A 50.0-ml volume of 0.15 m hbr is titrated with 0.25 m koh. calculate the ph after the addition of 14.0 ml of koh. express your answer numerically.

Answers

Answer 1
Final answer:

To find the pH after adding 14.0 mL of 0.25 M KOH to 50.0 mL of 0.15 M HBr, calculate the excess HBr and then use its concentration to determine the pH, assuming HBr dissociates completely as it's a strong acid.

Explanation:

To calculate the pH after the addition of 14.0 mL of 0.25 M KOH to 50.0 mL of 0.15 M HBr, we first need to determine if the reaction has reached the equivalence point. The millimoles of HBr initially present are calculated by multiplying the volume in liters by the molarity: 50.0 mL x 0.15 M = 7.5 mmol. Then, calculate the millimoles of KOH added: 14.0 mL x 0.25 M = 3.5 mmol.

Since we have more HBr than KOH, HBr is in excess and KOH is the limiting reactant. The excess amount of HBr is 7.5 mmol - 3.5 mmol = 4.0 mmol. The pH is determined by the concentration of the remaining HBr. To find this concentration, we take the remaining mmol of acid and divide by the total volume of the solution in liters (original acid solution plus the volume of KOH added).

The total volume after the addition of KOH is 50.0 mL + 14.0 mL = 64.0 mL or 0.064 L. The concentration of HBr is 4.0 mmol / 0.064 L. Now, convert mmol to mol by dividing by 1000, resulting in 4.0 x 10-3 mol / 0.064 L. Since HBr is a strong acid, it dissociates completely in water. The pH can be calculated using the formula pH = -log[H+], where [H+] is the concentration of hydronium ions, which is equal to the concentration of HBr.

Therefore, the pH is -log(4.0 x 10-3 / 0.064 L). Calculate this value to get the pH of the solution after the addition of 14.0 mL of KOH.

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Related Questions

The normal boiling point of 2-propanol, (ch3)2choh, is 83 ºc, while that of acetone, (ch3)2c=o, is 56 ºc. what is the principal reason for the greater boiling point of 2- propanol?

Answers

2-Propanol has higher boiling point than acetone due to the inter-molecular hydrogen bond formation of 2-propanol which increase the intermolecular forces between molecules and leads to higher boiling point in propanol while in case of acetone the intermolecular present between molecules are dipole-dipole interaction which has smaller effect when compared to hydrogen bond  

Calculate the change in entropy that occurs in the system when 48.6 g of water (h2o) vaporizes from a liquid to a gas at its boiling point (100.0 ∘c). the heat of vaporization is 40.7 kj/mol.

Answers

The water is vaporizing at 100°C. The vaporizing water is the system to be studied. Given the mass of water is 48.6 g. The chemical formula of water is H₂O. The molar mass of water is 18 g/ mol
moles of water are:
moles = mass / molar mass = 48.6 / 18 = 2.7 mol
The water is vaporizing, Thus, it gains the heat energy to vaporize
The amount of heat gained by the water is:
Q = n ΔH⁰ vap
   = 2.7 mol x 40.7 kJ/mol x 1000J / 1kJ = 109890 J
Q is the amount of heat lost or gained, n is the number of moles of water and ΔH⁰vap is heat of vaporization.
The expression for entropy change is:
ΔS = Q / T = 109890 J / 373 K = 294.6 J/K
The change in the entropy of the system to three significant digits is 295 J/K

When 48.6 g of water vaporizes at its boiling point (100.0 °C), the change in the entropy is 0.292 kJ/K.

First, we will calculate the change in the enthalpy (ΔH) when 48.6 g of water vaporizes considering the following relationships.

The heat of vaporization of water is 40.7 kJ/mol.The molar mass of water is 18.02 g/mol.

[tex]\Delta H = 48.06 g \times \frac{1mol}{18.02g} \times \frac{40.7kJ}{mol} = 109kJ[/tex]

Then, we will convert 100.0 °C (T) to Kelvin using the following expression.

[tex]T = K = \° C + 273.15 = 100.0\° C + 273.15 = 373.2 K[/tex]

Finally, we will calculate the change in the entropy (ΔS) for this process using the following expression.

[tex]\Delta S = \frac{\Delta H }{T} = \frac{109kJ}{373.2K} = 0.292kJ/K[/tex]

When 48.6 g of water vaporizes at its boiling point (100.0 °C), the change in the entropy is 0.292 kJ/K.

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PLEASE ANSWER FAST!!!

2/ Immediately after the Big Bang, the universe began to?
A/ cool down
B/ heat up
C/shrink
D/ condense

3/ A ____ Is a system of billions of stars and all of the planets that orbit around them?
A/ solar system
B/ universe
C/ galaxy
D/ planet

5/ Light energy moves in?
A/ waves
B/particles
C/ condensation
D/ plasma

6/ All of matter and energy in the universe was once contained in?
A/ a star
B/ a galaxy
C/ an element
D/ a singularity

Answers

2- Condense
3- ?
5- Waves
6- Star
I hope this helps! :)
For the question 1. It is A started to cool down

At a certain temperature and pressure, 0.20 mol of carbon dioxide has a volume of 3.1 L. A 3.1-L sample of hydrogen at the same temperature and pressure ____.

