Answer:
The value is [tex]x = 0.151 \ e [/tex]
Explanation:
From the question we are told that
The bond length is [tex]l = 1.93\ \r a = 1.93 *1 *10^{-10} =1.93 *10^{-10}\ m [/tex]
The bond dipole moment is [tex] \mu = 1.40 d = 1.40 * 3.33564 *10^{-30} = 4.6699 *10^{-30} \ C \cdot m[/tex]
Generally the dipole moment is mathematically represented as
[tex]\mu = Q * l[/tex]
Here Q is the partial negative charge on the bromine atom
So
[tex]Q = \frac{\mu}{ l}[/tex]
=> [tex]Q = \frac{4.6699 *10^{-30}}{ 1.93 *10^{-10} }[/tex]
=> [tex]Q = 2.42 *10^{-20} C [/tex]
Generally
1 electronic charge(e) is equivalent to [tex]1.60*10^{-19} C[/tex]
So x electronic charge(e) is equivalent to [tex]Q = 2.42 *10^{-20} C [/tex]
=> [tex]x = \frac{2.42 *10^{-20}}{1.60*10^{-19} }[/tex]
=> [tex]x = 0.151 \ e [/tex]
Blood plasma contains a total carbonate pool of 0.0252M.
(a) What is the HCO3"/CO2 ratio
(b) What is the concentration of each buffer component present at pH=7.4
(c) What would the pH be if 0.01M H" is added assuming that the excess CO2 is not
released.
(d) What would the pH be if 0.01M H is added assuming that the excess CO2 is released.
Answer:
a what is the HCO3"/CO2 ratio
typically, over 60% of the mass of a piece of bubble gum is made up of sugar. sugar is dissolved in the mouth and swallowed during chewing. if gum is chewed precisely until it loses its flavor, most if not all of the sugar content of the gum should be gone. how would this affect the mass of the gum? how will the mass of gum be affected by chewing time? how does the chewing time necessary to lose significant mass compare between different brands/types of gum? how large a role does saliva play in the processing of gum in the mouth?
Answer:
Hello your question is incomplete attached below is the complete question
A) How the mass will be affected by chewing time
The chewing time is an independent variable while the mass of the gum is a dependent variable and would change with increase in chewing time as seen in the regular gum the average mass increase above the average sugar mass before chewing
B) how the chewing time necessary to lose significant mass
The chewing time is neccessary to lose significant mass as seen in the sugar free chewing gum because the more time the gum is chewed the lesser the mass due to the absence/lose of sugar
C) The role saliva play in the processing of gum in the mouth
The saliva mixes up with the sugar contained in the chewing gum to increase its mass as seen when you compare the mean mass value of sugar contained in the regular chewing gum with the mean mass value of the gum after chewing for some time i.e 3.58 < 5.03 , 4.09
Explanation:
we will make two assumptions in order to resolve the given problem
next we have to calculate the mean(average) mass of the Chewing gums for each time interval
Regular gum :
initial mean mass = 5.96
mean mass after 1 minute = 5.03
mean mass after 2 minutes = 4.09
next we calculate the mean mass of sugar in the regular gum using this relationship :summation of( initial masses * 60/100 )/ 4 = 3.58 comparing it the the mass of the chewing gum at different time intervals i.e 1 minute and 2 minutes
sugar free gum :
initial mean mass = 2.98
mean mass after 1 minute = 2.59
mean mass after 2 minutes = 2.32
A) How the mass will be affected by chewing time
The chewing time is an independent variable while the mass of the gum is a dependent variable and would change with increase in chewing time as seen in the regular gum the average mass increase above the average sugar mass before chewing
B) how the chewing time necessary to lose significant mass
The chewing time is neccessary to lose significant mass as seen in the sugar free chewing gum because the more time the gum is chewed the lesser the mass due to the absence/lose of sugar
C) The role saliva play in the processing of gum in the mouth
The saliva mixes up with the sugar contained in the chewing gum to increase its mass as seen when you compare the mean mass value of sugar contained in the regular chewing gum with the mean mass value of the gum after chewing for some time i.e 3.58 < 5.03 , 4.09
How much energy (heat) is required to convert 248 g of water from 0 oC to 154 oC? Assume that the water begins as a liquid, that the specific heat of water is 4.184 J/goC over the entire liquid range, that the specific heat of steam is 1.99 J/goC, and the heat of vaporization of water is 40.79 kJ/mol.
