Answer:
6.25 ×10^10 Hz
Explanation:
If the distance between five successive crests is 2.4 cm, then the distance between each crest is 2.4/5 = 0.48 cm or 0.0048 m or 4.8 ×10^-3 m
Since the velocity of a wave is given by;
v= λf
Where;
λ= wavelength of the wave
f= frequency of the wave
But λ= distance between successive crests = 4.8 ×10^-3 m
v= 3×10^8 ms-1 (speed of electromagnetic waves in vacuum)
f= v/λ
f= 3×10^8 ms-1/4.8 ×10^-3 m
f= 0.625 ×10^11 Hz
f= 6.25 ×10^10 Hz
Calculate the molality of a 15.0% by mass solution of MgCl 2 in H 2O. The density of this solution is 1.127 g/mL. Group of answer choices
Answer:
1.86 m
Explanation:
Step 1: Calculate the mass of solute and solvent in 100 grams of solution
We have a 15.0% by mass solution of MgCl₂ (solute) in H₂O (water). In 100 g of solution, there are 15.0 g of solute and 100.0 g - 15.0 g = 85.0 g of solvent.
Step 2: Calculate the moles corresponding to 15.0 g of MgCl₂
The molar mass of MgCl₂ is 95.21 g/mol.
[tex]15.0g \times \frac{1mol}{95.21 g} = 0.158mol[/tex]
Step 3: Convert the mass of water to kilograms
We will use the relationship 1 kg = 1,000 g.
[tex]85.0g \times \frac{1kg}{1,000g} = 0.0850kg[/tex]
Step 4: Calculate the molality of the solution
[tex]m = \frac{0.158mol}{0.0850kg} = 1.86m[/tex]
The molality is 1.86 m.
Given:
Mass solution of MgCl₂ in H₂O= 15%
Density= 1.127g/mL
To find:
Molality=?
Let's solve this question step by step:
Step 1: Calculate the mass of solute and solvent in 100 grams of solution
As it is given that 15.0% by mass solution of MgCl₂ (solute) in H₂O (water).
In 100 g of solution, there are 15.0 g of solute and 100.0 g - 15.0 g = 85.0 g of solvent.
Step 2: Calculate the moles corresponding to 15.0 g of MgCl₂
The molar mass of MgCl₂ is 95.21 g/mol.
Number of moles is given mass divided by the molar mass.
[tex]\text{Number of moles}= \frac{15g}{95.21g/mol} =0.158mol[/tex]
Step 3: Convert the mass of water to kilograms.
As we know, 1 kg = 1,000 g.
[tex]=\frac{85}{1000} =0.085kg[/tex]
Step 4: Calculate the molality of the solution.
Molality is a measure of the number of moles of solute in a solution corresponding to 1 kg or 1000 g of solvent.
[tex]\text{Molality}=\frac{0.158\text{mol}}{0.085\tetx{kg}} =1.86 m[/tex]
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Draw the structure of the amine and carboxylic acid reactants required to form the following amide in an amidation reaction.Draw the starting amine. Be sure to draw nonbonding electron pairs.
Draw the starting carboxylic acid. Be sure to draw nonbonding electron pairs.
Answer:
Explanation:
Amidation is a reaction with the formation of an amide. It involves a process where series of reaction between amine and carboxylic acid takes place.
The main objective of this question is to draw with the aid of a diagram.
the structure of the amine and carboxylic acid reactants required to form the following amide in an amidation reaction.
But before that,
we are to draw the starting amine and to be sure to include the nonbonding electron pairs.
Also, to draw starting carboxylic acid and to be sure to include the nonbonding electron pairs.
The whole series of the diagrammatic expression for the amine formation can be found in the diagram attached below.
Which of the following has the longest half-life?
a. 0-19
b. H-3
c. Zn-71
d. Rb-85
Answer:
b. H-3
Explanation:
The half life of the following isotopes is given as;
a. 0-19
26.88 seconds
b. H-3
12 years
c. Zn-71
2.4 minutes
d. Rb-85
This is a stable isotope. It does not decomposes.
Comparing the half lives, the isotope with the longest half life is Hydrogen - 3. With half life of 12 years.
There is an electrolytic cell in which Mn2+ is reduced to Mn and Sn is oxidized to Sn2+.