Answers

Answer:
            A 3.1 L sample of hydrogen at the same temperature and pressure contains the same number of molecules.

Explanation:
                   As we know ;"1 mole of any gas at STP occupies exactly 22.4 L of volume". Therefore, CO₂ and H₂ gas having same number of moles (0.20 moles) will occupy exactly the same volume.

Now, converting moles into number of molecules,

# of Molecules  =  Moles × 6.022 × 10²³

Putting value of mole,

# of Molecules  =  0.2 mol × 6.022 × 10²³ molecules.mol⁻¹

# of Molecules  =  1.20 × 10²³ molecules

Hence, both gases will contain 1.20 × 10²³ molecules.

A 3.1 L sample of hydrogen at the same temperature and pressure as 0.20 mol of CO₂ will also contain 0.20 mol of hydrogen.

To answer the question, we need to understand the relationship between volume, temperature, and pressure for gases, as explained by the Ideal Gas Law.

The Ideal Gas Law is given by the equation PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature in Kelvin.

Since we are told that the volume sample of hydrogen is at the same temperature and pressure as the carbon dioxide, we can directly relate the two based on their moles.

For carbon dioxide (CO₂):

n = 0.20 mol

V = 3.1 L

This relationship tells us that 0.20 moles of CO₂ occupies 3.1 L at the given temperature and pressure.

Applying the same conditions to hydrogen (H₂), a 3.1 L sample of hydrogen gas will contain the same number of moles as the CO₂ under the same conditions:

0.20 mol of Hydrogen (H₂)

Therefore, 3.1 L of hydrogen at the same temperature and pressure will also contain 0.20 mol of hydrogen gas.

A supervisor spends a day inspecting a nuclear plant for potential radiation leaks. She has to move throughout the plant inspecting all the equipment and machinery. She needs to take two different radiation detection devices to help ensure her safety and to find radiation leaks. She needs the results immediately. Which two devices would be the best choices for the task?

Geiger counter and scintillation counter
Geiger counter and cloud chamber
cloud chamber and scintillation counter
film badge and scintillation counter

Answers

Answer:

Geiger counter and scintillation counter

Explanation:

The Geiger counter was invented by Hans Geiger in 1908 to measure the levels of radiation in bodies and the environment, so it is one of the indispensable equipment for the inspector to detect radiation leaks in a nuclear power plant. It contains a tube with argon, which ionizes by being crossed by alpha and beta particles of radiation, closing the electric circuit and triggering the counter.

Similarly, a scintillation detector is an apparatus used to detect ionizing radiation. When something in the environment has been reached by radiation, this detector emits a small ray of light, indicating the radiation contamination.

Answer:

A. Geiger counter and scintillation counter

Explanation:

When titrating a monoprotic strong acid with a weak base at 25°c, the
a.ph will be less than 7 at the equivalence point.
b.ph will be 7 at the equivalence point.
c.titration will require more moles of the base than acid to reach the equivalence point.
d.titration will require more moles of acid than base to reach the equivalence point.
e.ph will be greater than 7 at the equivalence point?

Answers

When a strong monoprotic acid is Titrated with a weak base at 25° ;

The pH will be less than 7 at the equivalence point ( A )

A monoprotic acid donates only a single proton in a titration experiment therefore at the equivalence point in an experiment involving the reaction between the strong monoprotic acid with a weak base, all the base ions will react, while the strong acid will have some unreacted ions  ( H⁺) left in the solution.

The unreactive protons of the strong monoprotic acid present in the solution will make the solution acidic therefore the pH of the solution will be less than 7 at the equivalence point.

Hence we can conclude that when a strong monoprotic acid is titrated with a weak base at 25°, the pH will be less than 7 at the equivalence point.

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Final answer:

In a titration of a monoprotic strong acid with a weak base, the pH will be less than 7 at the equivalence point because the conjugate acid of the weak base will slightly ionize, rendering the solution acidic at this point.

Explanation:

When titrating a monoprotic strong acid with a weak base at 25°C, the pH at the equivalence point will be less than 7. This is because the reaction at the equivalence point produces the conjugate acid of the weak base, which slightly ionizes in solution, contributing to an acidic pH. As outlined in resources such as LibreTexts, the equivalence point's pH depends on the strength of the acid and base involved in the titration. In the case of a strong acid with a weak base, the solution will be acidic because the weak base is not strong enough to fully neutralize the strong acid. Therefore, the correct answer to the question is a. pH will be less than 7 at the equivalence point. It is also important to note that the number of moles of base and acid required to reach the equivalence point depends solely on their stoichiometry and not on their strength, meaning one mole of acid will react with one mole of base to reach the equivalence point.

