Answer:
Zinc sulfide- more soluble in 0.10 M hydrochloric acid solution
Silver chloride- less soluble in 0.10 M hydrochloric acid solution
Lead iodide- same solubility in 0.10 M hydrochloric acid solution as in pure water
Silver hydroxide- same solubility in 0.10 M hydrochloric acid solution as in pure water
Explanation:
Solubility refers to the amount of a substance that dissolves in 1 dm^3 of solvent. Several factors influence the dissolution of solutes in a given solvent. Some substances dissolves at a low pH (in acid solutions) while other dissolve at a low pH (in basic solutions).
For Silver chloride, its solubility decreases in 0.10 M hydrochloric acid solution because of common ion effect when compared to its solubility in pure water.
What is the optimum pH to separate a mixture of lysine, arginine, and cysteine using electrophoresis
Answer:
9.5
Explanation:
Electrophoresis is a separation technique which involves the passage and the use of electric current in separating amino acid mixtures. The positively charged protein moves to the negative electrode while the negatively charged protein moves to the positive electrode which is in accordance to the law of attraction.
Proteins however perform optimally at the right temperature and pH. Studies shows that the optimum pH to separate a mixture of lysine, arginine, and cysteine using the electrophoresis method of separation is around 9.5
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Why does bromine (Br) have a smaller atomic radius than scandium (Sc)?
A. Scandium (Sc) has more energy shells than bromine (Br), making its atoms bigger.
B. Bromine (Br) has less neutrons than scandium (Sc), so it does not have as much mass.
C. Bromine (Br) has more electrons, which makes its energy sublevels more negative, which pulls them closer to the positively charged nucleus.
D. Electrons are added from right to left across the periodic table, so bromine (Br) does not have as many as scandium (Sc).
Answer:
Option (C) Bromine (Br) has more electrons, which makes its energy sublevels more negative, which pulls them closer to the positively charged nucleus.
Atomic radius is the size of the atom that measures from the center to the outermost electron. Bromine has a smaller atomic radius because of more number of electrons. Thus, option C is correct.
What is the periodic trend of radii?The elements are placed in the periods and groups in the periodic table based on the atomic number. Bromine belongs to group 17, period 4 in the periodic table, and scandium (Sc) to group 3, period 4.
In the periodic trend, the atomic radii of the element in the period decrease when moving from left-right. Here bromine and scandium both belong to the same period but differ in the group.
The number of electrons in bromine is more than the scandium element that in turn increases the force of attraction towards the nucleus making them have smaller radii.
Therefore, bromine has smaller radii.
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Reactants that are_____will appear in the rate law as the concentration of the reactant squared.
a. zero order.
b. first order.
c. second order.
d. fourth order.
Answer:
The correct answer is c. second order
Explanation:
In a rate law, second order means that the chemical rate is related with the reactant concentration raised to 2. For example, if the chemical equation is:
A ⇒ B
A is the reactant and it is a second order reaction, so the rate law will be:
r = [A]²
Suppose a balloon was released from the ground and rose to such a height that both the atmospheric pressure. Which statement is true?
A) The temperature change acts to decrease the balloon's volume and the pressure change acts to increase the balloon’s volume.
B) The temperature change acts to increase the balloon's volume and the pressure change acts to decrease the balloon’s volume.
C) Both the temperature and pressure changes act to increase the balloon's volume.
D) Both the temperature and pressure changes act to decrease the balloon's volume.
Answer:
The correct answer is option E) "none of the above".
Explanation:
This question is missing the following information "both the atmospheric pressure and temperature decreased"; as well as the last option, which is the correct one "none of the above". In this hypothetical scenario, a ballon has reach a height were the atmospheric pressure and the the temperature decreased. However, the change in the atmospheric pressure and the temperature would have no effect on the balloon's volume. The internal balloon's temperature and pressure tends to stabilize with the external conditions, therefore, the volume its not affected by the changes described in the question.
convert 0.237 μL to ml
Answer:
0.000237mL
Explanation:
0.237 x 10^-6L = 0.000000237L = mL
0.000237mL
When titrating 50.0 mL of 0.10 M H2SO4 with 0.10 M NaOH, how many mL of NaOH will you have added to reach the 1st equivalence point?
