If a light beam strikes a shiny object at a 90-degree angle, to what are the angle of incidence and the angle of reflection equal?

Answers

Answer 1

Answer:

Angle of incidence = angle of reflection = 0

Explanation:

Given that,

A light beam strikes a shiny object at a 90-degree angle.

The angle made by incident ray and normal to the mirror is called the angle of incidence and the angle made by the reflected ray and the normal is called the angle of reflection.

Here, the beam strikes normally. So, the angle of incidence is 0.

A/c to the law of reflection,

Angle of incidence = angle of reflection

So, the angle of reflection is also equal to 0.


Related Questions

A pure copper cube has an edge length of 1.76 cm. How many copper atoms does it contain? (volume of a cube = (edge length)^3; density of copper = 8.96 g/cm^3 )

Answers

I havent learnt this yet sry

The Fischer esterification mechanism is examined in this question. The overall reaction is: Benzoic acid, C H 3 O H and H C l react to form a methyl ester, H 2 O and H C l. Benzoic acid is a carboxylic acid bonded to a benzene ring. Identify the results or mechanism of each step.

Answers

Answer:

See explanation and image attached

Explanation:

Fischer esterification is a type of reaction used  to convert carboxylic acids to ester in the presence of excess alcohol and a strong acid which acts as a catalyst. Another final product formed in the reaction is water.

The mechanism for the fischer esterification of Benzoic acid and  C H 3 O H in the presence of HCl as the catalyst is shown in the image attached to this answer.

The final products of the reaction are methyl benzoate, water and H^+ as shown in the image attached.

The methyl ester, water, and the acid catalyst (HCl) are byproducts of the Fischer esterification process, which involves protonation, nucleophilic attack, elimination, and deprotonation processes.

Carbonyl oxygen protonation: The carbonyl oxygen of the carboxylic acid (benzoic acid) is protonated by the acid catalyst (HCl) in the first step. The protonation of the carbonyl carbon increases its electrophilicity and promotes the alcohol's nucleophilic assault. Attack by the alcohol's nucleophilic oxygen (methanol, CH3OH) on the protonated carboxylic acid's carbonyl carbon results in the formation of a tetrahedral intermediate. The acid catalyst also helps with this phase. Elimination of water: In the following step, the water molecule must be removed from the tetrahedral intermediate. The hydroxyl group (-OH) from the carboxylic acid and a hydrogen from the hydroxyl group of the alcohol are removed to create this water molecule. Deprotonation: A deprotonation occurs after the removal of water.

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In what kind of orbitals do the lone-pair electrons on the singly bonded oxygen of methyl acetate reside, and are they in the same plane as the group

Answers

Answer:

Both lone pairs reside in sp3 orbitals; they are co-planar with the CH3 group.

Explanation:

In the compound, methyl acetate, the  lone-pair of electrons on the singly bonded oxygen atom is accommodated in sp3 hybridized orbitals.

If we look at the compound, we will notice the both lone pair of electrons on oxygen are accommodated in sp3 orbitals which are co-planar with the  CH3 group  in the molecule.

3 enzimas presentes en nuestro organismo e indique en que procesos actúan

Answers

Answer:

ATP asa, Helicasa, Proteasa, ARN polimerasa

Explanation:

Las enzimas son un tipo de biomoleculas que se corresponden con las proteinas.

Al momento de referirse a ellas, se utiliza la terminación asa.

ATPasa → Sintetizando ATP para el funcionamiento celular

Helicasa → Abre las hebras de ADN permitiendo el paso de la horquilla para el proceso de replicación de ADN.

Proteasas → Enzimas que degradan proteinas mal plegadas, rompen los enlaces peptídicos.

ARN polimerasa → Sintesis de ARN mensajero a partir de ADN en el proceso de la Transcripción. Se la puede conocer a veces, como primasa.

Select all the true statements.
a. Ionic bonding is more prevalent for the higher oxidation states and covalent bonding is more prevalent for the lower states.
b. The highest oxidation state of elements in Groups 3A through 7B is +3.
c. In the transition series, atomic size across a period decreases at first but then remains relatively constant.
d. The transition elements in a period show a steady increase in electronegativity.

