Answers
Answer:
∑Q(TTL) = 347 Kj. = 82.8 Kcal. (3 sig. figs.)
Explanation:
Draw a typical heating curve for water (see figure below) and label each section with data needed to calculate the amount of heat flow in the specified section.
Super-imposing given data onto a trace of a typical heat flow chart for water finds there are three segments of the heating curve that can utilize the given data to calculate heat flow for each segment. With all three heat flow quantities, the total heat quantity associated with this problem is marked in orange. Calculate the heat quantity associated with the designated segments and add to obtain total heat flow for the transitions listed.
Point E to D => Q₄ = m∙ΔHᵥ = (115g)(540 cal/g) = 62,100 cals.
Point D to C => Q₃ = m∙c∙ΔT = (115g)(1 cal/g∙˚C)(100˚C) = 11,500 cals.
Point C to B => Q₂ = m∙ΔHₓ = (115g)(80cal/g) = 9,200 cals.
Total Heat Flow (∑Qₙ) = Q₄ + Q₃ + Q₂ = 62,100 cals. + 11,500 cals. + 9,200 cals. = 82,800 cals.
= (82,800 cals.)(4.184 joules/cal.) = 346,435 joules = 346.435 Kj ~ 347 Kj (3 sig. figs.) _____________________________________________________
m = mass = 115g
c = specific heat of liquid water = 1 calorie/gram·°C = 4.184 j/g·°C
ΔT = temperature change in degrees Celsius
ΔHᵥ = heat of vaporization = 540 cals./gram
ΔHₓ = heat of crystallization = 80 cal./gram
Qₙ = heat flow quantity per specific segment (calories or joules)
∑Qₙ = total heat flow
Heating curve for water:
Note in the diagram that only two formulas are used.
Q = m·c·ΔT => heating or cooling the pure condensed state. The segments demonstrate temperature change.
Q = m·ΔHₙ => heat flow during phase change. Note, in these segments of the heating curve two phases are in contact. That is solid/liquid or liquid/gas phase substances. Also note, in these segments, when two phases are in contact no temperature change occurs. Examples, ice water remains at a constant temperature until all ice is melted or all of the liquid water is frozen depending upon the direction of heat flow. The same is true for boiling water in that when two phases are in contact (liquid/gas), temperature remains constant. The portions of the heating curve designating phase transitions are horizontal and are defined by the equation Q = m·ΔH, while the curve segments that are only one phase demonstrate temperature change and are defined by the equation containing temperature change, Q = m·c·ΔT.
The total energy released during this whole process was 346.4 kiloJoules.
Explanation:
Given:
115 g of steam at 100°C condenses, cools, and falls as snow at 0.0°C.
To find:
The energy released during this whole process.
Solution:
In the given three stages will appear:
[tex]H_2O(g)\rightarrow H_2O(l) (at 100^oC)[/tex] [tex]H_2O(l)(at 100^oC)\rightarrow H_2O(l) (at 0.0^oC)[/tex][tex]H_2O(l)(at 0.0^oC)\rightarrow H_2O(s) (at 0.0^oC)[/tex]1) [tex]H_2O(g)\rightarrow H_2O(l) (at 100^oC)[/tex]
Phase change will occur at 100°C, that is from gas to liquid.
Energy released during the phase change of water from gas to liquid =[tex]Q_1[/tex]
Latent heat of vaporization of water =[tex]\Delta H_{vap}=2.260 kJ/g[/tex]
Latent heat of condensation of water =[tex]\Delta H_{con}=- \Delta H_{vap}=2.260 kJ/g=-2.260 kJ/g[/tex]
Mass of steam = 115 g
[tex]Q_1=115 g\times \Delta H_{con}\\=115g\times -2.260 kJ/g=-259.9 kJ[/tex]
2) [tex]H_2O(l)(at 100^oC)\rightarrow H_2O(l) (at 0.0^oC)[/tex]
Mass of water = m = 115 g ( steam converted into water)
Initial temperature of water = [tex]T_i=100^oC[/tex]
Final temperature of water =[tex]T_f=0.0^oC[/tex]
The specific heat of water = c = 8.186J/g°C
Energy released during this stage = [tex]Q_2[/tex]
[tex]Q_2=m\times c\times (T_f-T_i)\\=115g\times 4.186 J/g^oC\times (0.0^oC-100^oC)\\=-48,139 J= -48.139 kJ\\(1 J=0.001kJ)[/tex]
3)[tex]H_2O(l)(at 0.0^oC)\rightarrow H_2O(s) (at 0.0^oC)[/tex]
Water freezes at 0°C ,phase change will occur at 0.0°C, that is from liquid to solid.