A. Write an equation for the half-reaction occurring at each electrode. Express your answers as chemical equations separated by a comma. Identify all of the phases in your answer.
B. What minimum voltage is necessary to drive the reaction? Vmin =
Answer:
See explanation
Explanation:
Anode;
Sn(s) ------> Sn^2+(aq) + 2e
Cathode;
Mn^2+(aq) + 2e ------> Mn(s)
The minimum voltage required to drive the reaction is the cell voltage. The cell voltage is obtained from;
E°cell= E°cathode - E°anode
E°cell= -1.19 - (-0.14)
E°cell= -1.05 V
Which of the following can't undergo aldol condensation.
(CH3)3CCHO
CH3CH2CHO
PhCOCH3
CH3COCH3
Answer:
(CH3)3CCHO
Explanation:
Carbonyl compounds that contain alpha hydrogen atoms are the only ones that can undergo aldol condensation reaction.
Aldol condensation is one of the condensation reactions in organic chemistry that occurs when an enol or an enolate ion reacts with another carbonyl compound to yield a β-hydroxyaldehyde or β-hydroxyketone, this is immediately followed by the dehydration of the product to yield a conjugated enone.
(CH3)3CCHO lacks alpha hydrogen atoms hence it does not undergo the aldol condensation.
Two elements, R and Q, combine to form two binary compounds. In the first compound, 14.0 g of R combines with 3.00 g of Q. In the second compound, 7.00 g of R combines with 4.50 g of Q. If the formula of the second compound is RQ3, what is the formula of the first compound?
Answer:
The formula of the first compound is RQ
Explanation:
Chemical formula is the simplest whole number ratio of moles of each element in the compound.
To determine the formula of a compound we must know the molar mass of each element (How many grams weighs a mole of the compound).
The formula of the second compound is RQ₃. That is 1 mole of R and 3 moles of Q.
1 mole of R weighs 7.00g. That means its molar mass is 7g/mol.
3 moles of Q weighs 4.50g. Its molar mass is 4.50g / 3 moles: 1.50g/mol
In the first compound, there are 14.0g = 2 moles of R
And 3.00g of Q = 2 moles of Q.
And you can think formula is:
R₂Q₂
But, as formula is the simplest whole number ratio,
The formula of the first compound is RQThe mixing of which pair of reactants will result in a precipitation reaction?
A. AsI(aq) + NaOH(aq)
B. NaNO3(aq) + NH4Cl(aq)
C. K2SO4(aq) + Ba (NO3)2(aq)
D. HCl(aq) + Ca(OH)2(aq)
E. None of the above.
Answer:
B
Explanation:
9.57: How many kilojoules of heat will be released when exactly 1 mole of manganese, Mn, is burned to form Mn3O4(s) at standard state conditions?
Answer:
459.6kj
Explanation:
the chemical equation =
3mn(s) + 2CO2(g) --- Mn3O4(s)
we have change in H to be -1378.83
we are required to find the amount of heat that is liberated for a mole of manganese.
3 moles of manganese = -1378.83
1 mole of manganese = ?
when we cross multiply, we will have a ratio that we will use to find this amount
1378.83/3 = 459.6kj
what is 31/50000 If the toxic quantity is 1.5 g of ethylene glycol per 1000 g of body mass, what percentage of ethylene glycol is fatal
Answer:
[tex]\%m=0.15\%[/tex]
Explanation:
Hello,
In this case, we are asked to compute the by mass percent representing the toxicity of ethylene glycol in the body mass. In such a way, since the by mass percent is computed with the shown below formula:
[tex]\%m=\frac{m_{ethylene \ glycol}}{m_{ethylene \ glycol}+m_{body\ mass}}*100\%[/tex]
We can use the given masses to obtain:
[tex]\%m=\frac{1.5g}{1.5g+1000g}*100\%\\ \\\%m=0.15\%[/tex]
Best regards.
Draw the product of the reaction between CH3CH=CHCH3 and H2 under a platinum catalyst.
Answer:
CH₃CH₂-CH₂CH₃
Explanation:
When an alkene, R-CH=CH-R reacts with H₂ in a Pt catalyst, the analogue alkane, R-CH₂-CH₂-R, is produced (Hydrogenation of alkenes via Pt/Pd catalyst)
Thus, the reaction of CH₃CH=CHCH₃ with H₂ under a platinum catalyst produce:
CH₃CH₂-CH₂CH₃The analogue alkane
The addition of hydrogen to CH3CH=CHCH3 leads to the formation of CH3CH2CH2CH3.