Answer:
The total heat required is 691,026.36 J
Explanation:
Latent heat is the amount of heat that a body receives or gives to produce a phase change. It is calculated as: Q = m. L
Where Q: amount of heat, m: mass and L: latent heat
On the other hand, sensible heat is the amount of heat that a body can receive or give up due to a change in temperature. Its calculation is through the expression:
Q = c * m * ΔT
where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the change in temperature (Tfinal - Tinitial).
In this case, the total heat required is calculated as:
Q for liquid water. This is, raise 248 g of liquid water from O to 100 Celsius. So you calculate the sensible heat of water from temperature 0 °C to 100° CQ= c*m*ΔT
[tex]Q=4.184\frac{J}{g*C} *248 g* (100 -0 )C[/tex]
Q=103,763.2 J
Q for phase change from liquid to steam. For this, you calculate the latent heat with the heat of vaporization being 40 and being 248 g = 13.78 moles (the molar mass of water being 18 g / mol, then[tex]\frac{248 g}{18 \frac{g}{mol} } =13.78 moles[/tex] )Q= m*L
[tex]Q=13.78moles*40.79 \frac{kJ}{mol}[/tex]
Q=562.0862 kJ= 562,086.2 J (being 1 kJ=1,000 J)
Q for temperature change from 100.0 ∘ C to 154 ∘ C, this is, the sensible heat of steam from 100 °C to 154°C.Q= c*m*ΔT
[tex]Q=1.99\frac{J}{g*C} *248 g* (154 - 100 )C[/tex]
Q=25,176.96 J
So, total heat= 103,763.2 J + 562,086.2 J + 25,176.96 J= 691,026.36 J
The total heat required is 691,026.36 J
The specific heat can be defined as the amount of heat required to raise the temperature of one gram of substance by one degree Celsius. The total heat of the reaction has been 691.029 kJ.
What is heat of vaporization?The heat of vaporization is the amount of heat required to convert the liquid to the vapor state.
The water at 0 degree Celsius has been converted to the water at 100 degree Celsius. The 100 degree Celsius water vaporized to 100 degree steam. The 100 degree steam will be converted to the 154 degree Celsius.
The conversion of 0 degree Celsius water to 100 degree Celsius
[tex]Q_1=mc\Delta T[/tex]
Substituting the values of mass ([tex]m[/tex]), specific heat ([tex]c[/tex]), and change in temperature ([tex]\Delta T[/tex]):
[tex]Q_1=248\;\times\;4.184\;\times\;(100-0)\\Q_1=103,763.2\;\text J\\Q_1=103.7632\;\rm k\text J[/tex]
The amount of heat required to convert 100 degree Celsius water to 100 degree Celsius steam has been:
[tex]Q_2=mL[/tex]
Substituting the values of mass ([tex]m[/tex]), and heat of vaporization ([tex]L[/tex]):
[tex]Q_2=248\;\times\;40.79\\Q_2-562.0862\;\rm kJ[/tex]
The amount of heat required to convert 100 degree Celsius steam to 154 degree Celsius steam has been:
[tex]Q_3=mc\Delta T[/tex]
Substituting the values of mass ([tex]m[/tex]), heat of steam ([tex]c[/tex]), and change in temperature ([tex]\Delta T[/tex]):
[tex]Q_3=248\;\times\;1.99\;\times\;(154-100)\\Q_3=25,176.96 \;\text J\\Q_3=25.1796\;\rm kJ[/tex]
The total amount of heat in the reaction has been:
[tex]Q=Q_1+Q_2+Q_3[/tex]
Substituting the values for the total heat of the reaction:
[tex]Q=103.7632+562.0862+25.1796\;\rm kJ\\\textit Q=691.029\;kJ[/tex]
The total heat of the reaction for the conversion of water from 0 degree Celsius to 100 degree Celsius is 691.029 kJ.
