A buffer solution is 0.100 M in both HC7H5O2 and KC7H5O2 has a pH of 4.19. Which of the fo pH values would you expect from the addition of a small amount of a dilute solution of a strong base?
A. 5.79
b. 4.49
c. 3.69
d. 3.89

Explanation:

Zinc is the anode (solid zinc is oxidised). Silver is the cathode (silver ions are reduced).

By convention in standard cell notation, the anode is written on the left and the cathode is written on the right. So, in this cell: Zinc is the anode (solid zinc is oxidised). Silver is the cathode (silver ions are reduced).

0.1 mol of sodium hydroxide (NaOH) would react exactly with 7.5 g of the diprotic acid \(H_{2}\)A.

What is Molar Mass?

Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol). It is calculated by adding up the atomic masses of all the atoms in a molecule or the formula mass of all the ions in an ionic compound.

The balanced chemical equation for the reaction between diprotic acid, \(H_{2}\)A, and sodium hydroxide, NaOH, can be represented as follows:

2\(H_{2}\)A + 2 NaOH -> \(Na_{2}\)A + 2 \(H_{2}\)O

From the balanced equation, we can see that 2 moles of \(H_{2}\)A react with 2 moles of NaOH to produce 1 mole of \(Na_{2}\)A and 2 moles of water (\(H_{2}\)O).

First, we need to calculate the number of moles of \(H_{2}\)A in 7.5g using the formula:

moles = mass / molar mass

moles of \(H_{2}\)A = 7.5g / 150 g/mol = 0.05 mol

Since diprotic acid, \(H_{2}\)A, reacts in a 1:2 ratio with NaOH, we need to multiply the moles of \(H_{2}\)A by 2 to determine the moles of NaOH required for complete reaction:

Moles of NaOH = 2 * Moles of \(H_{2}\)A

Moles of NaOH = 2 * 0.05 mol

Moles of NaOH = 0.1 mol

0.1 mol of sodium hydroxide (NaOH) would react exactly with 7.5 g of the diprotic acid \(H_{2}\)A.

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Explanation:

Pure subsance is a substance that is made up of only one type of particle - each piece is the same throughout.

Being present before the reaction but not after means it's no the same (it couldve evaporated)

The volume of carbon dioxide produced is approximately (d) 11.9 liters.

To determine the amount of carbon dioxide (C\(O_2\)) produced when 37.8 grams of carbon disulfide (C\(S_2\)) reacts with excess oxygen gas (\(O_2\)), we need to use stoichiometry and the given balanced chemical equation:

C\(S_2\)(l) + 3\(O_2\)(g) → C\(O_2\)(g) + 2S\(O_2\)(g)

First, we calculate the number of moles of C\(S_2\) using its molar mass:

Molar mass of (C\(S_2\)) = 12.01 g/mol (C) + 32.07 g/mol (S) × 2 = 76.14 g/mol

Number of moles of (C\(S_2\)) = mass / molar mass = 37.8 g / 76.14 g/mol ≈ 0.496 mol

From the balanced equation, we can see that the stoichiometric ratio between (C\(S_2\)) and C\(O_2\) is 1:1. Therefore, the number of moles of C\(O_2\) produced will also be 0.496 mol.

Now we can use the ideal gas law to calculate the volume of C\(O_2\) at the given temperature and pressure. The ideal gas law equation is:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin.

Converting the temperature from Celsius to Kelvin:

T(K) = 28.85°C + 273.15 = 302 K

Using the ideal gas law:

V = nRT / P = (0.496 mol) × (0.0821 L·atm/mol·K) × (302 K) / (1.02 atm) ≈ 11.9 L

The correct answer is 11.9 liters.

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Answer:

potential of hydroxide

Explanation:

Answer:

13.7g

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

3Ca(OH)2(s) + 2H3PO4(aq) —> Ca3(PO4)2(aq) + 12H2O(l)

Step 2:

Determination of the masses of Ca(OH)2 and H3PO4 that reacted and the mass of Ca3(PO4)2 produced from the balanced equation.