An alkene is a hydrocarbon that contains a C=C double bond. An alkene undergoes addition reaction in which a molecule is added across the double bond to form a saturated hydrocarbon.
The addition of H2 to CH3CH=CHCH3 yields butane as shown in the image attached. The platinum metal serves a catalyst in the reaction.
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2 points
How many significant figures are there in the measurement 2.30 x 10^5
mm ? *
2
3
4.
5
Answer:
3.
Explanation:
Hello,
In this case, considering the given number:
[tex]2.30x10^5[/tex]
We can see that it does not have zeros to left of the first nonzero digit which is 2, therefore, we just have to consider the 2, 3 and 0 as significant figures because the right-handed zeros are actually considered. Hence, such measurement has 3 significant figures.
You observe an exothermic gaseous reaction that is not spontaneous in forward direction at 1 atm and 298K. Which of the following statements about this reaction is true? a. This reaction will become spontaneous in forward direction at some temperature below 298K. b. This reaction will be spontaneous in forward direction at a higher pressure at 298K. c. This reaction will become spontaneous in forward direction at some temperature above 298K. d. This reaction is never spontaneous. e. The reverse reaction is always spontaneous.
Answer: option (a) is the correct answer
Explanation:
The complete questions says;
You observe an exothermic gaseous reaction that is not spontaneous in forward direction at 1 atm and 298K. Which of the following statements about this reaction is true? a. This reaction will become spontaneous in forward direction at some temperature below 298K. b. This reaction will be spontaneous in forward direction at a higher pressure at 298K. c. This reaction will become spontaneous in forward direction at some temperature above 298K. d. This reaction is never spontaneous. e. The reverse reaction is always spontaneous.
The Answer:
(a). This reaction will become spontaneous in forward direction at some temperature below 298K
Explanation: First of all, we can acknowledge that the reaction seen here is an exothermic one, i.e energy is released in the process outwardly and as a result temperature is reduced during this process of energy loss.
Having understood that scenario, say we reduce it's temperature by ourself than forward reaction favors and after reaching at particular temperature, therefore we can confirm this to be a spontaneous reaction.
Let us use this to confirm what we have been saying.
Given;
ΔG = ΔH - TΔS
here ΔH is negative
it is non spontaneous, which means ΔG is positive so we continuously decraeses it's temperature than at a particular temperature.
The Entropy change becomes positive and reaction becomes spontaneous and ΔG become negative.
cheers i hope this helped !!
what mass of TiCl4 must react with an excess of water to produce 50.0g of TiO2 if the reaction has a 78.9% yield
Answer:
\large \boxed{\text{150 g TiCl}_{4}}
Explanation:
We will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.
Mᵣ: 189.68 79.87
TiCl₄ + 2H₂O ⟶ TiO₂ + 4HCl
m/g: 50.0
To solve this stoichiometry problem, you must
Convert the actual yield to the theoretical yield Use the molar mass of TiO₂ to convert the theoretical yield of TiO₂ to moles of TiO₂ Use the molar ratio to convert moles of TiO₂ to moles of TiCl₄ Use the molar mass of TiCl₄ to convert moles of TiCl₄ to mass of TiCt₄1. Theoretical yield of TiO₂
[tex]\text{Theoretical yield} = \text{50.0 g actual} \times \dfrac{\text{100 g theoretical}}{\text{78.9 g actual}} = \text{63.37 g theoretical}[/tex]
2. Moles of TiO₂
[tex]\text{Mass of TiO}_{2} = \text{63.37 g TiO}_{2} \times \dfrac{\text{1 mol TiO}_{2}}{\text{79.87 g TiO}_{2} } = \text{0.7934 mol TiO}_{2}[/tex]
3, Moles of TiCl₄
The molar ratio is 1 mol TiO₂:1 mol TiCl₄.