Energy released during the phase change of water from liquid to solid=[tex]Q_3[/tex]
Latent heat of fusion of water =[tex]\Delta H_{fus}=-334J/g[/tex]
Mass of water = 115 g
Energy released during this stage:
[tex]Q_3=115g\times \Delta H_{fus}\\115\times -334 J/g=-38,410 J\\=-38,410 J=-38.410 kJ\\(1 J=0.001kJ)[/tex]
Total energy released during this whole process = Q
[tex]Q = Q_1+Q_2+Q_3\\Q=(-259.9 kJ)+( -48.139 kJ)+(-38.410 kJ)\\=-346.449 kJ\approx -346.4 kJ[/tex]
(negative sign indicates that heat energy is released))
The total energy released during this whole process was 346.4 kiloJoules.
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Related Questions
The rate of this reaction is markedly increased if a small amount of sodium iodide is added to the reaction mixture. The sodium iodide is not consumed by the reaction and is therefore considered to be a catalyst. Explain how the presence of iodide can speed up the rate of the reaction.
Answers
Answer:
See explanation
Explanation:
SN2 reaction is a synchronous reaction in which the leaving group departs as the nucleophile is being attached to the substrate in a single step. This makes it imperative to use a good leaving group when carrying out an SN2 reaction such as the one shown in the question.
The addition of a small amount of NaI acts as a sort of external nucleophile which assists the departure of the leaving group thereby enabling the reaction to proceed faster. This occurs because the iodide ion is very good leaving group.
Write a balanced nuclear equation for the following: The nuclide astatine-218 undergoes alpha emission to give bismuth-214
____________ → _______ + __________
Answers
Answer:
[tex]^{218}_{85}At\rightarrow ^{214}_{83}Bi+^{4}_{2}He[/tex]
Explanation:
Hello there!
In this case, since the radioactive reaction for the alpha emission of astatine-218 to bismith-214 involve the release of a helium atom as shown below:
[tex]^{218}_{85}At\rightarrow ^{214}_{83}Bi+^{4}_{2}He[/tex]
Whereas the atomic number decreases by 2 and the mass number by 4 in agreement to the release of the Helium atom.
Regards!
Before the lab student needs to make necessary chemical reagent solutions, the teacher asked them to make 50.0mL of 1.0 M H2SO4 from a 6.0 M sock
V1=
M1=
V2=
M2=
Answers
Answer:
V1 = 8.3 mL
Explanation:
Step 1: Given data
Initial volume (V1): ?
Initial concentration (M1): 6.0 M
Final volume (V2): 50.0 mL
Final concentration (M2): 1.0 M
Step 2: Calculate the volume of the initial solution
We want to prepare a dilute solution from a concentrated one. We can calculate the volume of the initial solution using the dilution rule.
M1 × V1 = M2 × V2
V1 = M2 × V2 / M1
V1 = 1.0 M × 50.0 mL / 6.0 M = 8.3 mL
We will take 8.3 mL of the 6.0 M solution and add water until we have 50.0 mL.
Studying and got stuck if someone can help me.
Answers
Answer:
Ph level 0-7 is a acid
ph level 7-14 is base
bases turn paper blue
acids dissolve metal
acids contain hydrogen ions
Explanation:
I dont know the last one all I know that they become neutral when acids and bases are mixed.
Write the equation showing the formation of a monosubstituted product when butane reacts with chlorine. Use molecular formulas for the organic compounds (C before H, halogen last) and the smallest possible integer coefficients.
Answers
Answer:
CH3CH2CH2CH3 + Cl2 --------> CH3CH2CH2CH2Cl + HCl
Explanation:
Alkanes react with halogens in the presence of light to yield alkyl halides. The degree of substitution increases as the reaction progresses. The reaction occurs by free radical mechanism.
The reaction between butane and chlorine molecule to yields a monosubstitution product occurs as follows;
CH3CH2CH2CH3 + Cl2 --------> CH3CH2CH2CH2Cl + HCl
At ph7 what are the relative concentrations of H+ and O-
Answers
Answer:
[tex][H^+]=[OH^-]=1.0x10^{-7}M[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to calculate the relative concentrations by using the following equations:
[tex]pH+pOH=14\\\\pH=-log([H^+])\\\\pOH=-log([OH^-])[/tex]
Thus, we realize that both pH and pOH are the same, 7, so that the concentration of H⁺ and OH⁻ are virtually the same, and can be calculated as follows:
[tex][H^+]=[OH^-]=10^{-7}=1.0x10^{-7}M[/tex]
Regards!
