A crystallographer measures the horizontal spacing between molecules in a crystal. The spacing is 10.59 nm . What is the total width of a crystal in millimeters 105 molecules across?
Which of the following substances is a compound
A compound in chemistry is a substance that is formed when two or more elements are chemically bound together, with fixed ratios of the elements. Examples include water (H2O) and carbon dioxide (CO2).
Explanation:In chemistry, a compound is a substance formed when two or more elements are chemically bound together. A given compound will always contain the same elements in fixed ratios. For instance, water (H2O) is a compound because it is made of two hydrogen atoms and one oxygen atom. Similarly, carbon dioxide (CO2) is a compound composed of one carbon atom and two oxygen atoms. Therefore, the identification of a substance as a compound depends on its elemental composition.
A compound is a substance made up of two or more different elements chemically bonded together. From the given options, the substance that is a compound is sodium chloride (NaCl). This is because sodium chloride is formed by the chemical bond between sodium (Na) and chlorine (Cl). It has a fixed ratio of sodium to chlorine atoms, and its properties are different from those of the individual elements.
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How many liters of hcl measured at stp 4.00?
Suppose that the average speed (vrms) of carbon dioxide molecules (molar mass 44.0 g/mol) in a flame is found to be 1.90 105 m/s. what temperature does this represent? g
Identify the compounds that are soluble in both water and hexane. identify the compounds that are soluble in both water and hexane. 1-propanol and 1-pentanol 1-butanol and 1-pentanol ethanol and 1-butanol ethanol and 1-propanol methanol and 1-butanol
The compounds that are soluble in both water and hexane are usually polar compounds with a small size, such as ethanol and 1-butanol.
Explanation:The compounds that are soluble in both water and hexane are usually polar compounds with a small size. This is because water is a polar solvent and hexane is a nonpolar solvent. Polar compounds, such as alcohols, have a hydroxyl group that can interact with the polar water molecules, while their nonpolar alkyl chain allows them to dissolve in hexane.
Out of the given options, ethanol and 1-butanol can dissolve in both water and hexane. Larger alcohols such as 1-pentanol and alcohols with longer alkyl chains are typically less soluble in water because their hydrophobic alkyl chains dominate over the specific hydrophilic hydroxyl group.
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How many moles of n are in 0.215 g of n2o?
Final answer:
To find the number of moles of nitrogen in 0.215 g of N2O, calculate the molar mass of N2O (44 g/mol), then use the mass to calculate the moles of N2O. Finally, multiply by 2 since there are two nitrogen atoms per molecule of N2O, resulting in approximately 0.00978 moles of nitrogen.
Explanation:
To calculate the number of moles of nitrogen in 0.215 g of N2O, we first need the molar mass of N2O. The formula N2O contains two nitrogen atoms (N) and one oxygen atom (O). Since N has a molar mass of 14 g/mol, and O has a molar mass of 16 g/mol, the molar mass of N2O is calculated as:
[tex](2 mol N) times (14 g/mol) + (1 mol O) times (16 g/mol) = 28 g/mol + 16 g/mol = 44 g/mol.[/tex]
Next, we use the formula:
Number of moles (n) = mass (m) / molar mass (M)
n = 0.215 g / 44 g/mol
Therefore, the moles of N2O in 0.215 g is calculated as approximately:
n = 0.00489 moles of N2O
To find the number of moles of nitrogen (N) in N2O, we need to remember that each molecule of N2O contains two atoms of nitrogen, so we multiply the moles of N2O by 2:
n(N) = 0.00489 moles of N2O
n(N) = 0.00978 moles of nitrogen
Calculate the [OH–] if the hydrogen ion concentration is 3.64 x 10–8 M.
Using the ionic product of water, you can determine the hydroxide ion concentration [OH−] if you have the hydrogen ion concentration [H+]. In this case, [OH−] = 1.0 x 10^-14 / 3.64 x 10–8 M, results in [OH−] = 2.75 x 10^-7 M.
Explanation:To calculate the [OH−] if the hydrogen ion concentration is 3.64 x 10–8 M, first you need to understand the Ionic Product of water. It is the mathematical product of the concentrations of Hydrogen ions [H+] and Hydroxide ions [OH−], and its value is 1.0 x 10^-14 at 25 degrees Celsius. So, if you know the concentration of the hydrogen ions, you can calculate the concentration of the hydroxide ions using the equation [H+] x [OH−] = 1.0 x 10^-14.
