Answer:
The energy required to heat 1.30 kg of water from 22.4°C to 34.2°C is 64,121.2 J
Explanation:
Calorimetry is the measurement of the amount of heat that a body gives up or absorbs in the course of a physical or chemical process.
The sensible heat of a body is the amount of heat received or transferred by a body when undergoing a temperature variation (Δt) without there being a change in physical state. That is, when a system absorbs (or gives up) a certain amount of heat, it may happen that it experiences a change in its temperature, involving sensible heat. Then, the equation for calculating heat exchanges is:
Q = c * m * ΔT
Where Q is the heat or quantity of energy exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the variation in temperature (ΔT=Tfinal - Tinitial).
In this case:
[tex]c=4.18 \frac{J}{g*K}[/tex]m= 1.30 kg= 1,300 g (1 kg=1,000 g)ΔT= 34.2 °C - 22.4 °C= 11.8 °C= 11.8 °K Being a temperature difference, it is independent if they are degrees Celsius or degrees Kelvin. That is, the temperature difference is the same in degrees Celsius or degrees Kelvin.Replacing:
[tex]Q=4.18 \frac{J}{g*K}*1,300 g*11.8 K[/tex]
Q= 64,121.2 J
The energy required to heat 1.30 kg of water from 22.4°C to 34.2°C is 64,121.2 J
When scientists are ready to publish the result of their experiments why is it important for them to include a description of the procedure they used
Answer: So other scientist can replicate the experiment and see if they get the same results in other words, test reliability.
Explanation:
A compound (C_9H_9BrO_2) gives the following NMR data. Draw the structure of the compound.
'1^H-NMR: 1.39 ppm, t(3H); 4.38 ppm, q(2H); 7.57 ppm, d(2H); 7.90 ppm, d(2H)
13^C-NMR: 165.73; 131.56; 131.01; 129.84; 127.81; 61.18; 14.18
You do not have to consider stereo chemistry.
You do not have to explicitly draw H atoms.
Do not include lone pairs in your answer.
Answer:
ethyl 4-bromobenzoate
Explanation:
In this question, we can start with the Index of Hydrogen Deficiency (I.H.D):
[tex]I.H.D=\frac{2C+2+N-H-X}{2}=\frac{(2*9)+2+0-9-1}{2}~=~5[/tex]
This indicates, that we can have a benzene ring (I.H.D = 4) and a carbonyl group (I.H.D = 1), for a total of 5.
Additionally, in the 1H-NMR info, we have a triplet 1.39 (3H) followed by a doublet 4.38 (2H), this indicates the presence of an ethyl group ([tex]CH_3-CH_2-[/tex]). Also, in the formula, we have 2 oxygens if we have carbonyl group with 2 oxygens we have a high probability to have an ester group.
[tex]O=C-O-CH_2-CH_3[/tex]
Now, if we add this to the benzene ring and the "Br" atom that we have in the formula, we will have ethyl 4-bromobenzoate.
See figures 1 and 2 to further explanations.
I hope it helps!
Which of the following processes have a ΔS < 0? Which of the following processes have a ΔS < 0? carbon dioxide(g) → carbon dioxide(s) water freezes propanol (g, at 555 K) → propanol (g, at 400 K) methyl alcohol condenses All of the above processes have a ΔS < 0.
Answer:
All of the above processes have a ΔS < 0.
Explanation:
ΔS represents change in entropy of a system. Entropy refers to the degree of disorderliness of a system.
The question requests us to identify the process that has a negative change of entropy.
carbon dioxide(g) → carbon dioxide(s)
There is a change in state from gas to solid. Solid particles are more ordered than gas particles so this is a negative change in entropy.
water freezes
There is a change in state from liquid to solid. Solid particles are more ordered than liquid particles so this is a negative change in entropy.
propanol (g, at 555 K) → propanol (g, at 400 K)
Temperature is directly proportional to entropy, this means higher temperature leads t higher entropy.
This reaction highlights a drop in temperature which means a negative change in entropy.
methyl alcohol condenses
Condensation is the change in state from gas to liquid. Liquid particles are more ordered than gas particles so this is a negative change in entropy.