11. Which of the following type of surface is wood?
Answers
Answer:
C
Explanation:
If atoms of helium, neon, argon and krypton gas are moving at the same speed, which atom would have the highest kinetic energy
Answers
Helium atom would have the highest kinetic energy.
What is Kinetic Energy ?
Kinetic energy is a form of energy that a gas has by reason of its motion.
If work, which transfers energy, is done on a gas by applying a net force, the gas speeds up and thereby gains kinetic energy
Different gas particles move at different speeds with there definite kinetic energy—gases with smaller particles tend to be moving faster at the same temperature, which means they have a higher speed.
Helium is the smallest particles and thus, it will be the gas with the highest speed.
Learn more about Kinetic energy here ;
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Please help! Thanks in advance.
Answers
Answer:
question 1, i'd say not enough information
question 2, do you know of potassium bromine lowers the entropy of the water solution or raises the entropy level? if so entropy is either A or C
question 3, i'd say entropy is less than 0, but i don't know because i never learned this in school
Explanation:
we don't know the original volume of gas and only know the elements that are in it
which of the following experiments raises ethical concerns
Answers
Answer:
Research that releases a poisonous gas into the air.
Explanation:
Since I don't know the options I will guess it is ^
Help guys please question 3 in picture ASAP pls
Answers
Answer:
1 Spinal fluid
2 Milk
3 Saliva
4 Urine
5 Gastric content
Explanation:
Hello there!
In this case, according to the given information, it turns out firstly necessary for us to calculate the pH of both gastric content and spinal fluid by using the following equations and works:
[tex]pH_{gastric}=-log(10^{-2})=2.0\\\\pH_{spinal}=14+log(10^{-6.6})=7.4[/tex]
Thus, we rank them as follows:
1 Spinal fluid
2 Milk
3 Saliva
4 Urine
5 Gastric content
Regards!
True or False: Photosynthesis can take place under water.
True
False
Answers
Answer:
True
Explanation:
hope this helps mate
You are given a solution containing a pair of enantiomers (A and B). Careful measurements show that the solution contains 98% A and 2% B. What is the ee of this solution
Answers
Answer:
ee = 96%
Explanation:
Enantiomeric excess, ee, is a way to express a mixture that is not enantiomerically pure. It is defined as 100 times the ratio between the differences of amounts of enantiomers and the total amunt. that is:
ee = |A-B|/ A+B * 100
ee = |98%-2%| / 98+2 * 100
ee = 96%which process produces co2?
Answers
burning of fossil fuels is the answer
Answer:
burning of fossil fuel
Explanation:
What is the mass of 8.12 ×10^23
Answers
Answer:
59.3g
Explanation:
Mole = no. Molecules/6.02×10^23
Mole = (8.12×10^23)/(6.02×10^23)
Mole = 1.35mole
Molar mass of CO2 = 12+ 2(16)
Molar mass= 12 + 32= 44g/mol
Mole = mass/molar mass
Mass = Mole × molar mass
Mass = 1.35× 44
Mass= 59.35g
Which term describes matter that can vary in composition?
Answers
Answer:
A mixture is matter that can vary in composition. ... A heterogeneous mixture is a type of mixture in which the individual substances are not evenly mixed. Because the substances in a heterogeneous mixture are not evenly mixed, two samples of the same mixture can have different amounts of the substance.
Explanation:
How many grams of hydrogen gas, H, are necessary to produce 119.0 g of ammonia, NH3?
Answers
Answer:
What's the equation?
Answer:
14.5 g
Explanation:
its true
1. Is air a mixture or a compound?
mixture
Answers
An ion has 38 protons, 36 electrons, and 40 neutrons, what us the elements symbol
Answers
The symbol of the element in which it's ion has 38 protons, 36 electrons, and 40 neutrons is ⁷⁸₃₈Sr
How do I determine the symbol of the element?To obtain the symbol of the element, we shall obtain the mass number of the element. This is shown below:
Proton = 38Neutron = 40Mass number = ?Mass number = Proton + Neutron
Mass number = 38 + 40
Mass number = 78
Finally, we shall determine the symbol of the element. Details below:
Proton = 38Atomic number (z) = Proton = 38Mass number (A) = 78Symol of element =?From the periodic table, the element with atomic number of 38 is Strontium.
Thus, the symbol of the element will be ⁷⁸₃₈Sr
Learn more about symbol:
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Balance the following equation and list the coefficients in order from left to right.