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What is the steps of the photosynthetic equation?
Answer: The photosynthesis equation is as follows: 6CO2 + 6H20 + (energy) → C6H12O6 + 6O2 Carbon dioxide + water + energy from light produces glucose and oxygen.
Explanation: Photosynthesis is comprised of two stages, the light-dependent reaction and the light-independent reactions, so that for the light suports energy.
Hope this helps:)
List the inner planets in order from the closest to th
Answer : The inner planets (in order of distance from the sun, closest to furthest) are Mercury, Venus, Earth and Mars. After an asteroid belt comes the outer planets, Jupiter, Saturn, Uranus and Neptune. The interesting thing is, in some other planetary systems discovered, the gas giants are actually quite close to the sun.
Answer:
Mercury
venus
earth
mars
jupiter
saturn
uranus
Neptune
Explanation:
14) Describe the Cloud Model.
For double-helix formation, change in Gibbs free energy, ΔG, can be measured to be −54 kJ⋅mol−1 (−13 kcal⋅mol−1) at pH 7.0 in 1 M NaCl at 25 °C (298 K). The heat released indicates an enthalpy change of -251 kJ/mol (-60 kcal/mol). For this process, calculate the entropy change for the system and the entropy change for the surroundings.
Answer:
Explanation:
Entropy change in the system : --
ΔG = −54 kJ⋅mol−1 (−13 kcal⋅mol−1) = −54 kJ⋅mol−1 (−13 x 4.2 kJ⋅mol−1)
= - 108.6 KJ / mol
ΔH = -251 kJ/mol (-60 kcal/mol) = -251 kJ/mol (-60 x 4.2 kJ/mol)
= - 503 KJ / mol
ΔG = ΔH - TΔS
ΔS = ( ΔH - ΔG ) / T
= - 503 + 108.6 / ( 273 + 25 ) KJ / mol k⁻¹
= - 1323.48 J / mol k⁻¹
Entropy change in the surrounding
+ 1323.48 J / mol k⁻¹
At liftoff, a space shuttle uses a solid mixture of ammonium perchlorate and aluminum powder to obtain great thrust from the volume change of solid to gas. In the presence of a catalyst, the mixture forms solid aluminum oxide and aluminum chloride and gaseous water and nitrogen monoxide.A. Write a balanced equation for the reaction, assign oxidation states for all atoms, and identify the reducing and oxidizing agents.B. How much aluminum oxide can you make when you react 150.0 g of ammonium perchlorate with 50.0 g of powdered aluminum
Answer:
A) 3NH₄ClO₄ + 3Al ---> Al₂O₃ + AlCl₃ + 6H₂O + 3NO
B) 150 g of ammonium perchlorate will produce 43.4 g of aluminum oxide
Explanation:
A)Balanced equation of reaction:
3NH₄ClO₄ + 3Al ---> Al₂O₃ + AlCl₃ + 6H₂O + 3NO
Oxidation states:
Nitrogen, N: from -3 to +2
Hydrogen, H: from +1 to +1
Chlorine, Cl: From +7 to -1
Oxygen, O: from -2 t0 -2
Aluminum, Al: from 0 to +3.
Oxidizing agent is ammonium perchlorate while the reducing agent is the aluminum powder.
B) molar mass of aluminum oxide = 102 g/mol; molar mass of ammonium perchlorate =117.5 g/mol
From the equation of reaction 3 moles of ammonium perchlorate reacts with 3 moles of aluminum to produce 1 mole of aluminum oxide;
that is 3 * 117.5 g of ammonium perchlorate reacts with 3 * 27 g of aluminum to produce 102 g/mol of aluminum oxide
150 g of ammonium perchlorate will react with (150 * 3*27) / (3 * 117.5) of aluminum = 34.47 g of Al
Ammonium perchlorate is the limiting reactant.
150 g of ammonium perchlorate will produce (150 * 102)/(3*117.5) g of aluminum oxide = 43.4 g of aluminum oxide
The branch of science that deals with chemicals and bonds are called chemistry. Chemistry deals with the physical and chemical behavior of the element.