Molar mass of Ca(OH)2 = 40 + 2(16 + 1) = 74g/mol

Mass of Ca(OH)2 from the balanced equation = 3 x 74 = 222g

Molar mass of H3PO4 = (3x1) + 31 + (16x4) = 98g/mol

Mass of H3PO4 from the balanced equation = 2 x 98 = 196g

Molar mass of Ca3(PO4)2 = (40x3) + 2[31 + (16x4)]

= 120 + 2[95] = 310g/mol

Mass of Ca3(PO4)2 from the balanced equation = 1 x 310 = 310g.

From the balanced equation above,

222g of Ca(OH)2 reacted with 196g of H3PO4 to produce 310g of Ca3(PO4)2.

Step 3:

Determination of the limiting reactant.

This is illustrated below:

From the balanced equation above,

222g of Ca(OH)2 reacted with 196g of H3PO4.

Therefore, 9.8g of Ca(OH)2 will react with = (9.8 x 196)/222 = 8.65g of H3PO4

From the above illustration, we can see that only 8.65g of H3PO4 out 9.8g given reacted completely with 9.8g of Ca(OH)2. Therefore, Ca(OH)2 is the limiting reactant and H3PO4 is the excess reactant.

Step 4:

Determination of the maximum mass of Ca3(PO4)2 produced from reaction.

In this case, the limiting reactant will be used as all of it is used up in the reaction.

The limiting reactant is Ca(OH)2 and maximum mass of Ca3(PO4)2 produced can be obtained as follow:

From the balanced equation above,

222g of Ca(OH)2 reacted to produce 310g of Ca3(PO4)2.

Therefore, 9.8g of Ca(OH)2 will react to produce = (9.8 x 310)/222 = 13.7g of Ca3(PO4)2.

Therefore, the maximum mass of Ca3(PO4)2 produced is 13.7g

Taking into account the reaction stoichiometry and limiting reagent, 13.68 grams of Ca₃(PO₄)₂ are formed when 9.80 g of Ca(OH)₂ and 9.80 g of  H₃PO₄ react.

In first place, the balanced reaction is:

3 Ca(OH)₂  + 2 H₃PO₄ → Ca₃(PO₄)₂ + 6 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

The molar mass of the compounds is:

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, it is possible to use the reaction stoichiometry of the reaction a simple rule of three as follows: if by stoichiometry 196 grams of H₃PO₄ reacts with 222 grams of Ca(OH)₂, if 9.80 grams of H₃PO₄ react with how much mass of Ca(OH)₂ will be needed?

\(mass of Ca(OH)_{2} =\frac{9.80 grams of H_{3} PO_{4}x222 grams of Ca(OH)_{2}}{196 grams of H_{3} PO_{4}}\)

mass of Ca(OH)₂= 11.1 grams

But 11.1 grams of Ca(OH)₂ are not available, 9.80 moles are available. Since you have less moles than you need to react with 9.80 grams of H₃PO₄, Ca(OH)₂ will be the limiting reagent.

Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 222 grams of Ca(OH)₂ form 310 grams of Ca₃(PO₄)₂, 9.80 grams of Ca(OH)₂ form how much mass of Ca₃(PO₄)₂?

\(mass of Ca_{3} (PO_{4} )_{2} =\frac{9.80 grams of Ca(OH)_{2}x310 grams of Ca_{3} (PO_{4} )_{2} }{222 grams of Ca(OH)_{2}}\)

mass of Ca₃(PO₄)₂= 13.68 grams

Then, 13.68 grams of Ca₃(PO₄)₂ are formed when 9.80 g of Ca(OH)₂ and 9.80 g of  H₃PO₄ react.