Answer:
50.0mL 0.10M NaOH
Explanation:
The chemical equation of H₂SO₄ with NaOH to reach the first equivalence point is:
H₂SO₄ + NaOH → HSO₄⁻ + Na⁺ + H₂O
Where 1 mole of the H₂SO₄ reacts per mole of NaOH
The initial moles of H₂SO₄ are:
50.0mL = 0.0500L × (0.10 mol / L) = 0.0050 moles of H₂SO₄
As 1 mole of the acid reacts per mole of NaOH, to reach the first equivalence point we need to add 0.0050 moles of NaOH. As molarity of NaOH is 0.10M, the volume that we need to add to reach 1st equivalence point is:
0.0050 moles NaOH ₓ (1L / 0.10 moles NaOH) = 0.050L NaOH 0.10M =
50.0mL 0.10M NaOHWhat will be the volume occupied by 2.5 moles of nitrogen gas exerting 1.75 atm of pressure at 475K?
Answer:
THE VOLUME OF THE NITROGEN GAS AT 2.5 MOLES , 1.75 ATM AND 475 K IS 55.64 L
Explanation:
Using the ideal gas equation
PV = nRT
P = 1.75 atm
n = 2.5 moles
T = 475 K
R = 0.082 L atm/mol K
V = unknown
Substituting the variables into the equation we have:
V = nRT / P
V = 2.5 * 0.082 * 475 / 1.75
V = 97.375 / 1.75
V = 55.64 L
The volume of the 2.5 moles of nitrogen gas exerted by 1.75 atm at 475 K is 55.64 L
Select the correct answer. What affects the efficiency of a heat engine? A. high friction between hot and cold reservoirs B. the heat difference between hot and cold reservoirs C. low friction between hot and cold reservoirs D. the temperature difference between hot and cold reservoirs
Explanation:
Efficiency of a Carnot heat engine depends on :
(D) The temperature difference between hot and cold reservoirs
43 milliliters of water weighs 43 g. what is the density of the water?
Answer:
[tex]\rho =1g/mL[/tex]
Explanation:
Hello,
In this case, since the density is defined as the ratio between the mass and the volume as shown below:
[tex]\rho =\frac{m}{V}[/tex]
We can compute the density of water for the given 43 g that occupy the volume of 43 mL:
[tex]\rho =\frac{43g}{43mL}=1g/mL[/tex]
Regards.
A constant current of 1.30 amps is passed through an electrolytic cell containing OsO2. How many hours will it take to produce 45.0 grams of osmium metal from this process
Answer:
9.8 h
Explanation:
From the question,
Os²⁻ - 2e⁻ = Os
From the equation above,
2 F of electron is needed to liberate 1 mole of osmium.
45 g of osmium contains 45/190 mole of osmium = 0.237 mole.
2 F ⇒ 1 mole.
2(0.237) F ⇒ 0.237 mole
0.474 F. will produce 45 g of osmium.
If 1 F = 96500 C ,
Then, 0.474 F = 45741 C,
But,
Q = it
t = Q/i....................... Equation 1
Given: Q = 45741 C, i = 1.30 A
Substitute into equation 1
t = 45741/1.3
t = 9.8 h.
How do you solve this ??
Answer:
Option D. 400 mmHg
Explanation:
The following data were obtained from the question:
Mole of He (nHe) = 0.04 mole
Mole of Ne (nNe) = 0.06 mole
Total pressure = 10³ mmHg
Partial pressure of He =.?
Next, we shall determine the total number of mole in the reaction vessel.
This can be obtained as follow:
Mole of He (nHe) = 0.04 mole
Mole of Ne (nNe) = 0.06 mole
Total mole =?
Total mole = nHe + nNe
Total mole = 0.04 + 0.06
Total mole = 0.1
Next, we shall determine the mole fraction of He.
This can be obtained as follow:
Mole fraction = mole of gas /total mole
Mole of He (nHe) = 0.04 mole
Total mole = 0.1
Mole fraction of He =.?
Mole fraction of He = nHe/total mole
Mole fraction of He = 0.04/0.1
Mole fraction of He = 0.4
Finally, we shall determine the partial pressure of He as follow:
Partial pressure = mole fraction x total pressure
Mole fraction of He = 0.4
Total pressure = 10³ mmHg
Partial pressure of He =.?
Partial pressure of He = 0.4 x 10³
Partial pressure of He = 400 mmHg.
Therefore, the partial pressure of He is 400 mmHg.
Which of the following is not a
trigonal planar molecule?
Select one:
a. AICI:
b. NH3
c. BF3
d. AlH3
Question 17
Using standard heats of formation, calculate the standard enthalpy change for the following reaction.