Plssssssssss answer this question
Answer:
Table salt: answer salt
Tea: answer acidic
Carbonated drinks: answer acidic
Baking powder: answer acid and base
Detergent: answer acid and basic
Alum: answer acidic
Explanation:
I hope this helps. Enjoy your day!
A gas within a piston–cylinder assembly undergoes an isothermal process at 400 K during which the change in entropy is −0.3 kJ/K. Assuming the ideal gas model for the gas and negligible kinetic and potential energy effects, evaluate the work, in kJ.
Answer:
W = -120 KJ
Explanation:
Since the piston–cylinder assembly undergoes an isothermal process, then the temperature is constant.
Thus; T1 = T2 = 400K
change in entropy; ΔS = −0.3 kJ/K
Formula for change in entropy is written as;
ΔS = Q/T
Where Q is amount of heat transferred.
Thus;
Q = ΔS × T
Q = -0.3 × 400
Q = -120 KJ
From the first law of thermodynamics, we can find the workdone from;
Q = ΔU + W
Where;
ΔU is Change in the internal energy
W = Work done
Now, since it's an ideal gas model, the change in internal energy is expressed as;
ΔU = m•C_v•ΔT
Where;
m is mass
C_v is heat capacity at constant volume
ΔT is change in temperature
Now, since it's an isothermal process where temperature is constant, then;
ΔT = T2 - T1 = 0
Thus;
ΔU = m•C_v•ΔT = 0
ΔU = 0
From earlier;
Q = ΔU + W
Thus;
-120 = 0+ W
W = -120 KJ
At 2000°C the equilibrium constant for the reaction 9_1.gif is 9_2.gif If the initial concentration of 9_3.gif is 0.200 M, what are the equilibrium concentrations of 9_4.gif and 9_5.gif?
Answer:
[tex][N_2]_{eq}=[H_2]_{eq}=0.09899M[/tex]
[tex][NO]_{eq}=0.00202M[/tex]
Explanation:
Hello,
In this case, for the given chemical reaction:
[tex]2NO\rightleftharpoons N_2+O_2[/tex]
We know the equilibrium constant and equilibrium expression:
[tex]Kc=2.4x10^3=\frac{[N_2][O_2]}{[NO]^2}[/tex]
That in terms of the reaction extent [tex]x[/tex] (ICE procedure) we can write:
[tex]2.4x10^3=\frac{x*x}{(0.2M-2*x)^2}[/tex]
In such a way, solving for [tex]x[/tex] by using a quadratic equation or solver, we obtain:
[tex]x_1=0.09899M\\x_2=0.1010M[/tex]
Clearly the solution is 0.09899M since the other value will result in a negative equilibrium concentration of NO. In such a way, the equilibrium concentrations of all the species are:
[tex][N_2]_{eq}=[H_2]_{eq}=x=0.09899M[/tex]
[tex][NO]_{eq}=0.2M-2*0.09899M=0.00202M[/tex]
Regards.
A filament for a light bulb needs to conduct electricity. Which of the elements listed below might be useful as a light bulb filament? Explain your thinking.
A. Tungsten, W
B. Sulfur, S
C. Bromine, Br
Answer:
A. tungsten
Explanation:
Tungsten is a material which high melting point ie. does not melt easily incase of high temperature
Answer:
option(A):Tungsten
Explanation:
tungsten has highest melting point.
What is silica gel commonly used for? A. Absorbing moisture to protect goods from damage. B. As insulation in buildings. C. As a lacquer on wood to make it water-resistant. D. A soft, flexible padding, such as on pen grips or mouse pads.
Answer:
A
Explanation:
Absorbing moisture to protect goods from damage. Hence, option A is correct.
What is silica gel?Silica gel is a desiccant, or drying agent, that manufacturers often place in little packets to keep moisture from damaging certain food and commercial products.
Silica Gel is a good drying agent for preventing corrosion, contamination, spoilage, and mould growth in many commodities and products due to its physical properties.
Learn more about silica gel here:
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Gaseous methane (CH4) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO) and gaseous water (H2O) . Suppose 0.963 g of methane is mixed with 7.5 g of oxygen. Calculate the minimum mass of methane that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits.
Answer:
0 g.