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The equilibrium constant for the reaction given that the equilibrium concentration of [PH₃] = 0.250 M, [H₂] = 0.580 M, and [P₄] = 0.750 M is 7.3

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

The equilibrium constant for the reaction can be obtained as shown below:

Equilibrium constant = [Product]ᵐ / [Reactant]ⁿ

Where

Equilibrium constant = [H₂]⁶[P₄] / [PH₃]⁴

Equilibrium constant = [(0.580)⁶ × 0.750] / (0.250)⁴

Equilibrium constant = 7.3

Thus, the equilibrium constant for the reaction is 7.3

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Answer:

false

Explanation:

The counter-contemporary mechanism Increasing salt consumption expanded sodium excretion, however additionally suddenly precipitated the kidney to preserve water.

Excess sodium become for this reason launched in focused urine. This technique of protective the body's water become so green that the guys without a doubt drank much less whilst their salt consumption become highest.

The counter-contemporary multiplier or the countercurrent mechanism is used to pay attention urine withinside the kidneys with the aid of using the nephrons of the human excretory system. The nephrons concerned withinside the formation of focused urine increase all of the manner from the cortex of the kidney to the medulla and are observed with the aid of using vasa recta.As already indicated, the loop of Henle is essential to the cappotential of the kidney to pay attention urine.

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Answer:

30.1 moles of CaC₂  are needed.

Explanation:

Given data:

Number of moles of CaC₂ needed = ?

Number of moles of H₂O = 60.2 mol

Solution:

Chemical equation;

CaC₂ + 2H₂O      →    C₂H₂ + Ca(OH)₂

Now we will compare the moles of CaC₂ and H₂O.

                           H₂O       :       CaC₂        

                             2          :          1

                           60.2      :        1/2×60.2 = 30.1 mol

30.1 moles of CaC₂  are needed.

The oxidation state of nitrogen (N) in NaNO3 is +5. option B

To determine the oxidation state of nitrogen (N) in sodium nitrate (NaNO3), we need to assign oxidation numbers to each element in the compound.

In NaNO3, we know that the sodium ion (Na+) has a +1 oxidation state because it is an alkali metal. Oxygen (O) typically has an oxidation state of -2 in compounds, and there are three oxygen atoms in NaNO3. Since the compound is neutral, the sum of the oxidation states must be zero.

Let's assume that the oxidation state of nitrogen is x. Therefore, we can set up the equation:

(+1) + x + (-2) * 3 = 0

Simplifying the equation:

+1 + x - 6 = 0

x - 5 = 0

x = +5

Therefore, the oxidation state of nitrogen (N) in NaNO3 is +5.

The oxidation state of an element indicates the number of electrons it has gained or lost in a compound. In this case, the nitrogen atom in NaNO3 has gained five electrons to achieve a stable oxidation state of +5.

It is important to note that oxidation states are formal charges and do not necessarily represent the actual distribution of electrons in a compound. They are assigned based on a set of rules and can be useful in understanding the reactivity and behavior of elements in chemical reactions.

Option B

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Answer:

C. Lightening

Answer:

C.

Explanation:

When the pH is 7 there is equal amount of H+ and OH so the solution added must be strong enough to nuetralize the acid. So option A is out B is wrong because adding a acidic solution to an acidic solution wont nuetralize it, D is wrong because if the 2 solution was already equal in both it would essentaly be water. ALthough water would raise the pH it would not nuetralize it to a even 7.

The intermolecular force between bromine and benzene is primarily a van der Waals force known as London dispersion forces.

London dispersion forces occur due to temporary fluctuations in electron distribution, creating temporary dipoles in molecules. In the case of bromine and benzene, both molecules are nonpolar, meaning they have no permanent dipole. However, they still experience London dispersion forces.

Benzene is a cyclic aromatic hydrocarbon with a hexagonal ring structure. It consists of delocalized π electrons above and below the plane of the molecule. Bromine is a halogen element with seven valence electrons, which forms diatomic molecules. In the solid or liquid state, bromine molecules exist as Br2.