H2(g) + C2H4(g)--------C2H6(g)
Answer:
[tex]\Delta _rH=-136.27kJ/mol[/tex]
Explanation:
Hello,
In this case, for the given reaction, the enthalpy of reaction is computed in terms of the enthalpies of formation as:
[tex]\Delta _rH=\Delta _fH_{C_2H_6}-\Delta _fH_{C_2H_4}-\Delta _fH_{H_2}[/tex]
Whereas hydrogen, ethene and ethane enthalpies of formation are 0 kJ/mol, 52.47 kJ/mol and -83.8kJ/mol respectively. Therefore, we compute:
[tex]\Delta _rH=-83.8kJ/mol-52.47kJ/mol-0kJ/mol\\\\\Delta _rH=-136.27kJ/mol[/tex]
Best regards.
What is the procedure for heating a metal to an exact but measured temperature?
Answer:
An unknown metal between 10-30 g is transferred to a test tube, place the tube in water that is boiling for ten minutes until it reaches a thermal equilibrium.
Explanation:
Write the net ionic equation for the precipitation of silver phosphate from aqueous solution:________
(Use the solubility rules provided in the OWL Preparation Page to determine the solubility of compounds.)
Answer: 3Ag⁺ (aq) + PO₄³⁻ (aq) → Ag₃PO₄ (s)
Explanation:
The question requires us to write down the net ionic equation for the precipitation of silver phosphate from aqueous solution:________
Note that we we told to:
(Use the solubility rules provided in the OWL Preparation Page to determine the solubility of compounds.)
Answer:
From the solubility rule,
Phosphate (PO₄³⁻) and Carbonates are insoluble except for NH₄⁺ and Group1A compounds.
According to the solubility rule, the precipitation of silver phosphates and that of the net ionic equations is given thus;
3AgNO³ (aq) + K₃PO₄ (aq) → Ag₃PO₄ (s) + 3KNO₃ (aq)
while the Net ionic equation is given;
3Ag⁺ (aq) + PO₄³⁻ (aq) → Ag₃PO₄ (s)
cheers i hope this helped !!
An unknown compound’s semicarbazone melts at 162-165 degree Celsius. The compound does not give a silver mirror in the Tollens’ test. On reaction with KI/I2, NaOH/H2O no straw yellow precipitate forms. Identify the unknown. Explain your reasoning.
Answer:
Benzophenone
Explanation:
Given that:
an unknown compound’s semicarbazone melts at 162° -165°
The compound does not give a silver mirror in the Tollens’ test.
On reaction with KI/I2, NaOH/H2O no straw yellow precipitate forms.
The objective is to identify the unknown compound.
Since the unknown compound result to a semicarbazone, we can deduce that the unknown compound is a carbonyl compound. A carbonyl compound is either an aldehyde or ketone in nature. Also, the absence of silver mirror in the Tollens’ test carried out in the reaction confirms that the compound is a ketone because ketones will never give a silver mirror in Tollens’ test.
Similarly, on reaction with KI/I2, NaOH/H2O no straw yellow precipitate forms, that is an iodoform test. That implies that a keto-methyl group is absent in the unknown compound.
Finally, since the unknown compound melting point is between 162° -165° and Benzophenone semicarbazone melting point is 164°.
We can conclude that the unknown compound is Benzophenone semicarbazone
2. Determine the molarity of the NaOH solution in each trial. a. Trial 1 Molarity: b. Trial 2 Molarity: 3. Calculate the average molarity of the NaOH solution. 4. Label the volumetric flask containing the NaOH solution with the average molarity.
Answer:
This question is incomplete
Explanation:
This question is incomplete as the actual values are not provided however you can solve the question yourself if you have the actual value using the formulas below.
The formula for Molarity is
Molarity = number of moles (n) ÷ volume (in dm³ or Litre)
If number of moles is not given, to calculate for number of moles
number of moles (n) = given mass of NaOH ÷ molar mass of NaOH
Please NOTE that the volume MUST be converted to dm³ (or litre) if not provided in dm³.
The unit for Molarity is mol/dm³ or mol/L or M (molar concentration).
Average molarity = (Trial 1 + Trial 2 + Trial 3)/ 3
Hydrogen reacts with an element to form a compound. Which element would have the most valence electrons and also be able to react with hydrogen? oxygen chlorine neon nitrogen
Answer:
Chlorine
Explanation:
Hello.
