 1000 ml Solution

So, in 2 liters of solution we will have 0.2 moles of HCl

This 0.2 moles of acid, as we stated before, will react with 0.1 moles of Mg(OH)2, so we need to calculate the amount of milk of magnesia that has this required quantity of moles.

With the molar mass of Mg(OH)2 we calculate the weight of the compound that represents the 0.1 moles needed to react with all the HCl present in solution:

1 mol Mg(OH)2 = 58.32 g

0.1 mol = 5.832 g

Now we need to determine what volume of the milk of magnesia solution has 5.832 g of Mg(OH)2 to react with the acid:

The concentration of milk of magnesia is 8 % (w/v). This means that we have 8 gr of Mg(OH)2 per 100 ml of solution.

8 gr Mg(OH)2 per 100 mL Solution

5.832 gr Mg(OH)2 = 72.9 mL of Milk of Magnesia

6 0

2 years ago

A mixture of CH4 and H2O is passed over a nickel catalyst at 1000 K. The emerging gas is collected in a 5.00L flask and is found

Harman [31]

Answer:

Kc =<u> 3.74*10⁻³ </u>

Kp = 25.21

Explanation:

Step 1: Data given

Temperature = 1000 K

Volume = 5.00 L

Mass of CO = 8.62 grams

Mass of H2 = 2.60 grams

Mass of CH4 = 43.0 grams

Mass of H2O = 48.4 grams

Kc = [CO]*[H₂]³ / ([CH₄]∙*H₂O])

Kp = p(CO)*p(H₂)³ / (p(CH₄)*p(H₂O) )

Step 2: The balanced equation

CH₄ + H₂O ⇄ CO + 3 H₂

Step 3: Calculate number of moles

The number of moles of each compund in the equilibrium mixture are:

Moles = mass / molar mass

n(CH₄) = 43.0g / 16g/mol = 2.688mol

n(H₂O) = 48.4g / 18g/mol = 2.689mol

n(CO) = 8.62g/28g/mol = 0.308mol

n(H₂) = 2.60g / 2g/mol = 1.3mol

Step 4: Calculate concentrations at equilibrium

So the equilibrium concentrations are:

Concentration = moles / volume

[CH₄] = 2.688mol/5L = 0.5376 M

[H₂O] = 2.689mol/5L = 0.5378M

[CO] = 0.308mol/5L = 0.0616M

[H₂) = 1.3mol/5L = 0.26M

Step 5: Calculate Kc

Kc = 0.0616 ∙ (0.26)³ / (0.5376∙0.5378) = <u>3.74*10⁻³ </u>

Step 5: Calculate partial pressure

Partial pressures in equilibrium can be found from ideal gas law:

p(X) = n(X)∙R∙T/V = [X]∙R∙T

=> p(CH₄) = [CH₄]∙R∙T = 0.5376mol/L * 0.082 06Latm/molK ∙ 1000K = 44.11 atm

p(H₂O) = [H₂O]∙R∙T = 0.5738mol/L * 0.082 06Latm/molK * 1000K = 44.13 atm

p(CO) = [CO]∙R∙T = 0.0616mol/L * 0.082 06Latm/molK * 1000K = 5.05atm

p(H₂) = [CO]∙R∙T = 0.26mol/L * 0.082 06Latm/molK * 1000K = 21.34atm

Step 5: Calculate Kp

Kp = p(CO)*p(H₂)³ / (p(CH₄)*p(H₂O) )

Kp = 5.05*21.34³ / (44.11*44.13 ) = 25.21

8 0

2 years ago