A newly discovered element has two isotopes. one has an atomic weight of 120.9038 amu with 57.25% abundance. the other has an atomic weight of 122.8831 amu. what is the atomic weight of the element?

We have that the he rates of which reactions are increased when the temperature is raised is  endothermic reactions only

Option A

Generally, 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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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.

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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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.

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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Answer:

The correct answer is D. Phosphorus is not a catalyst.

Explanation:

A catalyst is a substance that accelerates a chemical reaction at a certain temperature, but does not itself wear off the reaction. It is involved in the chemical reaction but is not the starting or end product of the reaction.

Catalysts create an alternative reaction pathway for the reaction with a lower activation energy. Activation energy is the energy required for the reaction to take place. However, the catalyst cannot cause a reaction which does not take place without the catalyst.

Depending on the phases in which the catalyst and reactants are present, one speaks of homogeneous or heterogeneous catalysts. Biochemical processes are catalyzed by enzymes, while heterogeneous catalysts are metals, such as platinum and nickel.

In a sample of acetic acid, the number of moles of oxygen can be calculated by understanding the molecular formula of acetic acid (CH3COOH) and using it to establish the ratio of carbon to oxygen. Given 96.5 moles of carbon in the sample, there would be 193 moles of oxygen.

To find the amount of moles of oxygen in a sample of acetic acid, we must first understand the molecular formula of acetic acid which is CH3COOH. This reveals that for each molecule of acetic acid, there are 2 atoms of carbon and 4 atoms of oxygen. Hence, in mole terms, for each mole of acetic acid, there are 2 moles of carbon and 4 moles of oxygen. If the analytical chemist has determined that there are 96.5 moles of carbon, this implies that there are 48.25 moles of acetic acid in the sample (since 2 moles of carbon correspond to each mole of acetic acid). Given the ratio of carbon to oxygen is 2:4 according to the molecular formula, we can multiply the number of moles of acetic acid by 4 to get the number of moles of oxygen, thus: 48.25 moles * 4 = 193 moles of Oxygen. Therefore, you can expect 193 moles of oxygen in the acetic acid sample.

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96.5 moles of carbon in the sample of acetic acid contains 96.5 moles of oxygen.

First, we need to understand the molecular formula of acetic acid, which is CH₃COOH. This means each molecule of acetic acid contains 2 carbon atoms, 4 hydrogen atoms, and 2 oxygen atoms.

Given that the chemist has determined there are 96.5 moles of carbon in the sample, we can determine the moles of acetic acid molecules by dividing the moles of carbon by 2 (since each molecule of acetic acid has 2 carbon atoms):

96.5 moles of C / 2 = 48.25 moles of acetic acid

Since each acetic acid molecule contains 2 oxygen atoms, we can find the total moles of oxygen by multiplying the moles of acetic acid by 2:

48.25 moles of acetic acid x 2 = 96.5 moles of oxygen

Thus, there are 96.5 moles of oxygen in the sample.

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.

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).

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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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'.

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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The compounds that are soluble in both water and hexane are usually polar compounds with a small size, such as ethanol and 1-butanol.

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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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

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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Answer 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.

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

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]

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.

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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Final answer:

1. OPTION A.

2. OPTION B.

Ethanol has a higher vapor pressure than water due to less extensive hydrogen bonding, and the surface tension order for the substances is mercury (Hg), water (H₂O), octane (C₈H₁₈), reflecting the strength of the intermolecular forces.

Explanation:

The observation that ethanol (C₂H₅OH) has a higher vapor pressure than water (H₂O) at the same temperature is primarily due to more hydrogen bonding in water. Water's ability to form stronger hydrogen bonds compared to ethanol results in fewer water molecules escaping the liquid and hence, a lower vapor pressure. Ethanol, though capable of hydrogen bonding, exhibits weaker intermolecular forces (IMFs) compared to water, allowing more ethanol molecules to escape into the vapor phase.

Regarding the surface tension question, the correct order for the surface tension of the substances listed would be Hg, H₂O, C₈H₁₈ based on the intermolecular forces present in each substance. Mercury (Hg) has a very high surface tension due to strong metallic bonding, water (H₂O) has high surface tension from hydrogen bonding, and octane (C₈H₁₈) has comparatively lower surface tension as it is primarily affected by weaker London dispersion forces.

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.

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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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).

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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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.

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.

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.