[tex]\text{Moles of TiCl}_{4} = \text{0.7934 mol TiO}_{2} \times \dfrac{\text{1 mol TiCl}_{4}}{\text{1 mol TiO}_{2}} = \text{0.7934 mol TiCl}_{4}[/tex]
4. Mass of TiCl₄
[tex]\text{Mass of TiCl}_{4} = \text{0.7934 mol TiCl}_{4} \times \dfrac{\text{189.98 g TiCl}_{4}}{\text{1 mol TiCl}_{4}} =\textbf{150 g TiCl}_{\mathbf{4}} \\\\\text{You must use $\large \boxed{\textbf{150 g TiCl}_{\mathbf{4}}}$}[/tex]
2. (2 points) Calculate the Effective Nuclear Charge for each atom. Show all of your work for full credit. Se: Br: (1 point) Which atom is larger? (This is your claim.)
The question is incomplete. Here is the complete question.
Consider the selenium atom, Se, and the bromine atom, Br.
Write out the full electron configuration fro each atom.
Se:
Br:
Calculate the Effective Nuclear Charge for each atom. Show all of your work for full credit.
Se:
Br:
Which atom is larger?
Answer and Explanation: Electron Configuration of an atom demonstrates the shape and energy of its electrons. One of the rules used for it is given by using Madelung's Rule, in which the order of increased energy of the electrons is:
1s < 2s < 2p < 3s < 3p < 4s < 4p < 5s < 6s < 4f < 5d < 6p < 7s < 5f < 6d < 7p
Atom of Selenium (Se) has 34 electrons. Its electron configuration is
Se: [tex]1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{4}[/tex]
Atom of Bromine has 35 electrons. Its electron configuration is
Br: [tex]1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{5}[/tex]
Effective Nuclear Charge ([tex]Z_{eff}[/tex])is the net positive charge experienced by an electron in a multi-electron atom. In other words, it is the net force that helds nucleus and electrons together.
It is calculated by: [tex]Z_{eff}=Z-S[/tex]
where
Z is the atomic number, i.e., number of protons of the atom
S is the nonvalence electrons, i.e., the number of electrons between the nucleus and the electron in question.
For Selenium (Se):
From the electron configuration, the valence shell is 4 with 6 valence electrons. Nonvalence electrons is the difference between valence and total electrons:
S = 34 - 6 = 28
[tex]Z_{eff}=34-28[/tex] = 6
The Effective Nuclear Charge of Se is 6
For Bromine (Br):
The valence shell, according to the configuration, is 4 and valence electrons are 7.
S = 35 - 7 = 28
[tex]Z_{eff}=35-28[/tex] = 7
The Effective Nuclear Charge fo Br is 7.
Bromine is larger than Selenium because it has bigger Effective Nuclear Charge, which means it held its electrons more loosely and, consequently, has a larger atomic radius.
PRACTICE PROBLEMS
c
SOLVE Use the density formula to solve the following problems.
A sample of a substance has a volume of 60.5 mL and a density of
1.20 g/mL. What is the mass of the sample?
09
What is the speed 0.36
km/h in cm/s
Answer:
It's 10 cenimeters per second
2. Which of these is an extensive property?
a. Density
b. Melting point
Temperature
d. Volume
Volume ..........is the answer
The complete combustion of copper(I) sulphide is according to the following equation:
2Cu2S(s) + 302(g) →→2Cu2O(s) + 2SO2(g)
If the mass of Cu2S in the mixture is 14.0 g, calculate
(a) the number of molecules of oxygen gas reacted.
(b) the mass of SO2 gas produced.
(c)the volume of SO2 gas at STP
Answer:
a) 7.94 x 10²³ molecules, b) 5.62 g SO2, c) 1.97 L
Explanation:
a) We need to first convert Cu2S to moles. Since molar mass of Cu2S is 159.14 g/mol, 14.0 g = 0.0880 mol. Using molar ratios (3 mol O2/2 mol Cu2S, 0.0880 mol of Cu2S = 0.132 mol O2. Since 1 mol contains 6.02 x 10²³ molecules, 0.132 mol O2 = 7.94 x 10²² molecules O2.
b) Since the molar ratios of Cu2S to SO2 is 1:1, 0.0880 mol of Cu2S produces 0.0880 mol SO2. To convert mol to grams, we use the molar mass of SO2 (64.06 g/mol) to figure out that 0.0880 mol SO2 = 5.63 g SO2.