Explanation:
Hello,
In this case, since the reaction between methane and oxygen is:
[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]
If 0.963 g of methane react with 7.5 g of oxygen the first step is to identify the limiting reactant for which we compute the available moles of methane and the moles of methane consumed by the 7.5 g of oxygen:
[tex]n_{CH_4}=0.963gCH_4*\frac{1molCH_4}{16gCH_4}=0.0602molCH_4\\ \\n_{CH_4}^{consumed}=7.5gO_2*\frac{1molO_2}{32gO_2}*\frac{1molCH_4}{2molO_2} =0.117molCH_4[/tex]
Thus, since oxygen theoretically consumes more methane than the available, we conclude the methane is the limiting reactant, for which it will be completely consumed, therefore, no remaining methane will be left over.
[tex]left\ over=0g[/tex]
Regards.
How many grams of Br are in 445 g CaBr2 ?
Explanation:
The mass of bromine is 79.904, and since there are two of them in the Calcium Bromide molecule, we'll multiply it by 2 to get 159.808. Dividing that by the full mass of the molecule then multiply it by 100 will give you the answer.
An actacide tablet containing Mg(OH)2 (MM = 58.3g / (mol)) is titrated with a 0.100 M solution of HNO3. The end point is determined by using an indicator. Based on 20.00mL HNO3 being used to reach the endpoint, what was the mass of the Mg * (OH) in the antacid tablet? * 0.0583 g 0.583 5.83 g 58.3 g
Answer:
0.0583g
Explanation:
The equation of the reaction is;
2HNO3(aq) + Mg(OH)2(aq) -------> Mg(NO3)2(aq) + 2H2O(l)
From the question, number of moles of HNO3 reacted= concentration × volume
Concentration of HNO3= 0.100 M
Volume of HNO3 = 20.00mL
Number of moles of HNO3= 0.100 × 20/1000
Number of moles of HNO3 = 2×10^-3 moles
From the reaction equation;
2 moles of HNO3 reacts with 1 mole of Mg(OH)2
2×10^-3 moles reacts with 2×10^-3 moles ×1/2 = 1 ×10^-3 moles of Mg(OH)2
But
n= m/M
Where;
n= number of moles of Mg(OH)2
m= mass of Mg(OH)2
M= molar mass of Mg(OH)2
m= n×M
m= 1×10^-3 moles × 58.3 gmol-1
m = 0.0583g
For each of the following reactions calculate the mass (in grams) of both the reactants that are required to form 15.39g of the following products.
a. 2K(s) + Cl2(g) → 2Cl(aq)
b. 4Cr(s) + 302(g) → 2Cr2O3(s)
c. 35r(s) + N2(g) → SraNa(s)
Answer:
a.
[tex]m_K=8.056gK\\ \\m_{Cl_2}=4.028gCl_2[/tex]
b.
[tex]m_{Cr}=10.51gCr\\ \\m_{O_2}=4.851gO_2[/tex]
c.
[tex]m_{Sr}=13.88gSr\\\\m_{N_2}=1.479gN_2[/tex]
Explanation:
Hello,
In this case, we proceed via stoichiometry in order to compute the masses of all the reactants as shown below:
a. [tex]2K+Cl_2\rightarrow 2KCl[/tex]
[tex]m_K=15.36gKCl*\frac{1molKCl}{74.55gKCl}*\frac{2molK}{2molKCl}* \frac{39.1gK}{1molK}=8.056gK\\ \\m_{Cl_2}=15.36gKCl*\frac{1molKCl}{74.55gKCl}*\frac{1molCl_2}{2molKCl}* \frac{70.9gCl_2}{1molCl_2}=4.028gCl_2[/tex]
b. [tex]4Cr+ 3O_2\rightarrow 2Cr_2O_3[/tex]
[tex]m_{Cr}=15.36gCr_2O_3*\frac{1molCr_2O_3}{152gCr_2O_3l}*\frac{4molCr}{2molCr_2O_3}* \frac{52gCr}{1molCr_2O_3}=10.51gCr\\ \\m_{O_2}=15.36gCr_2O_3*\frac{1molCr_2O_3}{152gCr_2O_3l}*\frac{3molO_2}{2molCr_2O_3}* \frac{32gO_2}{1molCr_2O_3}=4.851gO_2[/tex]
c. [tex]3Sr(s) + N_2(g) \rightarrow Sr_3N_2[/tex]
[tex]m_{Sr}=15.36gSr_3N_2*\frac{1molSr_3N_2}{290.86gSr_3N_2}*\frac{3molSr}{1molSr_3N_2}* \frac{87.62gSr}{1molSr}=13.88gSr\\\\m_{N_2}=15.36gSr_3N_2*\frac{1molSr_3N_2}{290.86gSr_3N_2}*\frac{1molN_2}{1molSr_3N_2}* \frac{28gN_2}{1molN_2}=1.479gN_2[/tex]
Regards.