The correct answer to the question is [tex]3NH_4ClO_4 + 3Al ---> Al_2O_3 + AlCl_3 + 6H_2O + 3NO[/tex]
B) 150 g of ammonium perchlorate will produce 43.4 g of aluminum oxide
What is the balanced reaction?A balanced equation is an equation for a chemical reaction in which the number of atoms for each element in the reaction and the total charge are the same for both the reactants and the products. In other words, the mass and the charge are balanced on both sides of the reaction.Oxidation states:Nitrogen, N: from -3 to +2Hydrogen, H: from +1 to +1Chlorine, Cl: From +7 to -1Oxygen, O: from -2 t0 -2Aluminum, Al: from 0 to +3.The oxidizing agent is ammonium perchlorate while the reducing agent is the aluminum powder.
B) molar mass of aluminum oxide = 102 g/mol; molar mass of ammonium perchlorate =117.5 g/mol. From the equation of reaction 3 moles of ammonium perchlorate reacts with 3 moles of aluminum to produce 1 mole of aluminum oxide;
That is[tex]3 * 117.5[/tex] g of ammonium perchlorate reacts with [tex]3 * 27[/tex]g of aluminum to produce 102 g/mol of aluminum oxide. 150 g of ammonium perchlorate will react with [tex]\frac{(150 * 3*27) }{ (3 * 117.5)}[/tex] of aluminum = 34.47 g of Al
Ammonium perchlorate is the limiting reactant.
150 g of ammonium perchlorate will produce (150 * 102)/(3*117.5) g of aluminum oxide
= 43.4 g of aluminum oxide.
Hence, the correct answer is 4.34g.
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A student measures the mass of a substance as 1.7132 kg and it’s volume as 0.65 L. What is the density of the substance in g/mL? Round your answer to the correct number of significant figures.
Answer:
[tex]\rho =2.6g/mL[/tex]
Explanation:
Hello,
In this case, considering that the density is defined as:
[tex]\rho =\frac{m}{V}[/tex]
Thus, since 1 kg equals 1000 g and 1 L equal 1000 mL, the required density in g/mL turns out:
[tex]\rho =\frac{1.7132kg}{0.65L}*\frac{1000g}{1kg}*\frac{1L}{1000mL} \\\\\rho=2.6g/mL[/tex]
Take into account that since 0.65 L has two significant figures, the result is also shown with two significant figures.
Regards.
Which type of rock is a foliated metamorphic rock?
A. granite
B. marble
C. quartzite
D. slate
Answer:
D. Slate
Explanation:
Because when it come to think of it then you see that slate is arranged with many parallel layers just like a foliated metamorphic rock.
P.S I took the test and got a 100
Slate is a type of foliated metamorphic rock formed when rocks are subjected to heat,high pressure,etc.
What are metamorphic rocks?Metamorphic rocks are defined as a type of rock which is substantially changed from igneous or sedimentary rocks.They are formed when rocks are subjected to high heat, pressure,hot minerals and even combination of these factors.
Metamorphic rocks are formed by the process of metamorphism which does not melt the rocks rather they are transformed into denser and compact rocks.
New minerals are created by the rearrangement of of minerals or by the reaction of fluids withe the rocks.Metamorphic rocks are often smeared and folded.
Common metamorphic rocks are marble,gneiss,quartzite ,etc.There are two types of metamorphic rocks :foliated and non-foliated metamorphic rocks.
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An object has a mass of 33g and a volume of 11ml. What is it's density?
Answer:
3g/ml
Explanation:
Density = [tex]\frac{33g}{11ml}[/tex]
33 divided by 11 = 3
Which of the following isn't an example of numerical data?
A.Height
B.Weight
C.Gender
D.Temperature
Answer:
Gender
Explanation:
h. Identity of unknown metal
Answer:
you can identify an unknown substance by measuring its density and comparing your results to a list of known densities. Density=mass/volume. Assume that you have to identify an unknown metal. You can determine the mass of the metal on a scale.
Why do you think Democritus and dalton are grouped together
Democritus and Dalton are grouped together because they had similar opinion about the existence of atom.