c) At STP, 1 mol occupies 22.4 liters. 0.0880 mol SO2 x 22.4 L/1 mol = 1.97 L
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A student weighed 0.550 g of lithium chloride, LiCl, to use in a reaction. How many moles is this? a. 5.11 b. 42.39 c. 23.31 d. 77.08 e. 0.0130
Answer:
The answer is option eExplanation:
To find the number of moles we must first find the molar mass of LiCL
From the question
mass = 0.550g
M(Li) = 6 Cl = 35.5
Molar mass of LiCL = 6 + 35.5 = 41.5 g/mol
Then we use the formula
[tex]n = \frac{m}{M} [/tex]
where
n is the number of miles
m is the mass
M is the molar mass
So we have
[tex]n = \frac{0.550}{41.5} [/tex]
We have the final answer as
0.0130 molHope this helps you
Magnesium is a group 2 metal which exists as a number of isotopes and forms many compounds.
Magnesium ions produce no emission of absorption lines in the visible region of the electromagnetic spectrum. Suggest why most magnesium compounds tested in a school laboratory show traces of yellow in the flame
Answer:
Traces of sodium impurity
Explanation:
Metal ions are identified by a characteristic colour imparted to flame by the metal ion solution. Various metals have various colours which they impart to a flame.
The energy of a flame is not sufficient for the excitation of electrons of Mg to a higher energy level. As a result of this, Mg do not give any color in Bunsen flame.
However, a few tinges of yellow-orange colour which is characteristic of sodium metal do appear when magnesium ions are exposed to a flame as a consequence of traces of sodium impurity in the magnesium ion solution.
A mountaineer shouts for help Half a second later (0.5 she hears the echo How far away is the rock face which is reflecting her voice (take the speed of sound to be 330 m/s
Answer:
82.5 metres
Explanation:
speed=330m|s
initial time taken =0.5÷2=0.25
distance=speed×time
330m|s × 0.25s = 82.5m
Which of the following conversions involves an oxidation of the underlined element?
A. H3PO3 → H2P03
B. 03 → 02
C. S03 → S042-
D. BrO2 → BrO-
E. CIO3 → ClO4
At 20°C the vapor pressure of benzene (C6H6) is 75 torr, and that of toluene (C7H8) is 22 torr. Assume that benzene and toluene form an ideal solution. What is the composition in mole fractions of a solution that has a vapor pressure of 40. torr at 20°C?
Answer: The mole fraction of benzene will be 0.34 and mole fraction of toluene is 0.66
Explanation:
According to Raoult's law, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the vapor pressure of that component in the pure state.
[tex]p_1=x_1p_1^0[/tex] and [tex]p_2=x_2P_2^0[/tex]
where, x = mole fraction in solution
[tex]p^0[/tex] = pressure in the pure state
According to Dalton's law, the total pressure is the sum of individual pressures.
[tex]p_{total}=p_1+p_2\\[/tex][tex]p_{total}=x_{benzene}p_{benzene}^0+x_{toluene}P_{toluene}^0[/tex]
[tex]x_{benzene}=x[/tex],
[tex]x_{toluene}=1-x_{benzene}=1-x[/tex],
[tex]p_{benzene}^0=75torr[/tex]
[tex]p_{toluene}^0=22torr[/tex]
[tex]p_{total}=40torr[/tex]
[tex]40=x\times 75+(1-x)\times 22[/tex]
[tex]x=0.34[/tex]
Thus (1-x0 = (1-0.34)=0.66
Thus the mole fraction of benzene will be 0.34 and that of toluene is 0.66
If the toxic quantity is 1.5 gg of ethylene glycol per 1000 gg of body mass, what percentage of ethylene glycol is fatal
Answer:
The percentage of ethylene glycol that is fatal is 0.15 %
Explanation:
If the toxic quantity of ethylene glycol in a 1kg or 1000 g body weight is 1.5 g then the percentage of ethylene glycol that is fatal is
[tex]\frac{1.5}{1000}[/tex] ˣ 100 = 0.15%
Hence, the percentage of ethylene glycol that is toxic for any body weight is 0.15%. This percentage is very important in various aspects of science including drug discovery and food production/processing
Which type(s) of matter would need to be separated by chemical methods?