What is the density of a 10 kg mass that occupies 5 liters?
( pls need help)
Answer: d=2000 g/L
Explanation:
Density is mass/volume. The units are g/L. Since we are given mass and volume, we can divide them to find density. First, we need to convert kg to g.
[tex]10kg*\frac{1000g}{1kg} =10000 g[/tex]
Now that we have grams, we can divide to get density.
[tex]d=\frac{10000g}{5 L}[/tex]
d=2000g/L
Spinocerebellar Ataxia
Will a precipitate (ppt) form when 20.0 mL of 1.1 × 10 –3 M Ba(NO 3) 2 are added to 80.0 mL of 8.4 × 10 –4 M Na 2CO 3?
Answer:
A precipitate will form, BaCO₃
Explanation:
When Ba²⁺ and CO₃²⁻ ions are in an aqueous media, BaCO₃(s), a precipitate, is produced following its Ksp expression:
Ksp = 5.1x10⁻⁹ = [Ba²⁺] [CO₃²⁻]
Where the concentrations of the ions are the concentrations in equilibrium
For actual concentrations of a solution, you can define Q, reaction quotient, as:
Q = [Ba²⁺] [CO₃²⁻]
If Q > Ksp, the ions will react producing BaCO₃, if not, no precipitate will form.
Actual concentrations of Ba²⁺ and CO₃²⁻ are:
[Ba²⁺] = [Ba(NO₃)₂] = 1.1x10⁻³ × (20.0mL / 100.0mL) = 2.2x10⁻⁴M
[CO₃²⁻] = [Na₂CO₃] = 8.4x10⁻⁴ × (80.0mL / 100.0mL) = 6.72x10⁻⁴M
100.0mL is the volume of the mixture of the solutions
Replacing in Q expression:
Q = [Ba²⁺] [CO₃²⁻]
Q = [2.2x10⁻⁴M] [6.72x10⁻⁴M]
Q = 1.5x10⁻⁷
As Q > Ksp
A precipitate will form, BaCO₃
Need help with chemistry questions
Answer:
1. oxidation
2. reduction
3. oxidation
4. oxidation
Explanation:
Oxidation and Reduction in terms of hydrogen
Oxidation and Reduction with respect to Hydrogen Transfer. Oxidation is the loss of hydrogen. Reduction is the gain of hydrogen.Oxidation and Reduction in terms of Oxygen
Oxidation and Reduction with respect to Oxygen Transfer. Oxidation is the gain of Oxygen. Reduction is the loss of Oxygen.You find a clean 100-ml beaker, label it "#1", and place it on a tared electronic balance. You add small amount of unknown solid and place the
beaker with its contents on the balance. The recorded data is:
mass of the empty, clean beaker #1: 74.605 g
mass of the beaker #1 with the white solid: 74.896 g
Using the Law of Conservation of Mass, what is the mass of the unknown solid you placed in beaker #1?
Answer:
the mas is .291 g
Explanation:
the mass of a object does not change. so when added the substance the beaker. you had the mass of both objects together. you know the mass of the beaker and you know the mass of both. since mass does not change. the beakers mass is still 74.605g. the mass of both objects is 74.896. all you have to do is subtract the mass of the beaker from the total mass. 74.896 - 74.605 equals .291g. so the mass of the unknown substance Is .291g
Does the amount of methanol increase, decrease, or remain the same when an equilibrium mixture of reactants and products is subjected to the following changes?
a. the catalyst is removed
b. the temp is increased
c. the volume is decreased
d. helium is added
e. CO is added
Answer:
a. Methanol remains the same
b. Methanol decreases
c. Methanol increases
d. Methanol remains the same
e. Methanol increases
Explanation:
Methanol is produced by the reaction of carbon monoxide and hydrogen in the presence of a catalyst as follows; 2H2+CO→CH3OH.
a) The presence or absence of a catalyst makes no difference on the equilibrium position of the system hence the methanol remains constant.
b) The amount of methanol decreases because the equilibrium position shifts towards the left and more reactants are formed since the reaction is exothermic.
c) If the volume is decreased, there will be more methanol in the system because the equilibrium position will shift towards the right hand side.
d) Addition of helium gas has no effect on the equilibrium position since it does not participate in the reaction system.
e) if more CO is added the amount of methanol increases since the equilibrium position will shift towards the right hand side.
which of the following is an acid?