Who is Democritus?Democritus is an ancient Greek philosopher who is remembered for his formulation of an atomic theory.
Who is Dalton?Dalton is an english chemist who is known for introducing the theory of atom to chemistry.
What is an atom?An atom is the smallest unit of matter.
According to Democritus's theory of atom, all matter is made up of tiny particles called atoms and according to Dalton all matter is made up of tiny indivisible particles called matter.
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2. A student wishes to find the density of an irregularly shaped rock. He finds the mass to be 12.34
g. He fills a graduated cylinder to the 50.0 mL mark and drops in the rock. The fluid level rises to
54.5 mL. What is the density of the rock? (Show all work, with units, for credit.)
Answer:
The answer is
2.74 g/mLExplanation:
The Density of a substance can be found by using the formula
[tex]density = \frac{mass}{volume} [/tex]
From the question
mass of rock = 12.34 g
volume = final volume of water - initial volume of water
volume = 54.5 mL - 50 mL = 4.5 mL
So the density is
[tex]density = \frac{12.34}{4.5} \\ = 2.744444444...[/tex]
We have the final answer as
2.74 g/mLHope this helps you
what density will an object have if the mass of it ia 46g and it takes up 561 mL of space?
An example of International System of Measurement (SI) unit is:
A) pound <<
B) mile
C) liter
D) ounce
Answer:
the answer is A pound, I'm guessing
explain why metals above zinc in activity series of metals are extracted from the ores by the electrolysis?
Answer:
Metals above zinc in the reactivity series are too reactive to be extracted by heat- ing with carbon or carbon monoxide gas. In this extraction process the metal ions are reduced and the carbon is oxidised. Both iron and copper can be extracted from the ore by heating with carbon.
Explanation:
ok?
0.73 grams of toluene was reacted with 2.0 grams of potassium permanganate in presence of 7.0 mL of 6 Molar potassium hydroxide and 30 mL of water. After refluxing for 1 hour the reaction mixture was treated with 6 Molar sulfuric acid to pH~ 2.0 followed by oxalic acid. On cooling this solution in an ice bath 0.633 grams of pure benzoic acid was obtained. Calculate the % yield of benzoic acid in this reaction.
Answer:
The value is [tex]k =66\%[/tex]
Explanation:
From the question we are told that
The mass of toluene [tex] m_t =0.73 \ g [/tex]
The mass of potassium permanganate is [tex] m =2.0 \ g [/tex]
The volume of potassium hydroxide V = 7.0 mL
The concentration of potassium hydroxide C = 6 M
The mass of benzoic acid is [tex]m_b = 0.633 \ g[/tex]
Generally the % yield of benzoic acid is mathematically represented as
[tex]k = \frac{m_b}{Z} * 100[/tex]
Here Z is the theoretical yield which is mathematically represented as
[tex]Z = \frac{E}{W} * m_t [/tex]
Here W is the molecular weight of product (benzoic acid) with value
W = 92.14 \ g
E is the molecular weight of reactant (toluene)with a constant value of
E = 122.12 g
So
[tex]Z = \frac{122.12 }{92.14} * 0.73 [/tex]
=> [tex]Z = 0.968 \ g [/tex]
So
[tex]k = \frac{0.633}{0.968} * 100[/tex]
=> [tex]k =66\%[/tex]
Which of the following is true concerning ψ²? A) ψ² describes the probability of finding an electron in space. B) ψ² describes the exact path of electron motion in an orbital. C) ψ² describes the electronic structure of the atom according to the Bohr model. D) ψ² describes the exact volume of an atom. E) ψ² describes the size of an atom.
Answer:
A) ψ² describes the probability of finding an electron in space.
Explanation:
The Austrian physicist Erwin Schrödinger formulated an equation that describes the behavior and energies of submicroscopic particles in general.
The Schrödinger equation incorporates both particle behavior, in terms of mass m, and wave behavior, in terms of a wave function ψ, which depends on the location in space of the system (such as an electron in an atom).