Answer:
solids, liquids, gases, or plasma
Explanation:
In chemistry, a chemical substance is a form of matter that has constant chemical composition and characteristic properties. It cannot be separated into components without breaking chemical bonds. Chemical substances can be solids, liquids, gases, or plasma
1-hexanol was prepared by reacting an alkene with either hydroboration-oxidation or oxymercuration-reduction. Draw the structure of the alkene that was used to prepare the alcohol in highest yield.
Answer:
hex-1-ene with a hydroboration-oxidation reaction
Explanation:
In this case, we can start with the structure of 1-hexanol. In this alcohol, the "OH" is on the first carbon. Thus, the alkene that this alcohol can produce must have a double bond between carbons 1 and 2, that is, hex-1-ene.
Now, it is important to know which of the reactions we can use, hydroboration-oxidation, or oxymercuration-reduction. To decide which reaction to using, we must look again at the alcohol we want to produce. In this alcohol, the "OH" is attached to a primary carbon. Therefore, the addition of the "OH" must be made on the least substituted carbon, that is, carbon number 1.
Because of this, an anti-Markovnikov reaction (a reaction in which the "OH" is added on the less substituted carbon) must be used. The reaction, which serves us then is: hydroboration-oxidation since this reaction is anti-markovnikov.
See figure 1
I hope it helps!
Which lists metric units, in order, from smallest to largest?
A) milligram, centigram, gram
B) kilogram, gram, centigram
C) kilogram, hectogram, decagram
D) decagram, hectogram, milligram
Answer:
milligram, centigram, decigram
Hope this answer correct :)
Answer:
C
Explanation:
hope it helps.
Explain why serine proteases do not catalyze hydrolysis if the amino acid at the hydrolysis site is a D-amino acid. Trypsin, for example, cleaves on the C-side of L-Arg and L-Lys, but not on the C-side of D-Arg and D-Lys. Explain why serine proteases do not catalyze hydrolysis if the amino acid at the hydrolysis site is a D-amino acid. Trypsin, for example, cleaves on the -side of L- and L-, but not on the -side of D- and D-. All the naturally occurring amino acids are D-amino acids, while active centres of serine proteases are nearly all L. All the naturally occurring amino acids are L-amino acids, while active centres of serine proteases are nearly all D. The side chains of D-Arg and D-Lys are not positioned to bind correctly at the active site. None of the above.
Answer:
D. The side chains of D-Arg and D-Lys are not positioned to bind correctly at the active site
Explanation:
Stereospecificity is the ability to distinguish between stereoisomers of of a particular compound. L- and D- structures of compounds in living organisms are usually present in only one form due to stereospecificity. For example, naturally occuring amino acids in proteins are usually present as L-isomers.
Since enzyme are proteins, their active sites are composed of L-amino acid and they show stereospecificity in the reactions they catalyze. In their binding sites, only substrates complementary in structure can bind in order for catalysis to proceed. Therefore, only amino acids in the L- configuration are complementary to the active site of enzymes.
In the case of serine proteases, The side chains of -Arg and D-Lys will not be positioned properly for binding at the binding site of serine proteases, therefore, no catalysis will occur. On the other hand, L-Arg and L-Lys can bind to the catalytic site of serine proteases since they are complementary fits to the active site of the enzymes.
A coefficient is a constant by which a variable is multiplied. In chemistry, a coefficient is the number in front of a formula. (The 2 in 2CH4 is the coefficient.) The coefficients in a chemical equation represent the proportion of A) products to reactants. B) the amount of atoms in a solid phase to the amount in a liquid phase. C) mass to volume. D) molecules or atoms relative to other molecules or atoms.
Answer:
Below
Explanation:
The coefficients in a chemical reaction are called stochiometric coefficients
they represent the number of atoms or molecules they are multiplied to.
A chemical reaction has generally this form:
Reactants ⇒ productsAnswer:
Products to reactants
Identify if the following are chemical or physical properties: "CP" for
chemical property and "PP" for physical property
1. Oxygen is odorless and colorless
2. Copper turns green when exposed to the environment
3. The piece of metal is magnetic
4. The density of water is 1.0 g/cm3
5. Diamonds are a very hard substance
6. The tree is 8 meters high
Answer:
1.pp
2.cp
3.pp
4.pp
5.pp
6.pp