What is the volume of oxygen occupied by 2 moles at 1.3 atm pressure and 300 K? Use PV = nRT.

Answers

The  volume   of oxygen gas that occupied by  2 moles at 1.3 atm pressure and 300 k is calculated  using the ideal gas equation that is Pv =nRT

P(pressure) = 1.3 atm
R(gas  constant) =0.082 l.atm/mol.k
n (moles)= 2 moles
T(temperature) =300k
V(volume)=?

by making  the v the subject of the formula V =nRT/P

=(2 moles x 0.082 l.atm/mol.k x300 K)/ 1.3 atm= 37.85 Liters
we can use the ideal gas law equation to find the volume occupied by O₂
PV = nRT
where P - pressure - 1.3 atm x 101 325 Pa/atm = 131 723 Pa
V - volume 
n - number of moles - 2 mol 
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 300 K
Substituting these values in the equation 
131 723 Pa x V = 2 mol x 8.314 Jmol⁻¹K⁻¹ x 300 K
V = 37.9 L
volume occupied is 37.9 L

What do oxidation reactions produce during the electrolysis of water?

Answers

Final answer:

Oxidation reactions during the electrolysis of water produce oxygen gas (O₂) at the anode. The process requires an electrolytic cell with an electrolyte such as sulfuric acid to facilitate the reaction, and it results in a stoichiometric ratio with twice the volume of hydrogen gas produced compared to oxygen gas.

Explanation:

During the electrolysis of water, oxidation reactions occur at the anode, producing oxygen gas (O₂). The reaction involves water molecules losing electrons to form oxygen gas and hydrogen ions. The overall chemical reaction for the oxidation process at the anode can be represented as 2H₂O(l) → O₂(g) + 4H⁺ + 4e⁻. Oxygen gas is released at the anode, while hydrogen gas is produced at the cathode.

In an electrolytic cell with platinum electrodes and an electrolyte like sulfuric acid (H₂SO₄), water undergoes electrolysis to form these gases. It's worth noting that there are twice as many hydrogen atoms as oxygen atoms, and because they are both diatomic (exist as H₂ and O₂), twice the volume of hydrogen gas is produced compared to oxygen gas. This stoichiometric relationship is essential for understanding the mass and volume ratios of the gases produced during water electrolysis.

A glucose solution contains 55.8 g of glucose (c6h12o6) in 455 g of water. calculate the freezing point and boiling point of the solution. ( density of water = 1.00 g/ml, kb= 0.512 o c kg solvent/mol solute and 1.86°c kg/mol)

Answers

Answer is: the freezing point of the solution of glucose is -1.26°C and boiling point is 100.353°C.
m(H₂O) = 455 g ÷ 1000 g/kg = 0.455 kg.
m(C₆H₁₂O₆) = 55.8 g. 

n(C₆H₁₂O₆) = m(C₆H₁₂O₆)÷ M(C₆H₁₂O₆).
n(C₆H₁₂O₆) = 55.8 g ÷ 180.16 g/mol.
n(C₆H₁₂O₆) = 0.31 mol.
b(solution) = n(C₆H₁₂O₆) ÷ m(H₂O).
b(solution) = 0.31 mol ÷ 0.455 kg.
b(solution) = 0.68 mol/kg.
ΔTf = b(solution) · Kf(H₂O).
ΔTf = 0.68 mol/kg · 1.86°C·kg/mol.

ΔTf = 1.26°C.
Tf = 0°C - 1.26°C = -1.26°C.

ΔTb = b(solution) · Kb(H₂O).

ΔTb = 0.68 mol/kg · 0.52°C·kg/mol.

ΔTb = 0.353°C.

Tb = 100°C + 0.353°C.

The boiling point of the solution is 100.35 °c.

We know that;

ΔT = K m i

ΔT = freezing point depression

K = freezing constant

m = molality

i = Van't Hoff factor

Hence;

ΔT =  1.86°c kg/mol × 55.8 g/180 g/mol × 1/0.455 × 1

ΔT = 1.27 °c

Freezing point = 0 - 1.27 °c = - 1.27 °c

For boiling point;

ΔT = K m i

ΔT = 0.512 o c kg × 55.8 g/180 g/mol × 1/0.455 × 1

ΔT = 0.35 °c

Boiling point = 100 +  0.35 °c = 100.35 °c

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The ratio of oxygen-16 and oxygen-18 isotopes in plankton fossils in deep-sea sediments can be used to determine ________.

Answers

The ratio of oxygen-16 and oxygen-18 isotopes in plankton fossils in deep-sea sediments can be used to determine past temperatures.

Answer;

-past temperatures

The ratio of oxygen-16 and oxygen-18 isotopes in plankton fossils in deep-sea sediments can be used to determine past temperatures.