Substituting the given value into the equation, we get:
3.64 x 10–8 M x [OH−] = 1.0 x 10^-14.
To solve for [OH-], divide both sides by 3.64 x 10-8M resulting to [OH−] = 1.0 x 10^-14 / 3.64 x 10–8 M = 2.75 x 10^-7 M
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The two-slit diffraction experiment shows how light can be treated as particles and how light waves carry the statistical information for the experiment. if we were to use a beam of electrons instead of light in the experiment, how would the results differ?
The double-slit experiment is a famous tool to illustrate concepts within quantum mechanics. In particular it demonstrates the concept of wave-particle duality. Use of a light wave demonstrates diffraction and interference, which is a typical wave behaviour. Surprisingly, use of a beam of electrons also yields an interference pattern, showing electrons can behave like waves.
Explanation:
There would be a optical phenomenon pattern almost like, however totally different from, that exploitation light-weight.Interference and optical phenomenon are the phenomena that distinguish waves from particles: waves interfere and split, particles don't.
Light bends around obstacles like waves do, and it's this bending that causes the one slit optical phenomenon pattern.
What is the [h3o+] at equilibrium of a 0.50 m weak acid (ha) solution if the ka of the acid is 4.6 × 10−4?
Answer : The concentration of [tex]H_3O^+[/tex] at equilibrium is, 0.015 M
Solution :
The balanced equilibrium reaction will be,
[tex]HA+H_2O\rightleftharpoons H_3O^++A^-[/tex]
The expression for dissociation constant of weak aciod will be,
[tex]k_a=\frac{[H_3O^+]\times [A^-]}{[HA]}[/tex]
where,
[tex]k_a[/tex] = dissociation constant of weak acid
Let the concentration of [tex]H_3O^+[/tex] and [tex]A^-[/tex] be 'x'
Now put all the given values in this expression, we get
[tex]4.6\times 10^{-4}=\frac{(x)\times (x)}{0.50}[/tex]
[tex]x=0.015M[/tex]
The concentration of [tex]H_3O^+[/tex] = [tex]A^-[/tex] = x = 0.015 M
Therefore, the concentration of [tex]H_3O^+[/tex] at equilibrium is, 0.015 M
While doing a distillation, Farha heats the mixture too fast, and the temperature rises too quickly. If the mixture is composed of two liquids, how will this likely affect the separation of the mixture?
A) The liquids will be separated in reverse order.
B) Both liquids will evaporate and escape into the air.
C) The liquid with the lower density will be collected before the liquid with the higher density.
D) Some liquid with the higher boiling point will be collected along with the liquid with the lower boiling point.
Answer is: D) Some liquid with the higher boiling point will be collected along with the liquid with the lower boiling point.
For example if we have mixture of water and alcohol ethanol, we can separate water and ethanol with distillation (process of heating and cooling).
Ethanol and water have different boiling points (physical property), ethanol has boiling point (78.37°C) and water (100°C).
When the mixture is quickly heated, water contains a lot of ethanol.
The temperature should stay more or less constant.
What are [Ba2+] and [CrO42−] in a saturated BaCrO4 solution if the Ksp of BaCrO4 is 1×10−10?
[Ba2+] =
[CrO42−] =
Final answer:
The concentrations of[tex]Ba^{2+} and CrO_4^{2-[/tex] in a saturated solution of BaCrO4 with a Ksp of 1×10^{−10} are both 1×10^{−5} M.
Explanation:
The student asked about the concentrations of [tex]Ba^{2+} and CrO_4^{2-[/tex] in a saturated solution of barium chromate (BaCrO4) given that the solubility product constant (Ksp) is 1×10−10. In a saturated solution, the ions Ba2+ and CrO42− would be present in equal molar amounts, as the dissolution of BaCrO4 produces one of each ion. Hence, the Ksp equation for this dissolution is [tex]K_{sp} = [Ba^{2+}][CrO_4^{2-}][/tex]. Since both ions are in a 1:1 ratio, we can set [tex][Ba^{2+}] = [CrO_4^{2-}] = x[/tex]. Therefore, Ksp = x·x = x2 = 1×10^−10, and solving for x gives x = √(1×10^−10) = 1×10^−5M. This is the concentration of both Ba2+ and [tex]CrO_4^{2-[/tex] in the saturated solution.
A student collected nitrogen by displacing water in a graduated cylinder. the atmospheric pressure was 738.9 mmhg; the height of the water remaining in the cylinder was 13.2 mm; the partial pressure of the water was 18.8 mmhg. determine the partial pressure of the nitrogen in the cylinder. the density of mercury is 13.6 g/m
What effect does the use of an uncalibrated thermometer have on the boiling point?