HF
KOH
LIF
NH3
Based on a pH scale, the chemical compound which is classified as an acid is: A. Hydrogen fluoride (HF).
What is a pH scale?A pH scale can be defined as a tool that is designed and developed for the measurement of the molar concentration of hydrogen ions (H⁺) in both acidic and alkaline (basic) solutions.
This ultimately implies that, a pH scale can be used to measure and specify the acidity, neutrality or basicity of any chemical solution.
In this scenario, the pH of each of the given chemical compound are as follows:
Hydrogen fluoride (HF): it has a pH of 1.0 but it's a weak acid.Potassium hydroxide (KOH): it has a pH of 13 and as such it's a strong base.Lithium fluoride (LIF): it has a pH of 7.5 and as such it's neutral.Ammonia (NH₃): it has a pH of 10 and as such it's a weak base.Read more on pH here: brainly.com/question/24233266
Given the following reaction and data, A + B → Products
Experiment A (M) B (M) Rate (M/s)
1 1.50 1.50 0.320
2 1.50 2.50 0.320
3 3.00 1.50 0.640
Required:
a. What is the rate law of the reaction?
b. What is the rate constant?
Answer:
a. Rate = k×[A]
b. k = 0.213s⁻¹
Explanation:
a. When you are studying the kinetics of a reaction such as:
A + B → Products.
General rate law must be like:
Rate = k×[A]ᵃ[B]ᵇ
You must make experiments change initial concentrations of A and B trying to find k, a and b parameters.
If you see experiments 1 and 3, concentration of A is doubled and the Rate of the reaction is doubled to. That means a = 1
Rate = k×[A]¹[B]ᵇ
In experiment 1 and to the concentration of B change from 1.50M to 2.50M but rate maintains the same. That is only possible if b = 0. (The kinetics of the reaction is indepent to [B]
Rate = k×[A][B]⁰
Rate = k×[A]b. Replacing with values of experiment 1 (You can do the same with experiment 3 obtaining the same) k is:
Rate = k×[A]
0.320M/s = k×[1.50M]
k = 0.213s⁻¹Calculate [OH-] given [H3O+] in each aqueous solution and classify the solution as acidic or basic. [H3O+] = 2.6 x 10-8 M
Answer:
To calculate the [OH-] in the solution we must first find the pOH
That's
pH + pOH = 14
pOH = 14 - pH
First to find the pH we use the formula
pH = - log [H3O+]From the question
[H3O+]= 2.6 × 10^-8 M
pH = - log 2.6 × 10^-8
pH = 7.6
pH = 8
So we pOH is
pOH = 14 - 8 = 6
To find the [OH-] we use the formula
pOH = - log [OH-]6 = - log [OH-]
Find antilog of both sides
[OH-] = 1.0 × 10^-6 MThe solution is slightly basic since it's pH is in the basic region and slightly above the neutral point 7
Hope this helps you
The standard entropy of a substance refers to its entropy at:__________.
a. absolute zero and 1 bar
b. 0°C and 1 bar
c. 25 °C and 1 bar
d. 25 °C and 0 bar
Answer:
b. 0°C and 1 bar
Explanation:
Hello,
In this case, the STP conditions are standard temperature and pressure sets of conditions for experimental measurements to be established to allow comparisons to be made between different sets of data, it means that a specific pressure and temperature is assigned to analyze the properties of a substance. Such conditions are strictly 0°C and 1 bar because a large number of physical, chemical and thermodynamic properties are measured at them, therefore the standard entropy of a substance refers to its entropy at: b. 0°C and 1 bar.
Best regards.