The London dispersion forces between bromine and benzene arise from the temporary shifts in electron density within their electron clouds. The π electrons in benzene induce temporary dipoles in the bromine molecules, resulting in attractive forces between them. These temporary dipoles continuously form and break due to electron movements, resulting in an overall attractive force between the bromine and benzene molecules.

While London dispersion forces are generally weaker than other intermolecular forces like dipole-dipole interactions or hydrogen bonding, they still contribute to the stability of the system and affect properties such as boiling points and melting points.

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The solubility of a substance, such as sugar, depends on several factors, including temperature, pressure, and the presence of other solutes. In this case, the student adds the same amount of sugar and tea bag to the same amount of hot water every day, but the temperature of the water could vary from day to day, affecting how much sugar dissolves.

According to the graph, the least amount of sugar dissolved on day 3, while the most sugar dissolved on day 5. The difference in temperature on these days could explain this variation. On day 3, the water may have been cooler, making it more difficult for the sugar to dissolve. On the other hand, on day 5, the water may have been hotter, which could have increased the solubility of the sugar.

To increase the amount of sugar that dissolves in the tea, the student could try using hotter water or stirring the sugar more vigorously to distribute it evenly throughout the water. Alternatively, the student could try adding the sugar gradually while stirring to give it more time to dissolve before adding more.

Answer:

I'm not 100% sure on this because mine hasn't been graded yet, but here are the answers I submitted.

Explanation:

Which day the least sugar dissolved and which day the most sugar dissolved:

Day 4 is when the least sugar dissolved, and Day 2 is when the most sugar dissolved.

What could have caused less sugar to dissolve on some days:

The temperature of the tea could have caused the sugar not to dissolve on some days.

What the student could do to his drink to make more sugar dissolve.

The student would need to add the sugar to the tea as soon as it's done boiling.  The Sugar will dissolve faster in a warmer tea due to more energy of movement.  

Be sure to stir the tea as you add the sugar.  Stirring the sugar into the tea speeds up the rate of dissolving by helping distribute the sugar particles throughout the tea.  

If the student were to use granulated sugar, those are smaller particles and have greater surface area.  Greater surface area allows for more contact between the tea and the sugar.

Netiquette is the acting or the behavior to communicate on the internet or online. Thus, option D is accurate.

The behavior or the manners of good conduct on the internet platforms and social media exhibited by a person is the online etiquette.

The word netiquette is a combination of the word internet and etiquette, which means the manner or conduct on online or internet platforms.

Therefore, option D. rules and manners to behave online is the correct meaning of netiquette.

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So, what's shown here is the ion product of pure water: that is, the product of the concentrations of hydronium and hydroxide ions in pure water at 25 °C. By this relation, if you know the [H₃O⁺], you can calculate the [OH⁻], and vice-versa.

Since [H₃O⁺] × [OH⁻] = 1.0 × 10⁻¹⁴, [OH⁻] = (1.0 × 10⁻¹⁴)/[H₃O⁺].

Substituting the given [H₃O⁺] as 1.25 × 10⁻² M:

[OH⁻] = (1.0 × 10⁻¹⁴)/(1.25 × 10⁻² M) = 8.0 × 10⁻¹³ M.

The pOH of the 0.001 M NaOH solution is approximately 3.

To determine the pOH of a solution, we need to know the concentration of hydroxide ions (OH-) in the solution.

In the case of a 0.001 M NaOH solution, we can assume that all of the NaOH dissociates completely in water to form Na+ and OH- ions. Therefore, the concentration of hydroxide ions in the solution is also 0.001 M.

The pOH is calculated using the equation:

pOH = -log[OH-]

Substituting the concentration of hydroxide ions, we have:

pOH = -log(0.001)

Using a calculator, we can evaluate the logarithm:

pOH ≈ 3

Therefore, the pOH of the 0.001 M NaOH solution is approximately 3.