Which of the following are true about marijuana
All of the given options are true about marijuana (option D)
Why is this correct?
The primary psychoactive component in marijuana, THC, disrupts intercellular communication within the brain, notably affecting regions associated with memory and learning. Studies indicate that marijuana consumption can result in challenges with concentration, memory retention, and acquiring new knowledge.
THC has the capacity to engage with the brain's fear and anxiety hub, potentially inducing sensations of paranoia, anxiety, and, in severe cases, panic attacks. This tendency is amplified among individuals with pre-existing anxiety conditions or those consuming high doses of THC.
While not as inherently addictive as certain other substances, marijuana still holds the potential for habit formation. Regular usage can lead to dependence, marked by withdrawal symptoms such as irritability, anxiety, and sleep disturbances upon cessation of use.
Complete question:
Which of the following are true about marijuana: A. It can impair learning and memory B. It can bring upon panic attacks or anxiety C. It can become addictive D. All of the above
A solution contains 0.10 m sodium cyanide and 0.10 m potassium hydroxide. solid zinc acetate is added slowly to this mixture. what is the formula of the substance that precipitates first?
From the solubility products of the solutes produced, the Ksp of {Zn(OH)₂ is less than the Ksp of Zn(C_N)₂, Zn(OH)₂ precipitates first.
What is solubility product? Ksp of a solute?The solubility product, Ksp of a solute is the product of the ions produced when a solute dissociates into ions when dissolved in a solvent.
The higher the Ksp of a solute, the more soluble it is ain a solvent.
The equation of the reaction of zinc acetate with each of the solutions as well their solubility products is given below:
[tex] Zn(CH₃COO)₂(s) + 2KOH(aq) \rightarrow Zn(OH)₂(s) + 2CH₃COOK(aq) \\ [/tex]
[tex]Ksp \: {Zn(OH)₂}=1.2*10^{-17}[/tex]
[tex]Zn(CH₃COO)₂(s) + 2NaC \: N(aq) \rightarrow Zn(C \: N)₂(s) + 2CH₃COONa(aq) \\ [/tex]
[tex]Ksp \: {Zn(C \: N)₂}=2.6*10^{-13}[/tex]
Therefore, since the Ksp of Zn(OH)₂ is less than the Ksp of Zn(C_N)₂, Zn(OH)₂ precipitates first.
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What is the ph of a solution made by combining 157 ml of 0.35 m nac2h3o2 with 139 ml of 0.46 m hc2h3o2? the ka of acetic acid is 1.75 × 10-5?
A 2.50-l volume of hydrogen measured at â196 °c is warmed to 100 °c. calculate the volume of the gas at the higher temperature, assuming no change in pressure
To solve this we assume that the hydrogen gas is an ideal gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At a constant pressure and number of moles of the gas the ratio T/V is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:
T1 / V1 = T2 / V2
V2 = T2 x V1 / T1
V2 = (100 + 273.15) K x 2.50 L / (-196 + 273.15) K
V2 = 12.09 L
Therefore, the volume would increase to 12.09 L as the temperature is increased to 100 degrees Celsius.
The problem is about Charles's Law, which states the volume of a gas is directly proportional to its absolute temperature, while pressure is constant. With the data provided, using this concept in a formula, we calculate the volume of hydrogen to be roughly 12.16 Liters at 100°C.
Explanation:The question pertains to Charles's Law in Physics, specifically about gases. Charles's Law states that the volume of a given gas mass is directly proportional to its absolute temperature, provided that the pressure is kept constant.
For the problem given, the initial volume (V₁) of the hydrogen gas is 2.50 L, and its initial temperature is -196°C, which must be converted to Kelvin (K) by adding 273 (equating to 77K). The gas is warmed to 100°C, or 373K (T₂).
To calculate the volume of the gas at the higher temperature (V₂), we use the formula for Charles's Law: V₁/T₁ = V₂/T₂. Substituting in the provided values, we get (2.50 L / 77 K) = V₂ / 373K. Solving for V₂, we find that the volume of the hydrogen gas at 100°C is approximately 12.16 L.
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how much sodium chloride must be added to 100 mL of water so that its concentration is 20 parts per million assume that the density of water is 1.00g/ml
What is the IUPAC name for the following compound (problem C)
The mentioned compound's IUPAC name depends on its structure. It could be named as 'ethoxyethane' if it contains an ethoxy group attached to an ethane chain. Similarly, a molecule with chlorine atoms attached to the 2nd and 3rd carbon would be named '2,3-dichloropentane'.