Explanation;

-O-16 will evaporate more readily than O-18 since it is lighter, therefore; during a warm period, the relative amount of O-18 will increase in the ocean waters since more of the O-16 is evaporating.

-Hence, looking at the ratio of O16 to O18 in the past can give clues about global temperatures.

A student adds too much hcl during the titration. will the calculated ksp be too high, too low, or unaffected? why?

Answers

A student adds too much HCl during the titration, the calculated ksp be too high.

Reason
Ksp refers to solubility product. It is a measure of solute present in solution. Now, when a sparingly solution base, like calcium hydroxide, is titrated with an acid, like HCl. The reaction results in generation of salt (in present case CaCl2) and water. The solubility of salt is higher as compared to sparingly soluble base. So during the  course of reaction, Ca^2+ ions present in system will combine with Cl^- ions to form CaCl2. This will result in decreasing in conc. of Ca^2+ ions in solution. To compensate for this lose, more Ca^2+ ions from Ca(OH)2 will dissolve in solution. Thus, Ksp value will increase. 

What would happen to the rate of a reaction with rate law rate = k [NO]2[H2] if the concentration of NO were doubled?
A. The rate would be halved.
B. The rate would also be doubled.
C. The rate would not change.
D. The rate would be four times larger.

Answers

The answer will be ( D )

The rate would be four times larger.

when the law rate = K [NO]^2[H2]

so, for example when we assume that [NO] = 2 and K = 2 and [H2] = 1 

∴ law rate = 2* 2^2 * 1 = 8

and when the [NO] is doubled that means it becomes 4

∴ law rate = 2 * 4^2 *1 = 32 


when the first law rate = 8 and after [NO] doubled = 32 

the rate would be four-time larger

Answer:

The rate would be one-fourth. I just did it

Explanation:

Write the equilibrium-constant kp expression for the reaction a(g) + 2b(l) = 4c(g) + d(g)

Answers

Final answer:

For the reaction a(g) + 2b(l) = 4c(g) + d(g), the equilibrium constant expression using partial pressures, Kp, is Kp = (Pc)^4 (Pd) / (Pa).

Explanation:

To write the equilibrium-constant expression, Kp, for the reaction a(g) + 2b(l) = 4c(g) + d(g), we apply the principles of equilibrium for gases. Kp is an equilibrium constant calculated from partial pressures of gas-phase reactants and products at equilibrium. The liquids are not included in the Kp expression since their activities are considered constants under standard conditions and do not affect the equilibrium of gases.

The general form of an equilibrium constant expression for a reaction is Kp = (Pc)c (Pd)d / (Pa)a (Pb)b, where P represents the partial pressure of each gas, the lower-case letters right below the P are the chemical species, and the upper-case letters indicate the stoichiometric coefficients from the balanced equation.

Using the reaction given, we write the Kp expression as follows:

Kp = (Pc)4 (Pd) / (Pa)

Note that in our Kp expression, we do not include B since it is in the liquid state.

In the following reaction, is the nitrogen reduced or oxidized? 2Cu2O + 2NO 4CuO + N2

Answers

Nitrogen has reduced.

In the reaction, Cu₂O goes to CuO.
Oxidation state in 
         Cu₂O = +1                                                  CuO = +2
              2a + (-2) x 1 = 0                                            a + (-2) x 1 = 0
                              2a = +2                                                         a = +2
                                 a = 1
Hence, Cu has oxidized.

NO goes to N₂ by reducing its oxidation state from +2 to 0.
Oxidation state in 
         NO = +2                                                  
              a + (-2) x 1 = 0                                            
                              a = +2                                                         
                                 

In the following reaction, the nitrogen is reduced.
2Cu2O + 2NO → 4CuO + N2

If two nonmetals with the same electronegativity bond, what type of bond will form?
A. Metallic bond
B. Non polar ionic bond
C. Ionic bond
D. Non polar covalent bond

Answers

If two nonmetals with the same electronegativity bond, what type of bond will form then Non polar covalent bond (option D) will form.

Answer:

Two nonmetals with the same electronegativity will form a non polar covalent bond.

Explanation:

The type of bond between atoms is classified in 3 big groups:

Metallic bond: this type of bond only take place between metallic atoms like Cu, Al, Au, etc.Ionic bond: this type of bond is formed between ions, that means that it is necessary the presence of a cation (ion with positive charge) and and an anion (ion with negative charge) and when the atoms has a very high difference of electronegativity (more that 2), that makes the ionic bond always polar, because there will be always a positive pole (cation) and a negative pole (anion). This is common between a metal and a nonmetal, for example: sodium chloride (NaCl).Covalent bond: this type of bond occurs when atoms share one or more pairs of electrons, this happens between nonmetals, e.g.: the molecule of chlorine gas (Cl₂).

Apart from that, depending on the electronegativity difference, the covalent bonds are clasified in polar and non polar:

- Polar covalent bond: the difference of electronegativity is important but less than an ionic bond (between 0 and 2).

- Non polar covalent bond: this occurs when the atoms forming bonds have the same electronegativity.

So, analyzing the statement, if we have two nonmetals it is a covalent bond, and if the two nonmetals atoms have the same electronegativity the bond will be non polar.

Using oxidation and reduction half-reactions, balance the skeletal equation cl2o7(g) + h2o2(aq) → clo− 2 (aq) + o2(g) of the production of chlorite ions from dichlorine heptoxide. the reaction takes place in a basic solution. what is the smallest possible integer coefficient of o2 in the combined balanced equation?

Answers

Final answer:

In order to balance the equation, we identify the oxidation and reduction half-reactions and then combine them. The smallest possible integer coefficient of O₂ in the balanced equation is 10.

Explanation:

To balance the given equation cl2o7(g) + h2o2(aq) → clo− 2 (aq) + o2(g), we first have to identify the oxidation and reduction half-reactions. The oxidation half-reaction is: Cl₂O₇ → 2ClO₂⁻ + 5/2O₂. The reduction half-reaction is: H₂O₂ + 2e⁻ → 2OH⁻. Then, we combine the oxidation and reduction half-reactions: 4 Cl₂O₇ + H₂O₂ → 8 ClO₂⁻ + 2OH⁻ + 10O₂. The smallest possible integer coefficient of O₂ in the combined balanced equation is 10.

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Final answer:

The process of balancing redox reactions involves separating the reaction into two half-reactions, balancing them individually, and then combining them to ensure the total charges cancel out. In the given reaction, Cl2O7(g) + H2O2(aq) → ClO2−(aq) + O2(g), the smallest possible integer coefficient of O2 is 1 as one molecule of H2O2 produces one molecule of O2.

Explanation:

To balance the given redox reaction, Cl2O7(g) + H2O2(aq) → ClO− 2(aq) + O2(g), we split into two half-reactions, one for oxidation, and one for reduction. Assign oxidation numbers to identify which atoms have changed oxidation state during the reaction. The oxidation half-reaction can be identified as H2O2(aq) → O2(g) and the reduction half-reaction as Cl2O7(g)  → ClO− 2(aq).

Balance each of these half-reactions, first for atoms other than hydrogen and oxygen, then for oxygen, then for hydrogen, and lastly for charge. After balancing the half-reactions individually, combine them ensuring the total charges cancel out. The smallest integer coefficient of O2 in balanced equation results from the oxidation half-reaction, where one molecule of H2O2 produces one molecule of O2, thus the smallest possible integer coefficient of O2 is 1.

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How many total atoms are in 0.530 g of P2O5/12131141/dc2696ff?utm_source=registration

Answers

Final answer:

To find the number of atoms in 0.530 g of P2O5, you calculate the number of moles by dividing by the molar mass and then multiply by Avogadro's number, resulting in approximately 2.25 × 1021 atoms.

Explanation:

To determine the number of atoms in 0.530 g of P2O5, we need to calculate the number of moles and then multiply by Avogadro's number (6.022 × 1023 atoms/mol). The molecular weight of P2O5 is found by adding the atomic weights of its constituent atoms: 2P (2 × 30.973761 amu/atom) + 5O (5 × 15.9994 amu/atom), which gives us approximately 141.94 amu. Since molar mass is the mass of one mole of a substance, we can say P2O5 has a molar mass of about 141.94 g/mol.

Now, to find moles, we divide the given mass by the molar mass:
0.530 g / 141.94 g/mol = 0.00373 mol.
Next, we multiply the moles by Avogadro's number to get the total atoms:
0.00373 mol × 6.022 × 1023 atoms/mol = 2.25 × 1021 atoms.


Which is not a product of the fractional distillation of petroleum?

A) Gasoline
B) Jet Fuel
C) steel

Answers

Answer:
            Option-C, STEEL is not a product of the fractional distillation of petroleum.

Explanation:
                   Petroleum is the mixture of Hydrocarbons *carbon and hydrogen containing compounds) present beneath the Earth's surface. Petroleum is formed from the remains of animals and plants beneath earth's surface in an anaerobic conditions.
                   Petroleum contains from small hydrocarbons (gases) to medium (liquids) and long chain hydrocarbons (Solids). These hydrocarbons are separated from each other by Fractional Distillation method (separation due to difference in boiling points)
                   Gasoline is a derivative of one of the fraction of petroleum used in internal combustion engines.
                   Jet Fuel is also derived from Kerosene and Naphtha fractions of petroleum.
                   While, Steel is inorganic Alloy (mixture of metals) composed of mainly Iron, Carbon and other elements.

The process to separate the various components of a liquid mixture is called a fractional distillation true false

Answers

DecantationDecantation is a process for the separation of mixtures, by removing a layer of liquid, generally one from which a precipitate has settled. The purpose may be either to produce a clean decant, or to remove undesired liquid from the precipitate (or other layers). Separate liquids
so it is false

The answer is true. I just had this question on a test and I got it wrong for saying false.

What is the oxidation number of the chromium atom in k2cro4? what is the oxidation number of the chromium atom in ? +2 +6 -3 -7 +4?

Answers

(K⁺¹)2Cr(O⁻²)4

+1*2 +x-2*4=0
2+x-8=0
x=6
Oxidation number of Cr is +6.

Are the statements about hydrogen bonding of the compound below with water true or false? this compound can act as a hydrogen-bond donor. this compound can act as a hydrogen-bond acceptor

Answers

Here we have to judge the statement about the hydrogen bond is true or false for water molecule.

The both the statement about hydrogen bonding in water is true i.e. The water molecule can act as hydrogen-bond donor and acceptor.

The hydrogen bond is a weak interaction between a hydrogen atom and an electronegative atom.

In water (H₂O) there remains hydrogen atom (H) and also the electronegative atom oxygen (O). The donor property and acceptor property to produce hydrogen bond in water is shown in figure.

The donor and acceptor property of water is shown in the figure.

The correct answer is true.

Hope this helps! ;)


Which pair of atoms forms a nonpolar covalent bond? which pair of atoms forms a nonpolar covalent bond? na and cl c and o n and cl b and o?

Answers

Final answer:

Boron and Oxygen, due to their similar electronegativity can form a nonpolar covalent bond. This type of bond is formed when electrons are shared equally between atoms. Sodium and Chlorine, with their contrasting electronegativities, would rather form an ionic bond.

Explanation:

The atom pairs that form a nonpolar covalent bond from the options given would be B and O. Boron (B) and Oxygen (O) come from nonmetals with similar electronegativity, hence they share electrons equally forming a nonpolar covalent bond.

Nonpolar covalent bonds form when two atoms share electrons equally, meaning the electrons spend an equal amount of time around each atom. This could involve two atoms of the same element, like O₂, or atoms of different elements with similar electronegativity, like CH4 (methane).

On the contrary, atoms like Sodium (Na) and Chlorine (Cl), have a large difference in electronegativity, leading to an ionic bond instead of a covalent bond. In such cases, a clearly positive (cation) or negative (anion) charge develops on the atoms.

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The correct answer is: E) None of the above  atoms form a nonpolar covalent bond.

For a nonpolar covalent bond to form, the atoms involved must have similar electronegativities, typically seen in diatomic molecules of the same element

Given options:

A) Na and Cl - This forms an ionic bond.

B) C and O - This forms a polar covalent bond.

C) N and Cl - This forms a polar covalent bond.

D) B and O - This forms a polar covalent bond.

Full question

Which pair of atoms forms a nonpolar covalent bond?

A) Na and Cl  

B) C and O  

C) N and Cl  

D) B and O

E) None of the above.

What is the volume of a balloon of gas at 842 mm Hg and -23° C, if its volume is 915 mL at a pressure of 1,170 mm Hg and a temperature of 24°C?

Answers

Answer:
             V₂  =  1070 mL or 1.07 L

Solution:

Data Given;
                  P₁  =  1170 mmHg

                  V₁  =  915 mL

                  T₁  =  24 °C  +  273 K  =  297 K

                  P₂  =  842 mmHg

                  V₂  =  ?

                  T₂  =  - 23 °C  +  273 K  =  250 K

According to Ideal gas equation,

                       P₁ V₁ / T₁  =  P₂ V₂ / T₂

Solving for V₂,

                       V₂  =  P₁ V₁ T₂ / P₂ T₁

Putting Values,

                       V₂  = (1170 mmHg × 915 mL × 250 K) ÷ (842 mmHg × 297 K)

                       V₂  =  1070 mL or 1.07 L

What is the total number of molecules in a 0.5-mole sample of the He gas?

Answers

Avogadro’s constant tells us the number of atoms in a mole of any substance: 6.022x10^23. A mole of water, oxygen, sodium - they all have this many atoms in a 1 mole sample.

A 0.5 mole sample of helium gas would contain 0.5 x (6.022 x 10^23) atoms, or 3.011 x 10^23 atoms. Still a big amount.

Answer :  The total number of molecules in 0.5 mole of sample of He gas are, [tex]3.011\times 10^{23}[/tex]

Explanation : Given,

Moles of sample of He gas = 0.5 mole

As we know that,

1 mole of gas contains [tex]6.022\times 10^{23}[/tex] number of molecules of gas

As, 1 mole of He gas contains [tex]6.022\times 10^{23}[/tex] number of molecules of He gas

So, 0.5 mole of He gas contains [tex]0.5\times (6.022\times 10^{23})=3.011\times 10^{23}[/tex] number of molecules of He gas

Therefore, the total number of molecules in 0.5 mole of sample of He gas are, [tex]3.011\times 10^{23}[/tex]

A student has a 2.123 g sample of a hydrated salt. she heats it to drive off all the water. after heating to constant mass, the salt weighs 1.861 g. the anhydrous salt is copper(ii) phosphate. what is the empirical formula of the hydrated salt?