Answers

Answer:

In the transition series, atomic size across a period decreases at first but then remains relatively constant.

The transition elements in a period show a steady increase in electronegativity.

Explanation:

In considering the transition series, we observe that atomic sizes of the elements decreases first and subsequently remain constant. The reason for the initial decrease in atomic size is the increase in nuclear charge across the period. After the first few elements in the period, the atomic size remains relatively constant due to shielding effect of the inner d electrons which opposes the increase in effective nuclear charge.

It is also observed that electro negativity increases smoothly across the period for the transition series. Consequently, the transition series become less electro positive across the period.

Sodium hydrogen carbonate, also called baking soda, is an active ingredient in some antacids used for the relief of indigestion. Determine the percent of carbon in sodium hydrogen carbonate.

Answers

Answer:

14.30%

Explanation:

Step 1: Given data

Chemical formula of sodium hydrogen carbonate: NaHCO₃

Step 2: Determine the mass of C in 1 mole of NaHCO₃

The is 1 atom of C in 1 molecule of NaHCO₃ and the molar mass of C is 12.01 g/mol. Then, there are 12.01 g of C in 1 mole of NaHCO₃.

Step 3: Determine the mnolar mass of NaHCO₃

M(NaHCO₃) = 1 × M(Na) + 1 × M(H) + 1 × M(C) + 3 × M(O)

M(NaHCO₃) = 1 × 22.98 g/mol + 1 × 1.01 g/mol + 1 × 12.01 g/mol + 3 × 16.00 g/mol = 84.00 g/mol

Step 4: Determine the mass percent of C in NaHCO₃

We will use the following expression.

%C = mC / mNaHCO₃ × 100%

%C = 12.01 g / 84.00 g × 100% = 14.30%

Some glucose produced by gluconeogenesis is stored in the body as glycogen. Order the steps of glycogen synthesis.

a. Pyrophosphatase converts PPi and water into two Pi
b. Glycogen synthase adds a glucose unit from UDP-glucose to glycogen, producing a larger glycogen molecule and UDP
c. Glvcogen synthase removes a glucose unit from a glycogen molecule producing a smaller glycogen molecule and IJDP
d. ADP-glucose pyrophosphorylase catalyzes the reaction of glucose-I -phosphate and ATP to ADP-glucose and PPi
e. UDP-glucose pyrophosphorylase catalyzes the reaction of glucose-I-phosphate and UTP to UDP-glucose and PPi

Answers

Answer:

e. UDP-glucose pyrophosphorylase catalyzes the reaction of glucose-I-phosphate and UTP to UDP-glucose and PPi

a. Pyrophosphatase converts PPi and water into two Pi

b. Glycogen synthase adds a glucose unit from UDP-glucose to glycogen, producing a larger glycogen molecule and UDP

Explanation:

Glycogen synthesis or glycogenesis is the process of synthesis of glycogen molecules from glucose molecules in living organisms. Glycogen is a polysaccharide storage form of glucose and helps to store excess glucose in the body form use when required by the body.

The synthesis of glycogen involves sugar nucleotides. Sugar nucleotides are compounds in which a sugar molecule is attached to a nucleotide through phosphate ester bond, resulting in the activation of the sugar molecule. The sugar nucleotides then are used as substrates for the polymerization of the monosaccharide sugars into disaccharides, oligosaccharides and polysaccharides.

In the synthesis of glycogen, glucose-6-phosphate from phosphorylation of free glucose by hexokinase is first isomerized to glucose-1-phosphate by phosphoglucomutase.

Glucose-1-phosphate is then converted to UDP-glucose by its reaction with UTP catalyse by UDP-glucose pyrophosphorylase. The reaction is favoured by the rapid hydrolysis of PPi produced to two molecules of inorganic phosphate by the enzyme pyrophosphatase.

Glycogen synthase then adds a glucose unit from UDP-glucose to a growing chain of glycogen, producing a larger glycogen molecule and free UDP.

What is the molar mass of H2SO4?
(Molar mass of H = 1.0079 g/mol; S = 32.065 g/mol; O = 15.999 g/mol)

Answers

Answer: 98.08 g/mol

Explanation:

i just took this test and got it right

Answer:

A

Explanation:

What volume of water must be added to 10.5 mL of a pH 2.0 solution of HNO3 in order to change the pH to 4.0 g

Answers

Answer:

[tex]V'=1040ml[/tex]

Explanation:

From the question we are told that:

Initial Volume [tex]V_1=10.5mL[/tex]

Initial Aciditity of [tex]HN0_3 pH_1=2.0g[/tex]

Finial Aciditity of [tex]HN0_3 pH_2=4.0g[/tex]

Generally the equation for Acidity &Volume Relationship is mathematically given by

[tex]N_1V_1=N_2V_2[/tex]

Therefore

[tex]V_2=\frac{N_1V_2}{N_2}[/tex]

[tex]V_2=\frac{10^{-2}*10.5}{10^{-4}}V_2=1050ml[/tex]

Therefore

Change in Water volume is

[tex]V'=V_2-V_1[/tex]

[tex]V'=1050ml-10ml[/tex]

[tex]V'=1040ml[/tex]

help i’ll give u points

Answers

Answer:

60ug

Explanation:

If it has experienced two half lives, that means it has been halved twice. in that case, to undo it, just multiply it by two twice. 0.15ug * 2 = 0.30ug. 0.30ug * 2 = 0.60ug. Hope this helps.

A monoprotic weak acid when dissolved in water is 0.66% dissociated and produces a solution with a pH of 3.04. Calculate the Ka of the acid. g

Answers

Answer:

Ka = 6.02x10⁻⁶

Explanation:

The equilibrium that takes place is:

HA ⇄ H⁺ + A⁻Ka = [H⁺][A⁻]/[HA]

We calculate [H⁺] from the pH:

pH = -log[H⁺][H⁺] = [tex]10^{-pH}[/tex][H⁺] = 9.12x10⁻⁴ M

Keep in mind that [H⁺]=[A⁻].

As for [HA], we know the acid is 0.66% dissociated, in other words:

[HA] * 0.66/100 = [H⁺]

We calculate [HA]:

[HA] = 0.138 M

Finally we calculate the Ka:

Ka = [tex]\frac{[9.12x10^{-4}]*[9.12x10^{-4}]}{[0.138]}[/tex] = 6.02x10⁻⁶

How does the neutron number in each compare

Answers

Answer:

Neutrons are all identical to each other, just as protons are. Atoms of a particular element must have the same number of protons but can have different numbers of neutrons.

Explanation:

Since the vast majority of an atom's mass is found its protons and neutrons, subtracting the number of protons (i.e. the atomic number) from the atomic mass will give you the calculated number of neutrons in the atom. In our example, this is: 14 (atomic mass) – 6 (number of protons) = 8 (number of neutrons).

When aqueous solutions of AgNO3 and MgCl2 are mixed, a precipitate forms. What is the correct formula for the precipitate

Answers

Answer:

AgCl

Explanation:

The reaction that takes place is:

2AgNO₃ (aq) + MgCl₂ (aq) → Mg(NO₃) (aq) + 2AgCl (s)

The precipitate (meaning a solid substance) formed is silver chloride, AgCl.

All salts formed with silver and a halogen (F, Cl, Br, I) are insoluble in water, meaning that when working with aqueous solutions they will be precipitates.

Which additional product balances the reaction h2so4+2naoh>na2so4

Answers

Answer:

H 4so8 is the answer of balance the reaction

what will happen to the initial rate of the reaction if the concentration of iron(III) chloride is tripled

Answers

Answer:

Concentration and reaction rate is directly proportional

Explanation:

Basically, concentration is directly proportional to the concentration, so the higher the concentration, the higher the reaction rate.

Once the concentration is increased, more chemical will be added hence the reaction rate will increase

Calculate the number of cations and anions in each of the following compounds. Enter your answers in scientific notation. (a) 6.42 g of KBr:

Answers

Answer: The number of cations are [tex]3.24 \times 10^{22}[/tex] and number of anions are  [tex]3.24 \times 10^{22}[/tex] in 6.42 g of KBr.

Explanation:

The molar mass of KBr is (39.10 + 79.90) g/mol = 119.00 g/mol

Now, the dissociation equation for KBr is as follows.

[tex]KBr \rightarrow K^{+} + Br^{-}[/tex]

This means that 1 mole of KBr is forming 1 mole of [tex]K^{+}[/tex] (cation) and 1 mole of [tex]Br^{-}[/tex] (anion).

According to mole concept, 1 mole of every substance contains [tex]6.022 \times 10^{23}[/tex] atoms. Hence, number of cations present in 6.42 g KBr is calculated as follows.

[tex]No. of cations = Moles \times 6.022 \times 10^{23}\\= \frac{mass}{molar mass} \times 6.022 \times 10^{23}\\= \frac{6.42 g}{119.00 g/mol} \times 6.022 \times 10^{23}\\= 3.24 \times 10^{22}[/tex]

As according to the equation, there are equal number of moles of both cation and anions.

This means that the number of anions are also [tex]3.24 \times 10^{22}[/tex].

Thus, we can conclude that the number of cations are [tex]3.24 \times 10^{22}[/tex] and number of anions are  [tex]3.24 \times 10^{22}[/tex] in 6.42 g of KBr.

The number of cations and anions present in potassium bromide is 3.24 × 10²².

How we calculate atoms from moles?

In one mole of any substance 6.022 × 10²³ atoms of that substance is present and this is known as Avogadro's number.

KBr is a strong electrolyte means it fully dissociates into their constitute ions. So, the number of moles of produced ions is equal to the moles of KBr and dissociation is represented as:

KBr → K⁺ + Br⁻

From this it is clear that 1 mole of cation and 1 mole of anion is produced from 1 mole of KBr. Moles of KBr will be calculated as:

n = W/M, where

W = given mass = 6.42g

M = molar mass = 119

n = 6.42 / 119 = 0.053 moles

No. of cations and anions present in 0.053 moles = 0.053 × 6.022 × 10²³ = 3.24 × 10²².

Hence, 3.24 × 10²² is the no. of cations and anions.

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Show the complete ionic equation and net ionic equation for all the equations below, then state whether or not a precipitate (insoluble compound) will form. To receive full credit, you must show ALL your work.
Cacl2(aq) + K2co3(aq) + -------->
Bacl2(aq) + MgSO4(aq) + -------->
AgNO3(aq) + Kl(aq) →
Nacl(aq) + (NH4)2Cro4(aq) →

Answers

Answer:

(a): Precipitate of calcium carbonate will form.

(b): Precipitate of barium sulfate will form.

(c): Precipitate of silver iodide will form.

(d): Precipitate of sodium chromate will form.

Explanation:

Complete ionic equation is defined as the equation in which all the substances that are strong electrolytes present in an aqueous state and are represented in the form of ions.

Net ionic equation is defined as the equations in which spectator ions are not included.

Spectator ions are the ones that are present equally on the reactant and product sides. They do not participate in the reaction.

(a):

The balanced molecular equation is:

[tex]CaCl_2(aq)+K_2CO_3(aq)\rightarrow 2KCl(aq)+CaCO_3(s)[/tex]

The complete ionic equation follows:

[tex]Ca^{2+}(aq)+2Cl^-(aq)+2K^+(aq)+CO_3^{2-}(aq)\rightarrow 2K^+(aq)+2Cl^-(aq)+CaCO_3(s)[/tex]

As potassium and chloride ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

[tex]Ca^{2+}(aq)+CO_3^{2-}(aq)\rightarrow CaCO_3(s)[/tex]

Precipitate of calcium carbonate will form.

(b)

The balanced molecular equation is:

[tex]BaCl_2(aq)+MgSO_4(aq)\rightarrow MgCl_2(aq)+BaSO_4(s)[/tex]

The complete ionic equation follows:

[tex]Ba^{2+}(aq)+2Cl^-(aq)+Mg^{2+}(aq)+SO_4^{2-}(aq)\rightarrow Mg^{2+}(aq)+2Cl^-(aq)+BaSO_4(s)[/tex]

As magnesium and chloride ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

[tex]Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)[/tex]

Precipitate of barium sulfate will form.

(c):

The balanced molecular equation is:

[tex]AgNO_3(aq)+KI(aq)\rightarrow KNO_3(aq)+AgI(s)[/tex]

The complete ionic equation follows:

[tex]Ag^{+}(aq)+NO_3^-(aq)+K^+(aq)+I^{-}(aq)\rightarrow K^+(aq)+NO_3^-(aq)+AgI(s)[/tex]

As potassium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

[tex]Ag^{+}(aq)+I^{-}(aq)\rightarrow AgI(s)[/tex]

Precipitate of silver iodide will form.

(d):

The balanced molecular equation is:

[tex]2NaCl(aq)+(NH_4)_2CrO_4(aq)\rightarrow 2NH_4Cl(aq)+Na_2CrO_4(s)[/tex]

The complete ionic equation follows:

[tex]2Na^{+}(aq)+2Cl^-(aq)+2NH_4^+(aq)+CrO_4^{2-}(aq)\rightarrow 2NH_4^+(aq)+2Cl^-(aq)+Na_2CrO_4(s)[/tex]

As ammonium and chloride ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

[tex]2Na^{+}(aq)+CrO_4^{2-}(aq)\rightarrow Na_2CrO_4(s)[/tex]

Precipitate of sodium chromate will form.

Please help me I need help with this questions I’m very confused fused as to what the answer is please

Answers

Answer: This was because the experiment showed that a substance could emit radiation even while it was not exposed to light.

A student performs an experiment three times. The results are 4.52 g/mL, 4.54 g/mL, and 4.39 g/mL. Which of the following best describes these results with only the information given in this question?

a. high accuracy
b. high accuracy and high precision
c. high accuracy and low precision
d. high precision

Answers

I’m not sure but I believe the answer is D high precision

The pressure exerted by a gas container depends on

Answers

it’s due to the average number of collisions of gas molecules with the container walls per unit time. As such, pressure depends on the amount of gas (in number of molecules), its temperature, and the volume of the container.

This portion of the stratosphere if known as the _______.
A.
mesosphere
B.
tropopause
C.
thermosphere
D.
ozone layer

Answers

The answer to this question is B- tropopause

A researcher accidentally spilled some sand into his beaker containing crystalline compound X. Knowing what he did about melting points, he was not worried about obtaining the melting point of compound X with a few grains of sand present. Why

Answers

Answer:

See explanation

Explanation:

If we look at the structure of X, we will discover that X is an organic compound. The compound will certainly have a much lower melting point than sand.

This is because, sand is composed of high melting point inorganic materials.

Hence, even though the researcher accidentally spilled some sand into the the beaker containing the crystalline substance X, he does not need to be perturbed since the melting point of X is much lower than that of sand

Which type of macromolecule carries genetic information from parents to
their children?
A. Nucleic acids
B. Carbohydrates
C. Proteins
D. Lipids

Answers

Answer: Its A nucleic acids

Explanation:

Nucleic acids, and DNA in particular, are key macromolecules for the continuity of life. DNA bears the hereditary information that's passed on from parents to children, providing instructions for how (and when) to make the many proteins needed to build and maintain functioning cells, tissues, and organisms.

Nucleic acids carry genetic information from parents to their children.

What are Nucleic acids?

Nucleic acids are biopolymers, macromolecules, essential to all known forms of life.[1] They are composed of nucleotides, which are the monomers made of three components: a 5-carbon sugar, a phosphate group and a nitrogenous base. The two main classes of nucleic acids are deoxyribonucleic acid (DNA) and ribonucleic acid (RNA). If the sugar is ribose, the polymer is RNA; if the sugar is the ribose derivative deoxyribose, the polymer is DNA.

Nucleic acids are naturally occurring chemical compounds that serve as the primary information-carrying molecules in cells and make up the genetic material. Nucleic acids are found in abundance in all living things, where they create, encode, and then store information of every living cell of every life-form on Earth. In turn, they function to transmit and express that information inside and outside the cell nucleus to the interior operations of the cell and ultimately to the next generation of each living organism. The encoded information is contained and conveyed via the nucleic acid sequence, which provides the 'ladder-step' ordering of nucleotides within the molecules of RNA and DNA. They play an especially important role in directing protein synthesis.

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