Answer:
Explanation:
Ba(s) + Mn²⁺ (aq,1M) → Ba²⁺ (aq,1M) + Mn(s)
Ba⁺²(aq) +2e → Ba(s) , E° = −2.90 V
Mn⁺²(aq) +2e → Mn(s), E⁰ =0.80 V
Anode reaction :
Ba(s) → Ba⁺²(aq) +2e E° = −2.90 V
Cathode reaction :
Mn⁺²(aq) +2e → Mn(s) E⁰ =0.80 V
Cell potential = Ecathode - Eanode
Ecell = .80 - ( - 2.90 )
Ecell = 3.7 V .
equilibrium constant ( K ) :
Ecell = .059 log K / n
n = 2
3.7 = .059 log K / 2
log K = 125.42
K = 2.63 x 10¹²⁵ .
Free energy change :
ΔG = - n F Ecell
= - 2 x 96500 x 3.7
= 714100 J
= 7.141 x 10⁵ J .
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
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
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.
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:
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 NaOHPerform the following calculations and round according to the rules for significant figures. Just type in the number as a number (ex: 44)
8.20cm + 2cm (assume that the number is already labelled with the unit of cm)=
PLEASE HELP
Answer:
10 cm (Approx)
Explanation:
Given:
8.20 cm + 2 cm
Find:
Round of the following number
Computation:
⇒ 8.20 cm + 2 cm
⇒ 10.20 cm
We know that 0.20 < 0.50
So,
⇒ 10.20 cm = 10 cm (Approx)
Using VSEPR model, how is the molecular arrangement about the central atom (molecular geometry) for XeF4?
Answer:
Square planar.
Explanation:
Hello,
In this case, it can be demonstrated that the central atom Xenon in XeF₄ has a sp³d² hybridization which means that its geometry is likely to be octahedral. Nevertheless Applying by applying the valence shell electron pair repulsion (VSEPR) model to in order to minimize the repulsion among lone pairs, bond pairs and lone pair-bond pairs, we we realize repulsion is actually minimized when lone pairs are anti to one another and the fluorine atoms are in equatorial position, for that reason, the corrected and properly exhibited geometry or molecular arrangement of the compound turns out square planar. You can verify it on the attached picture.
Regards.
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
The following initial rate data are for the reaction of hypochlorite ion with iodide ion in 1M aqueous hydroxide solution: OCI+r → or +CI
Experiment [OCI] M I(-M) Rate (M/s)2
1 3.48 x 10-3 5.05 x 10-3 1.34 x 10-3
2 3.48 x 10-3 1.01 x 10-2 2.68 x 10-3
3 6.97 x 10-3 5.05 x 10-3 2.68 x 10-3
4 6.97 x 10-3 1.01 x 10-2 5.36 x 10-3
What is the rate law for this reaction?
a. Rate = [1-1
b. Rate = K[OCI-11-2
c. Rate = K[OCI,20-1
d. Rate = K[OCH
e. Rate = K[OC1-20-1
Answer:
Rate = k [OCl] [I]
Explanation:
OCI+r → or +CI
Experiment [OCI] M I(-M) Rate (M/s)2
1 3.48 x 10-3 5.05 x 10-3 1.34 x 10-3
2 3.48 x 10-3 1.01 x 10-2 2.68 x 10-3
3 6.97 x 10-3 5.05 x 10-3 2.68 x 10-3
4 6.97 x 10-3 1.01 x 10-2 5.36 x 10-3
The table above able shows how the rate of the reaction is affected by changes in concentrations of the reactants.
In experiments 1 and 3, the conc of iodine is constant, however the rate is doubled and so is the conc of OCl. This means that the reaction is in first order with OCl.
In experiments 3 and 4, the conc of OCl is constant, however the rate is doubled and so is the conc of lodine. This means that the reaction is in first order with I.