In this case, for the given elements, we are able to know that oxygen has 6 valence electrons, chlorine 7, nitrogen 5 and neon 8, therefore neon is not able to react as it already has 8 valence electrons. Besides, the element having the most valence electrons is chlorine and its reaction with hydrogen forms hydrogen chloride as shown below:
[tex]H_2(g)+Cl_2(g)\rightarrow 2HCl(g)[/tex]
Therefore, the required element is chlorine.
Regards.
Chlorine element would have the most valence electrons and also be able to react with hydrogen
The atomic number of hydrogen is 1 and its electronic configuration is [tex]1s^{1}[/tex] .
The atomic number of oxygen is 8 and its electronic configuration is [tex]1s^{2} 2s^{2} 2p^{4}[/tex]. The valence electron in oxgen is 6.
The atomic number of chlorine is 17 and its electronic configuration is [tex]1s^{2} 2s^{2} 2p^{6}3s^{2} 3p^{5}[/tex]. The valence electron in chlorine is 7.
The atomic number of neon is 10 and its electronic configuration is [tex]1s^{2} 2s^{2} 2p^{6}[/tex]. The valence electron in neon is 8.
The atomic number of nitrogen is 7 and its electronic configuration is [tex]1s^{2} 2s^{2} 2p^{3}[/tex]. The valence electron in nitrogen is 5.
Except Neon, all three atoms will react with hydrogen as the configuration of neon is a stable electronic configuration.
Chlorine has the greatest number of valence electrons after neon that is 7. So, chlorine would have the most valence electrons and also be able to react with hydrogen as follows:-
[tex]H_2+Cl_2\rightarrow2HCl[/tex]
Hence, the correct answer is chlorine.
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Which of the following statements is true? A. The process of meiosis forms daughter cells which are genetically identical to their parent cells. B. Mitosis results in the formation of two haploid gametes which can then combine to form a diploid daughter cell. C. During the process of meiosis, haploid cells are formed. After fertilization, the diploid number of chromosomes is restored. D. The daughter cells formed during mitosis are genetically similar to, though not identical to, their parent cell.
Answer:
Mitosis results in the formation of two haploid gametes which can then combine to form a diploid daughter cell
Explanation:
Meiosis creates different cells
Answer:
During the process of meiosis, haploid cells are formed. After fertilization, the diploid number of chromosomes is restored.
Explanation:
How much heat will be absorbed by a 50.3 g piece of aluminum (specific heat = 0.930 J/g･°C) as it changes temperature from 23.0°C to 67.0°C?
A 50.3 g piece of aluminum (specific heat = 0.930 J/g･°C) will change its temperature from 23.0°C to 67.0°C while absorbing 2.06 × 10³ J of heat.
When a material absorbs heat, its temperature increases. We can calculate the amount of heat (Q) absorbed using the following expression.
[tex]Q = c \times m \times \Delta T[/tex]
where,
c: specific heatm: massΔT: change in the temperature[tex]Q = c \times m \times \Delta T\\Q = \frac{0.930J}{g. \° C } \times 50.3 g \times (67.0 \° C - 23.0 \° C) = 2.06 \times 10^{3} J[/tex]
A 50.3 g piece of aluminum (specific heat = 0.930 J/g･°C) will change its temperature from 23.0°C to 67.0°C while absorbing 2.06 × 10³ J of heat.
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The molecule below, 2-acetyl-1-pyrroline (2-AP), is associated with the aroma and flavor of white bread and basmati rice. In higher concentrations, it gives cooked popcorn its characteristic buttery aroma and flavor. Its odor is very intense: We can detect 2-AP in water at concentrations below one nanogram/liter. The buttery aspect derives from a hydrolysis reaction. Identify the functional group sensitive to hydrolysis, write out a hydrolysis mechanism under acidic conditions, and give the structure of the product.
Answer:
OLA IQESOXPJMXXXXOX{OADUOOAOOOOOSCDCSXWAXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
Explanation:
The potential of a voltaic cell (Ecell) is calculated based on several variables. Identify from the list below the values you would need to determine the cell potential.
a. number of electrons, n
b. density of metal, rho
c. coefficients from balanced redox equation (various)
d. temperature, T
e. Gibbs free energy change, ΔG
f. enthalpy change, ΔH
g. reduction potential, Ered
Answer:
number of electrons, n
Gibbs free energy change, ΔG
temperature, T
coefficients from balanced redox equation (various)
Explanation:
The standard electrode potential of a cell can be obtained from;
∆G°= -nFE°cell
Also;
E°cell= RTlnK/nF
Where;
∆G°= standard free energy of the cell
n= number of electrons transferred
F= Faraday constant
E°cell= standard cell potential
R= universal gas constant
T= temperature
K= equilibrium constant
Which metal(s) can be oxidized with a Pb2+ solution but not with a Cd2+ solution? (Hint: The reactions are occurring under standard conditions.)
A. Cr
B. Ni
C. Zn
D. Sn
Answer:
B. Ni
D. Sn
Explanation:
Electrode Potential is the potential difference set up between an element and a solution of its ion. It is a measure of the tendency of an element to form ions.
The electrode potentials vary from one metal ion or metal system to another and the value depends on:
concentration of ions in the solution
the temperature at which the measurement is made , and
the overall energy change.
When two half-cells are joined together through a salt bridge, the e.m.f (electromotive force) of the cell formed is the algebraic difference between the two electrode potentials.
However, the set up in which chemical energy is converted to electrical energy is known as an Electrochemical cell. It consists of two half cells ;
an oxidation half-cell reaction
a reduction half cell reaction.
From the information given:
the standard reduction potential for each metal under standard conditions in the electrochemical series is as follows :
[tex]E^0 _{Pb} = -0.126 \ V[/tex]
[tex]E^0 _{Cr} = - 0.74 \ V[/tex]
[tex]E^0 _{Ni} =- 0.23 \ V[/tex]
[tex]E^0 _{Zn} =- 0.76 \ V[/tex]
[tex]E^0 _{Sn} = -0.13 \ V[/tex]
[tex]E^0 _{Cd} = - 0.40 \ V[/tex]
We will realize that Ni and Sn have reduction values in between Pb and Cd.
Thus , Ni can be oxidized by Pb2+ solution but not with a Cd2+ solution
The metal(s) that can be oxidized with a [tex]Pb^{2+}[/tex] solution
B. Ni
D. Sn
What is Electrode potential?It is the potential contrast set up between a component and an answer of its particle. It is a proportion of the inclination of a component to shape particles. The cathode possibilities shift from one metal particle or metal framework to another and the worth relies upon:
Centralization of particles in the arrangement, the temperature at which the estimation is made , and the general energy change.
Whenever two half-cells are combined through a salt scaffold, the e.m.f (electromotive power) of the cell shaped is the mathematical contrast between the two terminal possibilities.
However, the set up in which chemical energy is converted to electrical energy is known as an Electrochemical cell. It consists of two half cells ;
an oxidation half-cell reaction and a reduction half cell reaction.
On seeing the values of electrode potential from electrochemical series we observe that Ni and Sn have reduction values in between Pb and Cd.
Thus , Ni and Sn can be oxidized by [tex]Pb^{2+}[/tex] solution but not with a [tex]Cd^{2+}[/tex]solution.
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When the pH of the solution equals the pKa of an indicator, the solution will have an intermediate color. Estimate the pKa of bromocresol green, the indicator that made the transition in the acidic region of the titration.
Answer:
pKa bromocresol green is ≅4.5
Explanation:
The bromocresol green is a chemical indicator used in titrations with equivalence point at pH's between 3 and 5.
Is an indicator that, in acidic region is yellow, and in basicic region is blue. The intermediate color is green (at pH≅ 4.5).
As at the intermediate color of the indicator pKa = pH,
pKa bromocresol green is ≅4.5Give the net ionic equation for the reaction (if any) that occurs when aqueous solutions of H2SO4 and KOH are mixed.
a. H+(aq) + OH -(aq) → H2O(l)
b. 2K+(aq) + SO42-(aq) → K2SO4(s)
c. H+(aq) + OH-(aq) + 2K+(aq) + SO42-(aq) → H2O(l) + K2SO4(s)
d. H2^2+(aq) + OH-(aq) → H2(OH)2
e. No reaction occurs.
Answer:
a. H⁺(aq)+ OH⁻(aq) → H₂O(l)
Explanation:
First, we will write the molecular equation for the reaction between H₂SO₄ and KOH.
H₂SO₄(aq) + 2 KOH(aq) → K₂SO₄(aq) + 2 H₂O(l)
The complete ionic equation includes all the ions and the molecular species.
2 H⁺(aq) + SO₄²⁻(aq) + 2 K⁺(aq) + 2 OH⁻(aq) → 2 K⁺(aq) + SO₄²⁻(aq) + 2 H₂O(l)
The net ionic equation includes only the ions that participate in the reaction and the molecular species.