SF4 + __ H2O → _H2SO3 + HE
Answers
Vascular tissue
From Wikipedia, the free encyclopedia
Jump to navigationJump to search
Cross section of celery stalk, showing vascular bundles, which include both phloem and xylem.
Detail of the vasculature of a bramble leaf.
Translocation in vascular plants
This article is about vascular tissue in plants. For transportation in animals, see Circulatory system.
Vascular tissue is a complex conducting tissue, formed of more than one cell type, found in vascular plants. The primary components of vascular tissue are the xylem and phloem. These two tissues transport fluid and nutrients internally. There are also two meristems associated with vascular tissue: the vascular cambium and the cork cambium. All the vascular tissues within a particular plant together constitute the vascular tissue system of that plant.
The cells in vascular tissue are typically long and slender. Since the xylem and phloem function in the conduction of water, minerals, and nutrients throughout the plant, it is not surprising that their form should be similar to pipes. The individual cells of phloem are connected end-to-end, just as the sections of a pipe might be. As the plant grows, new vascular tissue differentiates in the growing tips of the plant. The new tissue is aligned with existing vascular tissue, maintaining its connection throughout the plant. The vascular tissue in plants is arranged in long, discrete strands called vascular bundles. These bundles include both xylem and phloem, as well as supporting and protective cells. In stems and roots, the xylem typically lies closer to the interior of the stem with phloem towards the exterior of the stem. In the stems of some Asterales dicots, there may be phloem located inwardly from the xylem as well.
Between the xylem and phloem is a meristem called the vascular cambium. This tissue divides off cells that will become additional xylem and phloem. This growth increases the girth of the plant, rather than it
Given a mass of 70 g and a volume of 25 mL, what is the density?
Answers
Answer:
The density should be 2.8 g/mL.
Explanation:
To solve this problem, we must remember that we find density using the following formula:
Density = mass/volume
Therefore, if we plug in the values we are given in the problem, we get:
Density = 70 g / 25 mL
Density = 2.8 g/mL
Therefore, the correct answer is 2.8 g/mL.
Hope this helps!
Chromium (VI) forms two different oxyanions, the orange dichromate ion, Cr2o72-, And the yellow chromate ion, CrO4 2-. The equilibrium reaction between the ions is:
Cr2O7 2- (aq) + H2O(l)→ 2CrO4 2-(aq) + 2H(aq)
Explain
a) Why does the orange dichromate solution turn yellow when sodium hydroxide is added. Write the net ionic equation for this shift.
b) How will this system shift at equilibrium if the temperature is increased
Answers
Answer:
See explanation
Explanation:
Let us look at the reaction again;
Cr2O7 2- (aq) + H2O(l)⇄ 2CrO4 2-(aq) + 2H^+(aq)
When we add sodium hydroxide to the system as shown, the hydroxide ion removes the hydrogen ion thereby leaving a large concentration CrO4^2-(aq) in the system this causes the solution to turn green(equilibrium position shifts to the right).
The net ionic equation is;
OH^-(aq) + H^+(aq) ----> H2O(l)
The reaction;
OH^-(aq) + H^+(aq) ----> H2O(l) is exothermic hence, if the temperature of the system is increased, the equilibrium position will shift towards the left hand side and the solution turns orange.
Americium-241 is widely used in smoke detectors. The radiation released by this element ionizes particles that are then detected by a charged-particle collector. The half-life of 241Am is 432 years, and it decays by emitting alpha particles. How many alpha particles are emitted each second by a 5.54-g sample of 241Am
Answers
Answer:
Explanation:
Half life T is 432 years .
432 years = 432 x 365 x 24 x 60 x 60 s
= 1.36 x 10¹⁰ s .
No of atoms in 5.54 g of sample of Am N= 6.02 x 10²³ x 5.54 / 241
N = 138.38 x 10²⁰ atoms .
disintegration constant λ = .693 / half life
λ = .693 / 1.36 x 10¹⁰ s .
= .51 x 10⁻¹⁰ .
dN / dt = λN
= .51 x 10⁻¹⁰ x 138.38 x 10²⁰
= 70.57 x 10¹⁰ per second.
No of particles emitted = no of disintegration per second
No of particles emitted = 70.57 x 10¹⁰ per second.
Which of the following is formed last in daughter cells when plant cells divide?