The probability of finding the electron in a certain region in space is proportional to the square of the wave function, ψ². According to wave theory, the intensity of light is proportional to the square of the amplitude of the wave, or ψ². The most likely place to find a photon is where the intensity is greatest, that is, where the value of ψ² is greatest. A similar argument associates ψ² with the likelihood of finding an electron in regions surrounding the nucleus.
A 250ml aqueous solution contains 45.1microgram of pesticide.express the pesticide concentration in:
1 .weight percent
2.parts per thousand
3.parts per million
please help its urgent
Answer:
1. 1.80x10⁻⁵ (w/w %).
2. 1.80x10⁻⁴ parts per thousand
3. 0.18 parts per million
Explanation:
The solution contains 45.1μg / 250mL.
1. Weight percent (100 times mass in grams of solute per gram of solution, as there are 250mL of water = 250g):
45.1x10⁻⁶g / 250g * 100 =
1.80x10⁻⁵ (w/w %)2. Parts per thousand (mg of solute per g of solution).
45.1μg * (1x10⁻³mg / 1μg) = 0.0451mg.
0.0451mg / 250g =
1.80x10⁻⁴ parts per thousand3. Parts per million (μg of solute per g of solution):
45.1μg / 250g =
0.18 parts per millionWhich of the scientists reponsible for cell theory would be the most likely to write a book titled cells come from cells
Answer:
Matthias Schleiden and Theodor Schwann. However, many other scientists like Rudolf Virchow contributed to the theory.
Explanation:
Answer:Theodor Shwann
Explanation:
For which initial concentration of chromate anion would[Ag +] = 6.0 x 10^ -6 M and cause the solution to begin to preciitate Ag2CrO4(s)? Where Ksp = 9.0 x 10^-12
Ag2CrO4 --> 2Ag+ + CrO4(-2)
a. 0.08
b. 0.11
c. 0.21
d. 0.25
Answer:
d. 0.25.
Explanation:
Hello,
In this case, since the equilibrium repression for the considered chemical reaction is:
[tex]Ksp=[Ag^+]^2[CrO_4^{2-}][/tex]
For a concentration of silver of 6.0x10⁻⁶ we need a concentration of chromate anion that makes the reaction quotient greater than the solubility product, thus, we write:
[tex][CrO_4^{2-}]=\frac{Ksp}{[Ag^+]^2} =\frac{9.0x10^{-12}}{(6.0x10^{-6})^2}[/tex]
[tex][CrO_4^{2-}]=0.25M[/tex]
It means that at concentrations of chromate anion of 0.25 M or more, the reaction quotient Q becomes greater than the solubility product which means that precipitation will begin occurring, therefore, answer is d. 0.25.
Best regards.
If a soft-drink bottle whose volume is 1.10 L is completely filled with water and then frozen to -10 ∘C, what volume does the ice occupy? Water has a density of 0.997 g/cm3 at 25 ∘C; ice has a density of 0.917 g/cm3 at -10 ∘C.
Answer:
The ice occupies a volume of 1.196 liters at -10 ºC.
Explanation:
We must remember that density ([tex]\rho[/tex]), measured in grams per cubic centimeters, is the ratio of mass ([tex]m[/tex]), measured in grams, to occupied volume ([tex]V[/tex]), measured in cubic centimeters, that is:
[tex]\rho = \frac{m}{V}[/tex]
We clear the mass within the formula:
[tex]m = \rho\cdot V[/tex]
The mass of the water inside the soft-drink bottle is: ([tex]\rho = 0.997\,\frac{g}{cm^{3}}[/tex] and [tex]V = 1100\,cm^{3}[/tex])
[tex]m =\left(0.997\,\frac{g}{cm^{3}} \right)\cdot (1100\,cm^{3})[/tex]
[tex]m = 1096.7\,g[/tex]
There are 1096.7 grams of water filling the soft-drink bottle completely.