Explanation:The IUPAC name for the mentioned compound depends on its structure, but given the examples, if it’s a molecule made up of an ethoxy group attached to an ethane chain, then its IUPAC name would be ethoxyethane. If the molecule is a substituted alkane with chlorine atoms attached on the 2nd and 3rd carbon, the name would be 2,3-dichloropentane. It's also noteworthy that the IUPAC adopted new nomenclature guidelines in 2013 that require the place number of substituents to be put as an “infix” rather than a prefix, which should be considered as well.
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What would be the resulting molarity of a solution made by dissolving 25.4 grams of KOH in enough water to make a 985-milliliter solution? Show all of the work needed to solve this problem.
Write a net ionic equation for the overall reaction that occurs when aqueous solutions of potassium hydroxide and phosphoric acid are combined. assume excess base.
The net ionic equation for overall reaction is 2 KOH(aq) + H₃PO₄(aq) → K₂HPO₄(aq) + 2 H₂O(l)
Kalium hydroxide (KOH) and phosphoric acid (H₃PO₄) in water react to form potassium hydrogen phosphate (K₂HPO₄) and water. The balanced chemical equation for this reaction is:
2 KOH(aq) + H₃PO₄(aq) → K₂HPO₄(aq) + 2 H₂O(l).
In this process, potassium hydroxide contributes OH⁻ to phosphoric acid, which donates H⁺. The hydroxide and hydrogen ions produce water molecules. In the meantime, the remaining ions, K⁺ and HPO₄²⁻, create potassium hydrogen phosphate.
Since potassium hydroxide is a strong base and phosphoric acid is weak, the reaction completes, yielding the products. This net ionic equation simplifies the process by concentrating on chemical change species.
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The net ionic equation for the reaction between potassium hydroxide and phosphoric acid, with excess base, is H3PO4(aq) + 3OH-(aq) → 3H2O(l) + PO4^3-(aq).
Explanation:When aqueous solutions of potassium hydroxide (KOH) and phosphoric acid (H3PO4) are combined, the hydroxide ions (OH-) from the strong base KOH will react with the hydrogen ions (H+) from the weak acid H3PO4 to form water (H2O) and the phosphate ion (PO43-). Given the excess base, we will see the complete neutralization of H3PO4. The net ionic equation, considering H3PO4 does not fully dissociate, is as follows:
H3PO4(aq) + 3OH-(aq) → 3H2O(l) + PO43-(aq)
The key to writing the net ionic equation is recognizing that the potassium ions (K+) and excess hydroxide ions (OH-) remain in solution and thus are spectator ions, not part of the net ionic equation.
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What are the net ionic equations for:
Ni(No3)2(aq) + Na2S(aq) = NiS(s) + 2 NaNo3(aq)
KBr(aq) + NaNO3(aq) = KNO3(s) + NaBr(aq)
Li2SO4(aq) + BaCl2(aq) = BaSO4(s) + 2 LiCl(aq)
Explanation :
In the net ionic equations, we are not include the spectator ions in the equations.
Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.
(a) The given balanced ionic equation is,
[tex]Ni(NO_3)_2(aq)+Na_2S(aq)\rightarrow NiS(s)+2NaNO_3(aq)[/tex]
The ionic equation in separated aqueous solution will be,
[tex]Ni^{2+}(aq)+2NO_3^-(aq)+2Na^+(aq)+S^{2-}(aq)\rightarrow NiS(s)+2Na^+(aq)+2NO_3^-(aq)[/tex]
In this equation, [tex]Na^+\text{ and }NO_3^-[/tex] are the spectator ions.
By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.
The net ionic equation will be,
[tex]Ni^{2+}(aq)+S^{2-}(aq)\rightarrow NiS(s)[/tex]
(b) The given balanced ionic equation is,
[tex]NaNO_3(aq)+KBr(aq)\rightarrow KNO_3(s)+NaBr(aq)[/tex]
The ionic equation in separated aqueous solution will be,
[tex]Na^+(aq)+NO_3^-(aq)+K^+(aq)+Br^-(aq)\rightarrow KNO_3(s)+Na^+(aq)+Br^-(aq)[/tex]
In this equation, [tex]Na^+\text{ and }Br^-[/tex] are the spectator ions.
By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.