Which of the following types of electromagnetic radiation have higher frequencies than visible light and which have shorter frequencies than visible light? Sort them accordingly. ltems (6 items) (Drag and drop into the appropriate area below)
a. Gamma rays
b. Infrared radiation
c. Ultraviolet liht
d. X-rays
e. Microwaves
f. Radio waves
Answer:
Higher frequency than visible light - Ultraviolet light, X-rays, and Gamma rays
Lower frequency than visible light - Infrared radiation, microwaves, and Radio waves
Explanation:
The frequencies of electromagnetic radiations vary according to their wavelengths. The relationship between the frequency and wavelength of the waves is expressed such that:
λ = c/f, where λ = wavelength, c = speed of light, and f = frequency.
Thus, there is an inverse relationship between the wavelength and the frequency of electromagnetic waves.
The order of the electromagnetic waves based on their frequency from the lowest to the highest is radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma-rays
Hence, electromagnetic waves with higher frequencies than visible light include ultraviolet light, X-rays, and Gamma rays while those with lower frequencies include Infrared radiation, microwaves, and Radio waves.
Answer:
need points
Explanation:
11. How many oxygen atoms are there in one formula unit of Ca2+?
Answer:
NO3 has 3 atoms of oxygen in it per molecule (indicated by the subscript of 3) and (NO3)2 means you have 2 NO3 molecules, meaning you double the subscript to get 6 oxygen atoms.
Explanation:
what is the molality of a solution
Answer: The number of moles of a solute per kilogram of solvent
Explanation:
What is the correct IUPAC name for Ir(NO₂)₄
Answer
Iridium(IV)Nitrite
The correct IUPAC name of the Ir(NO₂)₄ compound is Iridium(IV)Nitrite.
What is the IUPAC name?Whether it's in a continuous chain or just a ring, the largest chain of carbons joined by a single bond serves as the basis for IUPAC nomenclature.
What is a compound?
A chemical compound would seem to be a substance that contains numerous similar molecules made of atoms from different elements joined by chemical bonds.
The given compound is Ir(NO₂)₄. It can be seen that 4 nitro group is attached with Ir and its coordination number is 4. Hence, the IUPAC name will be Iridium(IV)Nitrite.
To know more about IUPAC nomenclature.
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A compound is found to contain 30.45 % nitrogen and 69.55 % oxygen by weight. To answer the questions, enter the elements in the order presented above. 1. What is the empirical formula for this compound? 2. The molecular weight for this compound is 46.01 g/mol. What is the molecular formula for this compound?
Answer:
Empirical formulae is NO2
Molecular Formulae is NO2
You are working in a lab when radiation alarms go off. You are able to hide inside a steel cabinet, whose sides are about 1.5 inches thick, until the alarm goes off. Preliminary reports show that the radiation was weakly ionizing and negatively charged. Were you safe in the cabinet?
A. Yes, it was beta radiation and the steel was enough to block it.
B. Yes, it was alpha radiation and not harmful.
C. No, it was beta radiation and very dangerous.
D. No, it was gamma radiation and can only be blocked by a thick wall of lead.
Answer
A
Explanation:
Beta radiation is only very dangerous when it is ingested or if there is nothing in the way. The steel should protect you.
Yes, it was beta radiation and the steel was enough to block it. Hence, option A is correct.
What is beta radiation?Beta radiation is a current of electrons released at a rate exceeding the speed of light, with kinetic energy between 0.2 MeV and 3.2 MeV. Due to their lower mass which is approximately 5.5×[tex]10^{-4}[/tex] amu (9.130×[tex]10^(-24)[/tex], interactions between β-particles and the atoms of pass-through materials are much less frequent than the interactions between 5-007-particles: less than 200 ion pairs are typically formed in each centimetre of the air passage.
When we are working in a lab when radiation alarms go off and we are able to hide inside a steel cabinet, whose sides are about 1.5 inches thick, until the alarm goes off.
Preliminary reports show that the radiation was weakly ionizing and negatively charged. Yes, it was beta radiation and the steel was enough to block it.
Hence, option A is correct.
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(9443+45−9.9) (9443+45−9.9) ×8.4× 10 6
13. Arrange each group of units from smallest to
largest
a)km, mm, cm, m
b) mg, kg, g
C) L, mL
d) s, ms, min, h
Answer:
A. mm,cm,m,Km
B. mg, g, Kg
C. mL,L
D. ms, s, min, h
Explanation: