Answer:
Yes, the 4 Krabby patties are an example of high precision.
Explanation:
Precision is the closeness of the values obtained by applying them to the same parameter. To calculate the precision, the closeness of each of the values with the mean must be determined, all the values will be taken by subtracting the mean and the values will be taken into account to know which are the most accurate.
(12.12 + 11.9 + 12.2 + 12. 1) / 4 = 12.08
12.12 - 12.8 = 0.04
11.9 - 12.08 = - 0. 18
12.2 - 12.08 = 0.12
12.1 - 12.08 = 0.02
The most accurate precision is that of the fourth Krabby Pattie and the least precise is that of the second.
Explanation:
Step 1: Determine what precision and accuracy is
According to online, "precision is how close measure values are to each other, basically how many decimal places are at the end of a given measurement. Precision does matter. Accuracy is how close a measure value is to the true value. Accuracy matters too, but it's best when measurements are both precise and accurate."
Step 2: Determine which are displayed in this problem
We can see that most of these weight are really close to the original weight which means that the accuracy is pretty high. This means that our weights are fairly close to the original predicated amount.
We can also see that our precision is somewhat decent because most of the numbers bunch up around each other. Precision and accuracy don't have to occur at the same time. For example, a precise collection can be 15, 15.01, 14.99. Although they are far away from the original point, they are bunched up together giving a high precision.
How much energy has your body used, in joules, if your health indicates the 450 Calories were burned during your workout? How many Calories are in a dinner containing 3.5 x 106 joules?
Answer:
107.55 J, 88.67 cal
Explanation:
1 Joule = 0.239 Calories
450 cal x 0.239 = 107.55 J
3.5 x 106 = 371
371 J = 371 x 0.239 cal
371 J = 88.67 cal
For a chemical reaction to be spontaneous at all temperatures, which of the following conditions must be met?
A. ?S° > 0, ?H° > 0.
B. ?S° > 0, ?H° < 0.
C. ?G° > 0.
D. ?S° < 0, ?H° > 0.
E. ?S° < 0, ?H° < 0.
Answer:
B. S° > 0, ?H° < 0.
Explanation:
Expression for change in free energy is
ΔG = ΔH - TΔS
For a reaction to be spontaneous , ΔG should be negative . When we watch the relation above , we find that for ΔG will be negative at any temperature , if ΔH is negative and ΔS is positive . Then both the terms on the right hand side will be negative and then ΔG will become negative.
So option B is correct .
A solution of sulfuric acid contains 35.2% by mass of H2SO4 and has a density of 1.27 g/mL. What is the molarity of H2SO4 in this solution?
Answer:
[H₂SO₄] = 4.56 M
Explanation:
35.2 % by mass of H₂SO₄ means that in 100 g of solution, we have 35.2 grams of solute.
We convert the mass of solute to moles.
35.2 g / 98 g/mol = 0.359 moles
These moles are contained in 100 g of solution, so we use density to determine the volume.
1.27 g/mL = 100 g / volume
Volume = 100 g / 1.27 g/mL = 78.74 mL
Molarity is mmol /mL (either we can say, moles in 1L of solution).
We convert the moles to mmoles → 0.359 mol . 1000 = 359 mmoles
M = 359 mmoles/74.74 mL = 4.56 M
In the laboratory you dissolve 23.8 g of manganese(II) nitrate in a volumetric flask and add water to a total volume of 125 . mL. What is the molarity of the solution
Answer:
The correct answer is 1.06 M
Step-by-step explanation:
We have to calculate the molarity (M), which is:
M= moles solute/ 1 L solution.
The chemical formulae of manganese (II) nitrate is Mn(NO₃)₂. So, we first calculate its molecular weight (Mw) as follows:
Mw(Mn(NO₃)₂)= molar mass Mn + (2 x molar mass N) + (6 x molar mass N)= 55 g/mol + (2 x 14 g/mol) + (6 x 16 g/mol) = 179 g/mol
Then, with Mw we calculate the number of moles there is in the given mass of Mn(NO₃)₂:
moles Mn(NO₃)₂= mass/Mw= 23.8 g/(179 g/mol)= 0.133 mol
Now, we need the final volume in liters, so we convert the volume from mL to L:
125 mL x 1 L/1000 mL = 0.125 L
Finally, we divide the moles of Mn(NO₃)₂ into the volume in L, to obtain the molarity in mol/L:
M= 0.133 moles/0.125 L = 1.06 mol/L= 1.06 M
A 50 ml graduated cylinder is filled with 23.0 mL of water. Water is then added to the cyl- inder dropwise until 36.0 mL of water are in the cylinder. If it took 265 drops to increase the volume in the cylinder to 36.0 mL. How many drops are in 1 ml?
Answer:
20 drops
Explanation:
Step 1: Given data
Initial volume of water: 23.0 mLFinal volume of water: 36.0 mLDrops of water added: 265 dropsStep 2: Calculate the volume of water added
We calculate the volume of water added by subtracting the initial volume to the final volume.