The net ionic equation will be,
[tex]NO_3^-(aq)+K^+(aq)\rightarrow KNO_3(s)[/tex]
(c) The given balanced ionic equation is,
[tex]Li_2SO_4(aq)+BaCl_2(aq)\rightarrow BaSO_4(s)+2LiCl(aq)[/tex]
The ionic equation in separated aqueous solution will be,
[tex]2Li^+(aq)+SO_4^{2-}(aq)+Ba^{2+}(aq)+2Cl^-(aq)\rightarrow BaSO_4(s)+2Li^+(aq)+2Cl^-(aq)[/tex]
In this equation, [tex]Li^+\text{ and }Cl^-[/tex] are the spectator ions.
By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.
The net ionic equation will be,
[tex]SO_4^{2-}(aq)+Ba^{2+}(aq)\rightarrow BaSO_4(s)[/tex]
Balance the following equation with the smallest whole number coefficients. Choose the answer that is the sum of the coefficients in the balanced equation. Do not forget coefficients of "one."
Cr2(SO4)3 + RbOH Cr(OH)3 + Rb2SO4
(a) 10
(b) 12
(c) 13
(d) 14
(e) 15
The sum of the coefficients in the balanced chemical equation Cr2(SO4)3 + RbOH → Cr(OH)3 + Rb2SO4 is 24. No listed options match the sum.
Explanation:To balance the given equation, you must first count the number of each type of atom on both sides of the equation and then use coefficients to balance the numbers of each atom on both sides.
The correctly balanced chemical equation is 2Cr2(SO4)3 + 12RbOH → 4Cr(OH)3 + 6Rb2SO4.
Now, add these coefficients: 2 (for Cr2(SO4)3), 12 (for RbOH), 4 (for Cr(OH)3), and 6 (for Rb2SO4). The sum of the coefficients in the balanced equation is 2 + 12 + 4 + 6 = 24.
None of the options given match the sum of the coefficients in the balanced equation.
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Rank the size of a change in temperature of one degree Fahrenheit, one degree Celsius, and one kelvin. In other words, if a thermometer were to show that the temperature outside increased by these amounts, which change would feel the largest? If any of the options are the same magnitude, stack them above one another. Rank the sizes of one degree in each scale from largest to smallest. To rank items as equivalent, overlap them.
Why might a chemist add a buffer to a solution?
Chemists add buffers to solutions to stabilize pH levels, critical for maintaining the proper environment for chemical and biological processes. Buffers resist pH changes by balancing the effects of added acids or bases through weak acids or bases and their salts.
A chemist might add a buffer to a solution to maintain a stable pH when acids or bases are added. This is critical because many chemical reactions and biological processes require a consistent pH to function properly. Buffers work by utilizing a weak acid or base and their corresponding salts to neutralize added acids or bases without significantly changing the solution's pH. For instance, a buffer system can consist of acetic acid and sodium acetate; the acetic acid can react with any added base, while the sodium acetate reacts with any added acid, both acting to dampen fluctuations in pH levels.
In biological systems, for example, buffers help maintain conditions that are necessary for enzyme function and cellular processes. There are practical limits to buffer concentrations due to the potential formation of unwanted precipitates and the tolerance of the system to dissolved salts.
Another important role of buffers includes calibration in analytical chemistry. Adding substances like Na2SO4 to a buffer can adjust ions' concentrations that are crucial for the accuracy of a calibration curve, which is especially useful in detecting low concentrations of certain compounds.
Find the number of moles of water that can be formed if you have 182 mol of hydrogen gas and 86 mol of oxygen gas.
A voltaic cell is constructed with two silver-silver chloride electrodes, where the half-reaction is
The rates of which reactions are increased when the temperature is raised?
a. endothermic reactions only
b. both endothermic reactions and exothermic reactions
c. exothermic reactions only
d. depends on the value of delta h not the sign
We have that the he rates of which reactions are increased when the temperature is raised is endothermic reactions only
Option A
Endothermic ReactionGenerally, the rates of which reactions are increased when the temperature is raised is an Endothermic reaction only
This assertion lies on the basis of the fact that Endothermic reaction requires a net gain of energy for reaction
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NaCl + H2O → NaOH + Cl2 + H2 How many grams of chlorine gas, Cl2 are given off if 7.5 grams of sodium chloride, NaCl, are decomposed? Is this equation balanced?
Calculate the ph of a 0.20 m solution of kcn at 25.0 ∘c. express the ph numerically using two decimal places
The pH of a 0.20 M solution of KCN is [tex]\boxed{11.31}[/tex].