The rate law is given as;
Rate = k [OCl] [I]
Calculate the standard cell potential given the following standard reduction potentials: Al3++3e−→Al;E∘=−1.66 VAl3++3e−→Al;E∘=−1.66 V Ag++e−→Ag;E∘=0.799 V
Answer:
2.459 V
Explanation:
If we look at the data we have from the question closely, we will discover that aluminum has a very negative reduction potential. Recall that very negative reduction potentials are associated with strong reducing agents.
Similarly, silver has a positive reduction potential signifying that it is an oxidizing agent. Since the reducing agent is oxidized in a redox reaction and the oxidizing agent is reduced.
Hence;
E°cell= E°reduction process - E°oxidation process
E°cell= 0.799 -(-1.66)
E°cell= 2.459 V
The standard cell potential is 2.459 V
Standard cell potential:The standard cell potential is the potential difference between the cathode and anode.
The overall cell potential can be calculated by using the equation: [tex]E_0cell=E_0red-E_0oxid[/tex].
Aluminum has a very negative reduction potential. The negative reduction potentials are associated with strong reducing agents and Silver has a positive reduction potential signifying that it is an oxidizing agent. Since the reducing agent is oxidized in a redox reaction and the oxidizing agent is reduced.
Hence;
E°cell= E°reduction process - E°oxidation process
E°cell= 0.799 -(-1.66)
E°cell= 2.459 V
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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.
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.
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.
Which of these is the basic unit of mass?
L
kg
cm
ml
A galvanic (voltaic) cell has the generic metals X and Y as electrodes. X is more reactive than Y, that is, X more readily reacts to form a cation than Y does. Classify the descriptions by whether they apply to the X or Y electrode.
a. Anode
b. Cathode
c. Loses mass
d. Electrons in the wire flow away from it
e. Anions from the salt bridge flow toward it
f. Cations from the salt bridge flow toward it
g. Gains mass
f. Electrons in the wire flow toward it
Answer:
X
Anode
Loses mass
Electrons in the wire flow away from it
Anions from the salt bridge flow toward it
Y
Cathode
Gains mass
Electrons in the wire flow toward it
Cations from the salt bridge flow toward it
Explanation:
A galvanic cell consists of an anode, a cathode and a salt bridge. The cathode is the negative electrode. The cathode increases in mass due to mass deposition. The cathode becomes increasingly negative as the cell reaction progresses, hence cations from the salt bridge flow towards it.
The anode is the positive electrode. It becomes more positive as the cell reaction progresses hence anions from the salt bridge flow towards it. The anode is the electrode where oxidation reaction occurs hence electrons are lost here and flow towards the cathode. The anode looses mass as the cell reaction progresses due to oxidation.
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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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Identify the product of radioactive decay and classify the given nuclear reactions accordingly.
a) 23488Ra â 23086Rn + _____
b) 23892U â 23893Np + _______
c) 24294Pu â 23892U + ______
d) 146C â 147N + ______
e) 2412Mg â 2412Mg + ________
Answer: a) [tex]_{88}^{234}\textrm{Ra}\rightarrow _{86}^{230}\textrm{Ra}+_{2}^{4}\textrm{He}[/tex]
b) [tex]_{92}^{238}\textrm{Ra}\rightarrow _{93}^{238}\textrm{Np}+_{-1}^{0}{\beta}[/tex]
c) [tex]_{92}^{238}\textrm{Ra}\rightarrow _{93}^{238}\textrm{Np}+_{-1}^{0}{\beta}[/tex]
d) [tex]_{6}^{14}\textrm{C}\rightarrow _{7}^{14}\textrm{N}+_{-1}^{0}{\beta}[/tex]
e) [tex]_{12}^{24}\textrm{Mg}\rightarrow _{12}^{24}\textrm{Mg}+_{0}^{0}{\gamma}[/tex]
Explanation:
A balanced nuclear equation is one in which the atomic number and mass number remains same on both sides of the equation i.e the number of protons and neutrons remain same.