Answers

the mass of hydrated salt - 2.123 g
mass of anhydrous salt - 1.861 g
mass that has been reduced is the mass of water that has been heated and lost from the compound thereby making the salt anhydrous.
therefore mass of water lost - 2.123 - 1.861 = 0.262 g
number of moles of water lost - 0.262 g / 18 g/mol = 0.0146 mol 
number of moles of salt - 1.861 g / 380.6 g/mol = 0.00490 mol 
molar ratio of moles of water to moles of salt 
molar ratio = 0.146 mol / 0.00490 mol = 2.98 rounded off to 3
for every 1 mol of salt there are 3 moles of water
therefore empirical formula - Cu₃(PO₄)₂.3H₂O


Given the reaction zn(s) + pb(no3)2 (aq) = pb(s) +zn (no3)2 (aq) the oxidation number of zn metal is

Answers

 The oxidation number of Zn metal is "0",  because it is free element.

How many grams of silver would have to be dissolved in 1120?

Answers

Missing question: How many grams of silver would have to be dissolved in 1120 g of ethanol to lower the freezing point by 0.25ºC?Answer is: 15.18 grams of silver would have to be dissolved.
Kf(ethanol) = 1.99°C/m.
ΔT = Kf(ethanol) · b(solution).
0.25°C = 1.99°C/m · b(solution).b(solution) = 0.126 m.
m(ethanol) = 1120 g ÷ 1000 g/kg = 1.12 kg.
n(Ag) = m(ethanol) · b(solution).
n(Ag) = 0.126 m · 1.12 kg.
n(Ag) = 0.140 mol.
m(Ag) = n(Ag) · M(Ag).
m(Ag) = 0.140 mol · 107.87 g/mol.
m(Ag) = 15.18 g.

About 15.20g of silver would have to be dissolved in 1120 g of ethanol to lower the freezing point by 0.25°C.

To find out how many grams of silver need to be dissolved in 1120 g of ethanol to lower its freezing point by 0.25°C, we can use the formula for freezing point depression:

[tex]\Delta T_f = K_f \cdot m[/tex]

Where:

[tex]\Delta T_f[/tex]​ is the change in freezing point (0.25°C in this case), [tex]K_f[/tex] is the freezing point depression constant for ethanol (1.99°C/m), and m is the molality of the solution.

Step 1: Calculate Molality (m)

Rearranging the formula to solve for molality gives us: [tex]m = \frac{\Delta T_f}{K_f}[/tex]Substituting in the values: [tex]m = \frac{0.25 \textdegree C}{1.99 \textdegree C/m} \approx 0.1256 m[/tex]

Step 2: Calculate Moles of Silver Required

Now, we need to calculate how many moles of silver (Ag) are needed to achieve that molality. Molality is defined as moles of solute per kilogram of solvent. Since we have 1120 g of ethanol, this is equivalent to 1.120 kg.

Using the definition of molality:

[tex]m = \frac {\text{moles of Ag}}{\text{kg of Ethanol}} = 0.1256 \text{m}[/tex]

Thus:

Moles of Ag = m × kg of ethanol = 0.1256 m × 1.120 kg ≈ 0.1407 moles

Step 3: Convert Moles to Grams

Next, we convert moles of silver to grams. The molar mass of silver (Ag) is approximately 107.87 g/mol.Grams of Ag = moles of Ag × molar mass of AgGrams of Ag = 0.1407 moles × 107.87 g/mol ≈ 15.20 g

Therefore, approximately 15.20 grams of silver would need to be dissolved in 1120 g of ethanol to lower the freezing point by 0.25°C.

For the oxidation–reduction reaction equation 2na+s ⟶ na2s indicate how many electrons are transferred in the formation of one formula unit of product.

Answers

2Na⁰ +S⁰ ---> [tex]Na_{2}^{+1}S^{-2}[/tex]

[tex]2Na^{0} -2e^{-} ---\ \textgreater \ 2Na^{+1} S^{0} +2e^{-}----\ \textgreater \ S^{-2} 2 electrons are transferred from Na to S.[/tex]

The given reaction of metallic sodium with sulphur involves two electrons which are lost  from two sodium atoms and gained by the sulphur atom. Thus sodium atom oxidizes from 0 to +1 and sulphur reduces from 0 to -1.

What is redox reaction?

A redox reaction involves oxidation of one reactant species and reduction of other species. The species which loss or donate electrons are oxidized to higher oxidation states whereas, the species which gain one or more electrons are reduced to lower oxidation states.

Metals are electron rich and will lose electrons easily to a non-metal during chemical bonding. Here the valency of sulphur is two thus it needs to gain 2 electrons. One sodium donate one electrons and thus two sodium atoms are needed to react with sulphur.