H⁺(aq)+ OH⁻(aq) → H₂O(l)
You wish to set up an electrochemical cell that has a copper anode and an
Eºcell value > 1.00 V. What should your cathode and cathode solution be?
Consider the following reaction at equilibrium. What effect will increasing the pressure of the reaction mixture have on the system?
CuS(s) + O2(g) ? Cu(s) + SO2(g)
Consider the following reaction at equilibrium. What effect will increasing the pressure of the reaction mixture have on the system?
CuS(s) + O2(g) ? Cu(s) + SO2(g)
a) The equilibrium constant will increase.
b) The reaction will shift to the left in the direction of reactants.
c) The reaction will shift to the right in the direction of products.
d) No effect will be observed.
e) The equilibrium constant will decrease.
Answer:
Choice d. No effect will be observed as long as other factors (temperature, in particular) are unchanged.
Explanation:
The equilibrium constant of a reaction does not depend on the pressure. For this particular reaction, the equilibrium quotient is:
[tex]Q = \displaystyle \frac{[\mathrm{SO_2\, (g)}]}{[\mathrm{O_2\, (g)}]}[/tex].
Note that the two sides of this balanced equation contain an equal number of gaseous particles. Indeed, both [tex][\mathrm{SO_2\, (g)}][/tex] and [tex][\mathrm{O_2\, (g)}][/tex] will increase if the pressure is increased through compression. However, because [tex]\rm SO_2\, (g)[/tex] and [tex]\rm O_2\, (g)[/tex] have the same coefficients in the equation, their concentrations are raised to the same power in the equilibrium quotient [tex]Q[/tex].
As a result, the increase in pressure will have no impact on the value of [tex]Q\![/tex]. If the system was already at equilibrium, it will continue to be at an equilibrium even after the change to its pressure. Therefore, no overall effect on the equilibrium position should be visible.
We explored the copper cycle. In the first step, copper was oxidized by nitric acid to make a green solution. Water was then added to the solution and the color changed from green to blue. Is this result supported by the spectrochemical series?
Answer:
Yes, the reaction is supported by spectrochemical series.
Explanation:
When copper oxidizes with concentrated nitric acid, it produces copper ions; and the nitric acid is then reduced to nitrogen dioxide, a very poisonous gas which is brown in color and with an irritating odor.
[tex]$ Cu + 4HNO_{3} \rightarrow Cu(NO_{3})_{2} + 2NO_2+2H_2O$[/tex]
The copper ions ([tex]$Cu^{2+} $[/tex]) product which is initially coordinated to form nitrate ions from nitric acid and thus first gives the solution a green color.
Now when the solution is further diluted with water, the water molecules then displaces the nitrate ions around the copper ions and causes the solution to change its color to blue. The reaction is -
[tex]$ 3Cu + 8HNO_3 \rightarrow 3Cu(NO_3)_2+ 2NO + 4H_2O $[/tex]
Yes, the results are supported by spectrochemical series.
How does a lone pair distort the molecular shape?
O A. The lone pair forces bonding atoms farther from the nucleus.
B. The lone pair makes new bonds, changing the molecule's shape.
O C. The lone pair pushes bonding atoms closer to the nucleus.
O D. The lone pair forces bonding atoms away from itself.
SUBMIT
Answer:
D. The lone pair forces bonding atoms away from itself.
Explanation:
The lone pairs are found in the outermost shell thereby making sharing of electrons easier. Lone pairs which are found in a covalent bond creates a bond angle which makes it a determinant of the bond angle.
The lone pairs which are negatively charged repulses the bond pairs thereby creating a distortion in the shape of the molecule.
The lone pairs creates the distortion by forcing bonding atoms away from itself.
Express 450 x 10-9 m (wavelength of blue light) in decimal notation (i.e., express
the number without using scientific notation).
Answer:
0.00000045 m is the wavelength of blue light in decimal
0.00000045 m is the wavelength of blue light in decimal without using scientific notation.
What is scientific notation?
Scientific notation is defined as a way of expressing numbers which are too large or too small so that they can be easily written in decimal form. It can be referred to the scientific form or the scientific index form or even the standard form.
Base ten notation is used by scientists, engineers as it helps in simplification of arithmetic operations. It contains the significant figures which include all non-zero numbers , the zeroes between significant digits and zeroes which are needed to be significant.
In scientific notation, the base should always be ten. The exponent should be a non-zero integer and it can be either positive or negative.
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