General representation of an element is given as:
Z represents Atomic number
A represents Mass number
X represents the symbol of an element
a) [tex]_{88}^{234}\textrm{Ra}\rightarrow _{86}^{230}\textrm{Ra}+_{2}^{4}\textrm{He}[/tex]
b) [tex]_{92}^{238}\textrm{U}\rightarrow _{93}^{238}\textrm{Np}+_{-1}^{0}{\beta}[/tex]
c) [tex]_{94}^{242}\textrm{Pu}\rightarrow _{92}^{238}\textrm{Ra}+_{2}^{4}\textrm{He}[/tex]
d) [tex]_{6}^{14}\textrm{C}\rightarrow _{7}^{14}\textrm{N}+_{-1}^{0}{\beta}[/tex]
e) [tex]_{12}^{24}\textrm{Mg}\rightarrow _{12}^{24}\textrm{Mg}+_{0}^{0}{\gamma}[/tex]
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
The molecular formula of benzene is C6H6. what is the empirical formula of benzene?
I WILL GIVE BRAINLYEST 40+ POINTS
A is correct because the atomic number determines that there are two protons and electrons. The atomic mass minus the atomic number determines that there are two neutrons.
B is correct because the atomic number determines that there are two protons and neutrons. The atomic mass determines that there are four electrons in the electron shell.
D is correct because the atomic number determines that there are two protons and electrons. The atomic mass minus the atomic number determines the number of neutrons.
C is correct because the atomic number determines that there are two protons and electrons. The atomic mass determines that there are four neutrons in the nucleus.
Answer:
The correct answer is A.
Explanation:
Helium's Atomic Model has 2 Protons and 2 Neutrons, and on the outside it has 2 electrons circling around it.
Answer:
is A
Explanation:
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?
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 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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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.
If salt and water mixed together which is a solvent and which is the solute
Answer:
Solvent = Water
Solute = Salt
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)
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Explain how passive and active transport are similar and how they are different.
Answer:
Sample Response: Both passive and active transport move substances into and out of the cell across the cell membrane. However, passive transport does not require the use of energy because molecules move from an area of high concentration to an area of low concentration. Active transport does require the use of energy because molecules move from an area of low concentration to an area of high concentration.
Explanation:
Active and passive transport are the two main biological processes which play an important role in providing the nutrients, oxygen, water and other essential molecules to the cells along with removing waste materials.
What are active and passive transport?The active transport involves the movement of molecules across a membrane against a concentration gradient which requires energy. The passive transport is the movement of molecules or ions across the cell membrane without using energy.
The differences between active and passive transport are: Active transport is a rapid process, passive transport is a slow process. Active transport transpires in one direction whereas passive one transpires bidirectionally. Active transport requires carrier proteins and energy whereas passive one does not require both.
The similarities among them are: both transport system require concentration gradient, both helps in homeostatic regulation and allow entry and exit of ions.
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your friend makes the statement that " I think water boils at a higher temperature than ethanol," to which step of scientific method does this statement belong? scientific law hypothesis experiment observation
hypothesis
need 20 characters hope this helps
A 100.0 mL sample of 0.18 M HCl is titrated with 0.27 M KOH at 25°C. Determine the pH of the solution after the addition of 66.67 mL of the KOH titrant.
Answer:
pH = 7.0
Explanation:
The HCl reaction with KOH as follows:
HCl + KOH → H₂O + KCl
To find pH first we need to know how many moles of each reactant reacts as follows:
HCl = 0.1000L * (0.18 mol / L) = 0.018 moles HCl
KOH = 0.06667L * (0.27mol / L) = 0.018 moles KOH
That means all HCl reacts with all KOH and you will have in solution just water. The equilibrium of water is:
H₂O(l) ⇆ H⁺(aq) + OH⁻(aq)
K = 1x10⁻¹⁴ = [H⁺] [OH⁻]
As the amount of H⁺ = OH⁻:
1x10⁻¹⁴ = [H⁺]²
[H⁺] = 1x10⁻⁷M
as pH = -log [H⁺]
pH = 7.0Express 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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