Further Explanation:
pH is used to describe acidity or basicity of substances. Its range varies from 0 to 14. It is defined as negative logarithmof hydrogen ion concentration.
The expression for pH is mentioned below.
[tex]{\text{pH}} = - \log \left[ {{{\text{H}}^ + }} \right][/tex] …… (1)
Where [tex]\left[ {{{\text{H}}^ + }}\right][/tex] is the concentration of hydrogen ion.
Dissociation reaction of KCN is as follows:
[tex]{\text{KCN}} \to {{\text{K}}^ + } + {\text{C}}{{\text{N}}^ - }[/tex]
Cyanide ions thus formed can react with water to form HCN and [tex]{\text{O}}{{\text{H}}^ - }[/tex] as follows:
[tex]{\text{C}}{{\text{N}}^ - } + {{\text{H}}_{\text{2}}}{\text{O}} \rightleftharpoons {\text{HCN}} + {\text{O}}{{\text{H}}^ - }[/tex]
The relation between [tex]{{\text{K}}_{\text{w}}}[/tex], [tex]{{\text{K}}_{\text{b}}}[/tex] and [tex]{{\text{K}}_{\text{a}}}[/tex] is expressed by following relation:
[tex]{{\text{K}}_{\text{w}}} = {{\text{K}}_{\text{b}}} \cdot {{\text{K}}_{\text{a}}}[/tex] …… (2)
Where,
[tex]{{\text{K}}_{\text{w}}}[/tex] is the ionic product constant of water.
[tex]{{\text{K}}_{\text{b}}}[/tex] is the dissociation constant of base.
[tex]{{\text{K}}_{\text{a}}}[/tex] is the dissociation constant of acid.
The value of [tex]{{\text{K}}_{\text{w}}}[/tex] is [tex]{10^{ - 14}}[/tex].
The value of [tex]{{\text{K}}_{\text{a}}}[/tex] is [tex]4.9 \times {10^{ - 10}}[/tex].
Substitute these values in equation (2).
[tex]{10^{ - 14}} = {{\text{K}}_{\text{b}}}\left( {4.9 \times {{10}^{ - 10}}} \right)[/tex]
Solve for [tex]{{\text{K}}_{\text{b}}}[/tex],
[tex]{{\text{K}}_{\text{b}}} = 2 \times {10^{ - 5}}[/tex]
The expression for [tex]{{\text{K}}_{\text{b}}}[/tex] of HCN is as follows:
[tex]{{\text{K}}_{\text{b}}} = \dfrac{{\left[ {{\text{HCN}}} \right]\left[ {{\text{O}}{{\text{H}}^ - }} \right]}}{{\left[ {{\text{C}}{{\text{N}}^ - }} \right]}}[/tex] …… (3)
Consider x to be change in equilibrium concentration. Therefore, equilibrium concentrationof [tex]{\text{C}}{{\text{N}}^ - }[/tex], HCN and becomes (0.2 – x), x and x respectively.
[tex]{\text{2}} \times {\text{1}}{{\text{0}}^{ - 5}} = \dfrac{{{x^2}}}{{\left( {0.2 - x} \right)}}[/tex]
Solving for x,
[tex]x = 0.002[/tex]
Therefore concentration of hydroxide ion is 0.002 M.
The expression to calculate pOH is as follows:
[tex]{\text{pOH}} = - \log \left[ {{\text{O}}{{\text{H}}^ - }} \right][/tex] …… (4)
Substitute 0.002 M for [tex]\left[ {{\text{O}}{{\text{H}}^ - }} \right][/tex] in equation (4).
[tex]\begin{aligned}{\text{pOH}} &= - \log \left( {0.002{\text{ M}}} \right) \\&= 2.69 \\\end{aligned}[/tex]
The relation between pH and pOH is as follows:
pH + pOH = 14 …… (5)
Substitute 2.69 for pOH in equation (4).
[tex]{\text{pH}} + 2.69 = 14[/tex]
Solving for pH,
pH = 11.31
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Write the chemical equation responsible for pH of buffer containing and : https://brainly.com/question/8851686 Reason for the acidic and basic nature of amino acid. https://brainly.com/question/5050077Answer details:
Grade: High School
Subject: Chemistry
Chapter: Acids, base and salts
Keywords: pH, pOH, 11.31, 2.69, 14, 0.002 M, Kb, Kw, Ka, 10^-14, 2*10^-5.