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In the three-dimensional structure of methane (CH₄), the hydrogen atoms attached to a carbon atom are aligned at the corners of a tetrahedron.

The 3D methane has tetrahedral structure in which central atom is carbon which has four valance electron.

Each of the valence electron of carbon is shared by hydrogen atom as there are 4 valence electron in (CH₄) so there will be 4 hydrogen atom, and each hydrogen atom is aligned at the corner of tetrahedral structure of methane.

In 3D figure of methane (CH₄) the central atom (carbon) is bonded to four hydrogen atoms. Hence, the central atom (carbon) is having four electron pairs and all pairs are bonding pairs and lacks any lone pair of electron. According to Valence Shell Electron Pair Repulsion (VSEPR) Theory the central atom with four bonding pair electrons and zero lone pair electrons will always have a tetrahedral geometry with bond angles of 109°. Therefore, the shape of CH₄ is tetrahedral.

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Answer:

The difference between an ideal and a nonideal solution is given below:-

Explanation:

Ideal Solution:  

Non-ideal Solution:  

6.02 x \(10^{20\) formula units of MgCl₂ is equal to 0.1 moles of MgCl₂.

To convert formula units of MgCl₂ to moles of MgCl₂, we need to use Avogadro's number, which relates the number of formula units to the number of moles.

Avogadro's number (NA) is approximately 6.022 x 10^23 formula units per mole.

Given that we have 6.02 x 10^20 formula units of MgCl₂, we can set up a conversion factor to convert to moles:

(6.02 x 10^20 formula units MgCl₂) * (1 mol MgCl₂ / (6.022 x 10^23 formula units MgCl₂))

The formula units of MgCl₂ cancel out, and we are left with moles of MgCl₂:

(6.02 x 10^20) * (1 mol / 6.022 x 10^23) = 0.1 mol

Therefore, 6.02 x 10^20 formula units of MgCl₂ is equal to 0.1 moles of MgCl₂.

It's important to note that this conversion assumes that each formula unit of MgCl₂ represents one mole of MgCl₂. This is based on the stoichiometry of the compound, where there is one mole of MgCl₂ for every one formula unit.

Additionally, this conversion is valid for any substance, not just MgCl₂, as long as you know the value of Avogadro's number and the number of formula units or particles you have.

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Answer:

so so

Explanation:

this your question?? <_>

Answer:

110.78 K

Explanation:

First we convert 22.6 g of krypton gas (Kr) to moles, using its molar mass:

Then we use the PV=nRT formula, where:

2.71 atm * 0.904 L = 0.270 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * T

And solve for T:

Answer:

B.K2(OH)2 i think that is the answer

Answer:

\(2Fe +3Cl_{2} -> 2FeCl_{3}\)

Explanation:

In order to have a balanced chemical reaction, you need coefficients of 2, 3 and 2 before each substance in this equation. Chlorine is diatomic so it occurs as a molecular pair when a gas.

Answer: The new volume at different given temperatures are as follows.

(a) 109.81 mL

(b) 768.65 mL

(c) 18052.38 mL

Explanation:

Given: \(V_{1}\) = 571 mL,       \(T_{1} = 26^{o}C\)

(a) \(T_{2} = 5^{o}C\)

The new volume is calculated as follows.

\(\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{5^{o}C}\\V_{2} = 109.81 mL\)

(b) \(T_{2} = 95^{o}F\)

Convert degree Fahrenheit into degree Cesius as follows.

\((1^{o}F - 32) \times \frac{5}{9} = ^{o}C\\(95^{o}F - 32) \times \frac{5}{9} = 35^{o}C\)

The new volume is calculated as follows.

\(\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{35^{o}C}\\V_{2} = 768.65 mL\)

(c) \(T_{2} = 1095 K = (1095 - 273)^{o}C = 822^{o}C\)

The new volume is calculated as follows.

\(\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{822^{o}C}\\V_{2} = 18052.38 mL\)