Then, the water is frozen to -10 ºC and transformed into ice, the volume occupied by the ice which we can deduct from definition of density. That is:
[tex]V = \frac{m}{\rho}[/tex]
The volume occupied by the ice inside the soft-drink bottle is: ([tex]m = 1096.7\,g[/tex] and [tex]\rho = 0.917\,\frac{g}{cm^{3}}[/tex])
[tex]V = \frac{1096.7\,g}{0.917\,\frac{g}{cm^{3}} }[/tex]
[tex]V = 1195.965\,cm^{3}\,(1.196\,L)[/tex]
The ice occupies a volume of 1.196 liters at -10 ºC.
Please help
1. The atomic number tells us how many ____________ an element has.
protons
neutrons
electrons
atomic number tells us how many neutrons an element has
What is a good example of homeostasis?
Your body aging overtime
ice melting into water on a warm day (my guess)
weather patterns changing based on the season
a corspe presserved at absolute zero temperature
Answer:
it is:
a corpse preserved at absolute zero temperature
Explanation: hope it helped
A good example of homeostasis is a corpse preserved at absolute zero temperature. The correct option is D.
What is homeostasis?Homeostasis is the ability of the body to adapt according to the environment. The process of homeostasis is important for the survival of life. This term was given by Walter Cannon in 1933.
Homeostasis is of different types. They are thermoregulation, blood glucose regulation, calcium homeostasis, potassium homeostasis, osmoregulation, etc.
The human body keeps the temperature of the body remains stable every the environmental temperature is increasing or decreases. Because an optimum temperature is necessary for performing various functions.
Thus, the correct option is D. a corpse preserved at absolute zero temperature.
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John used an empty beaker with the mass of 53.126 g. In a separate jar, he had a mixture of salt and sand. He transferred two scoops of salt and sand mixture to the beaker. The overall mass was 65.449 g. As you know salt is soluble in water, he added enough water in the beaker with salt and sand mixture to completely dissolve the salt. He filtered the mixture and dried what was left in the filter. The mass of that residue was 8.200 g. Based on the experiment calculate the mass of salt, sand and the mixture (salt sand).
Answer:
[tex]m_{mixture}=12.323g[/tex]
[tex]m_{sand}=8.200 g[/tex]
[tex]m_{salt}=4.123g[/tex]
Explanation:
Hello,
In this case, since the mass of the empty beaker is 53.126 g and with the mixture is 65.449 g, we can compute the mass of the whole mixture (water, sand and salt) via subtraction:
[tex]m_{mixture}=65.449g-53.126g=12.323g[/tex]
In such a way, since the residue obtained via filtration was sand only as it is not soluble in water, we infer that the mass of sand is 8.200 g:
[tex]m_{sand}=8.200 g[/tex]
Finally, since the mixture is composed by salt and sand, the mass of salt is:
[tex]m_{salt}=m_{mixture}-m_{sand}=12.323g-8.200g\\\\m_{salt}=4.123g[/tex]
Best regards.
3. Classify each of the following as a physical or a
chemical property
a. Iron and oxygen form rust.
b. Iron is more dense than aluminum.
C. Magnesium burns brightly when ignited.
d. Oil and water do not mix.
e Mercury melts at -39°C.
The classification of the following elements as a physical or chemical property is shown below
a) iron and oxygen form rust
b)iron is more dense than aluminum
c) magnesium burns brightly when ignited
d)oil and water do not mix
e)mercury melts as -39 C
A. Chemical
B. Physical
C. Chemical
D. Physical
E. Physical
For better understanding let's explain what the statement that
Physical Property is simply known as any features of a material that can be observed or measured even without changing the composition of the substances in the materialChemical Property is commonly known as a features of matter that can be observed as it alters to a different type of matter.From the above we can therefore say that the answer The classification of the following elements as a physical or chemical property is shown below;
a) iron and oxygen form rust
b)iron is more dense than aluminum
c) magnesium burns brightly when ignited
d)oil and water do not mix
e)mercury melts as -39 C
A. Chemical
B. Physical
C. Chemical
D. Physical
E. Physical
is correct
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which choice is not an example of a molecule
Mn
O3
KOH
H2S
Answer:
The answer to your question is A | Mn
Solve for x, where M is molar and s is seconds. x=(5.3×103M−2s−1)(0.26M)3
Answer:
guys can u solve my ques plz