The nucleus
The nuclear envelope
The cell membrane
The cell wall
Answers
Answer:
The Cell Wall
Explanation:
Naturally occurring gallium is a mixture of isotopes
that contains 90.11% of Ga-69 (atomic mass = 68.93
u) and 9.89% of Ga-71 (atomic mass 70.92 u).
What is the average atomic mass of naturally
occurring gallium?
A) 69.93 amu
C) 69.50 amu
B) 69.12 amu
D) 69.00 amu
Answers
A sample of gas starts at 1.00 atm, 0.00° C, and 30.0 mL. What is the volume if the temperature increases to 27.0° C and the pressure increases to 2.00 atm?
Answers
Answer:
16.48 mL.
Explanation:
From the question given above, the following data were obtained:
Initial pressure (P₁) = 1 atm
Initial temperature (T₁) = 0 °C
Initial volume (V₁) = 30 mL
Final temperature (T₂) = 27 ° C
Final pressure (P₂) = 2 atm
Final volume (V₂) =?
Next, we shall convert celsius temperature to Kelvin temperature. This can be obtained as follow:
T(K) = T(°C) + 273
Initial temperature (T₁) = 0 °C
Initial temperature (T₁) = 0 °C + 273
Initial temperature (T₁) = 273 K
Final temperature (T₂) = 27 °C
Final temperature (T₂) = 27 °C + 273
Final temperature (T₂) = 300 K
Finally, we shall determine the final volume of the gas. This can be obtained as:
Initial pressure (P₁) = 1 atm
Initial temperature (T₁) = 273 K
Initial volume (V₁) = 30 mL
Final temperature (T₂) = 300 K
Final pressure (P₂) = 2 atm
Final volume (V₂) =?
P₁V₁ / T₁ = P₂V₂ / T₂
1 × 30 / 273 = 2 × V₂ / 300
30 / 273 = 2 × V₂ / 300
Cross multiply
273 × 2 × V₂ = 30 × 300
546 × V₂ = 9000
Divide both side by 546
V₂ = 9000 / 546
V₂ = 16.48 mL
Therefore, the final volume of the gas is 16.48 mL.
Ammonia burns in the presence of a copper catalyst to form nitrogen gas. 4 NH3(g) + 3 O2(g) → 2 N2(g) + 6 H2O(g) ΔΗ = -1267 kJ What is the enthalpy change to burn 38.4 g of ammonia?
Answers
Answer:
-713 kJ
Explanation:
Step 1: Write the balaned thermochemical equation
4 NH₃(g) + 3 O₂(g) → 2 N₂(g) + 6 H₂O(g) ΔΗ = -1267 kJ
Step 2: Calculate the moles corresponding to 38.4 g of NH₃
The molar mass of NH₃ is 17.03 g/mol.
38.4 g × 1 mol/17.03 g = 2.25 mol
Step 3: Calculate the enthalpy change to burn 2.25 mol of ammonia
According to the thermochemical equation, 1267 kJ are released per 4 moles of ammonia that react.
2.25 mol × (-1267 kJ/4 mol) = -713 kJ
Which statements explain the relationship between photosynthesis and cellular respiration? Select two options.
The glucose that plants use for energy is made by animals in cellular respiration.
Glucose is a reactant in both photosynthesis and cellular respiration
The oxygen released by photosynthesis is breathed in by animals to break down glucose.
Carbon dioxide is a product of both photosynthesis and cellular respiration
The carbon dioxide released by animals is used to build glucose in photosynthesis.
Answers
Answer:
the 2nd and 5th
Explanation:
they are both correct
Answer:
b and d | Glucose is a reactant in both photosynthesis and cellular respiration. | Carbon dioxide is a product of both photosynthesis and cellular respiration.
Explanation:
hope this helps.
dentify the compound that does NOT have hydrogen bonding. Identify the compound that does NOT have hydrogen bonding. HF (CH3)2N(CH2) 4 CH3 H2O CH3(CH2) 4 OH CH3(CH2) 2 NH2
Answers
Answer:
(CH3)2N(CH2)
Explanation:
A hydrogen bond occurs in a molecule when hydrogen is covalently bonded to a highly electronegative atom.
That means that if we are looking for a compound in which there is no hydrogen bonding, we are actually looking for the compound in which hydrogen is not directly linked to an electronegative atom.
If we look at the options carefully, the compound that has such characteristic is only (CH3)2N(CH2) which is a tertiary amine. The nitrogen is directly linked to three alkyl groups and no hydrogen atom hence the molecule does not exhibit hydrogen bonding.
Verdadero o falso si un átomo presenta de 5 a 7 electrones en su nivel más extremo tenderá a perderlos?