The oxidation reaction here is :

[tex]\rm 2 Na \rightarrow 2Na^{+} + 2 e^{-}[/tex]

Reduction of sulphur is written as:

[tex]\rm S + 2e ^{-} \rightarrow S^{2-}[/tex]

Therefore, the number of electrons involved in this oxidation -reduction reaction is 2.

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How many moles are equivalent to 2.50x1020 atoms of Fe?

Answers

Final answer:

To find the  moles equivalent to 2.50x10²⁰ atoms of Fe, you divide the given number of atoms by Avogadro's number (6.022x10²³ atoms/mole). This yields approximately 0.415 moles of Fe.

Explanation:

To calculate the number of moles equivalent to 2.50x10²⁰ atoms of Fe (Iron), you can use Avogadro's number, which is 6.022x10²³ atoms/mole. Let's divide the given no. with Avogadro's number. Let's do the computation:

2.50x10²⁰ atoms Fe * (1 mol Fe / 6.022x10²³ atoms Fe) = ~0.415 moles of Fe

This means that 2.50x10²⁰ atoms of Fe is equivalent to 0.415 moles. Avogadro's number is a fundamental constant in chemistry and is used to convert between the atomic scale and macroscopic scale.

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Final answer:

2.50x10^20 atoms of Fe are equivalent to approximately 4.15x10^-4 moles. This is calculated by dividing the number of atoms by Avogadro's number (6.022x10^23 atoms per mole).

Explanation:

To calculate how many moles are equivalent to 2.50x10^20 atoms of Fe, we use Avogadro's number, which states that one mole of any substance contains 6.022x10^23 elementary entities (like atoms). Therefore, to convert the number of atoms to moles, we divide the number of atoms by Avogadro's number.

In this case, (2.50x10^20) / (6.022x10^23), which equals approximately 4.15x10^-4 moles of Fe.

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Calculate the actual yield for the production of ammonia gas (nh3) from hydrogen and nitrogen gases if the percent yield is 68.2% and you begin with 2.00 kg of nitrogen gas

Answers

Percentage yield = (actual yield / theoretical yield) x 100%

Hence,
  actual yield = percentage yield x theoretical yield / 100

The balanced reaction equation for the production of ammonia is
N₂(g) + 3H₂(g) → 2NH₃(g)

The mass of N₂(g)  used = 2.00 kg = 2 x 10³ g
Molar mass of N₂(g) = 28 g mol⁻¹

moles (mol) = mass (g) / molar mass (g/mol)

Hence, moles of N₂(g) = 2 x 10³ g / 28 g mol⁻¹ 
                                     = 71.428 mol

Stoichiometric ratio between N₂(g) and NH₃(g) is 1 : 2

Hence, produced NH₃(g) moles = 71.428 mol x 2
                                                   = 142.856 mol

Molar mass of NH₃(g) = 17 g mol⁻¹
mass of NH₃(g) = 142.856 mol x 17 g mol⁻¹ = 2428.552 g

Hence, the theoretical yield = 2428.552 g

Then the actual yield of NH₃(g) produced = (68.2 x 2428.552 g) / 100
                                                                       = 1656.27 g

Final answer:

To determine the actual yield of ammonia gas, convert the mass of nitrogen gas to moles, calculate the theoretical yield using stoichiometry, and then apply the percent yield. The actual yield for a 68.2% percent yield from 2.00 kg of nitrogen is 1658.41 g NH3.

Explanation:

To calculate the actual yield of ammonia gas (NH3) production from nitrogen (N2) and hydrogen (H2) gases when given the percent yield and mass of nitrogen gas, we'll first need to convert the mass of nitrogen to moles, then calculate the theoretical yield of ammonia based on stoichiometry, and finally use the percent yield to find the actual yield.

Step-by-step Calculation:

Calculate moles of nitrogen: Molecular weight of N2 is 28.02 g/mol. 2.00 kg of N2 is 2000 g. Moles = 2000 g / 28.02 g/mol = 71.38 mol N2.

Using the balanced chemical equation (N2 + 3H2 → 2NH3), we see the stoichiometry is 1:2 for nitrogen to ammonia. So, moles of NH3 = 2 moles NH3/mole N2 × 71.38 mol N2 = 142.76 mol NH3.

Convert moles of NH3 to grams: Molecular weight of NH3 is 17.03 g/mol. The theoretical yield in grams = 142.76 mol NH3 × 17.03 g/mol = 2431.89 g NH3.

Calculate actual yield using the percent yield: Actual Yield = Percent Yield / 100 × Theoretical Yield = 68.2% / 100 × 2431.89 g = 1658.41 g NH3.

Therefore, the actual yield of ammonia when starting with 2.00 kg of nitrogen gas and a percent yield of 68.2% is 1658.41 g.

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