36.0 mL - 23.0 mL = 13.0 mL
Step 3: Calculate the number of drops in 1 mL
We will divide the number of drops of water added by the volume they represent.
265 drops / 13.0 mL = 20.4 drops/1 mL ≈ 20 drops/1 mL
. A sample of crude oil has a density of 0.87 g/mL. What volume (in liters) does a 3.6 kg sample of this oil occupy
Answer:
The volume is 4.13793 L
Explanation:
Density is a quantity that expresses the relationship between the mass and the volume of a body, so it is defined as the quotient between the mass and the volume of a body:
[tex]density=\frac{mass}{volume}[/tex]
Density is a characteristic property of every body or substance.
The most commonly used units of density are [tex]\frac{kg}{m^{3} }[/tex] or [tex]\frac{g}{cm^{3} }[/tex] for solids, and [tex]\frac{kg}{L}[/tex] or [tex]\frac{g}{mL}[/tex] for liquids and gases.
In this case, you know:
density= 0.87 [tex]\frac{g}{mL}[/tex]mass= 3.6 kg= 3,600 g (being 1 kg=1,000 g)volume= ?Replacing:
[tex]0.87\frac{g}{mL} =\frac{3,600 g}{volume}[/tex]
Solving:
[tex]volume =\frac{3,600 g}{0.87\frac{g}{mL}}[/tex]
volume= 4,137.93 mL
Being 1,000 mL=1 L, then volume= 4,137.93 mL= 4.13793 L
The volume is 4.13793 L
evidence and reasning on when puppies are going to be born
Answer:
Keep a close eye on her in case she starts giving birth to the next pup at the same time. A greenish/brown discharge may suggest a placenta has separated. If you see this, a puppy should be born within the next 2-4 hours.
Carl is approaching the Canadian border on a trip from Vermont to Montreal, Canada, and the gas in his tank is low. He wants to determine if it would be more cost efficient to buy gas before or after crossing the border. He consults Gas Buddy and finds that the cost of gas in the US is 2.21 USD per gallon, while just across the border it is $1.40 CAD per liter. The currency exchange rate is $0.750 USD per $1.000 CAD. How much more gas (in gallons) will he get for $20 USD if he buys the gas before crossing the border instead of buying it in Canada? (1.000 liter = 0.264 US gallons).
Answer:
4.02 gallons
Explanation:
From the illustration:
$0.750 USD = $1.00 CAD
Therefore,
$20 USD = 20 x 1/0.750 = $26.667 CAD
The cost of gas in the US is 2.21 USD per gallon.
20 USD will be able to get 20/2.21 = 9.05 gallons
The cost of gas across the border is 1.40 CAD per liter.
0.264 gallon = 1 liter
1 gallon = 1/0.264 = 3.788 liters
This means that 1 gallon = 3.788 x 1.40 = 5.303 CAD
Hence, 26.667 CAD will get 26.667/5.303 = 5.03 gallons
If he buys a $20 USD gas before crossing the border, Carl will get 9.05 gallons, but if he buys it after crossing the border, he will get 5.03 gallons. Therefore, if he buys the gas before crossing the border instead of buying it in Canada, he will get 9.05 - 5.03 = 4.02 gallons extra.
I know how to solve it with D=M/V and M1V1 however the answer isn’t correct. Help me please
Answer:
23.28 g of O2.
Explanation:
We'll begin by calculating the mass of hexane. This can obtain as follow:
Volume of hexane = 10 mL
Density of hexane = 0.66 g/mL
Mass of hexane =?
Density = mass /volume
0.66 = mass of hexane /10
Cross multiply
Mass of hexane = 0.66 x 10
Mass of hexane = 6.6 g
Next, we shall write the balanced equation for the reaction. This is given below:
2C6H14 + 19O2 —> 12CO2 + 14H2O
Next, we shall determine the masses of C6H14 and O2 that reacted from the balanced equation. This can be obtained as follow:
Molar mass of C6H14 = (12.01x6) + (1.008 x 14)
= 72.06 + 14.112
= 86.172 g/mol
Mass of C6H14 from the balanced equation = 2 x 86.172 = 172.344 g
Molar mass of O2 = 16x2 = 32 g/mol
Mass of O2 from the balanced equation = 19 x 32 = 608 g
From the balanced equation above,
172.344 g of C6H14 reacted with 608 g of O2.
Finally, we shall determine the mass of O2 needed to react with 10 mL (i.e 6.6 g) of hexane, C6H14. This can be obtained as follow:
From the balanced equation above,
172.344 g of C6H14 reacted with 608 g of O2.
Therefore, 6.6 g of C6H14 will react with = (6.6 x 608)/172.344 = 23.28 g of O2.
Therefore, 23.28 g of O2 is needed for the reaction.
You are the manager of a chemical stockroom, and find a bottle containing approximately one liter of a clear and colorless solution of unknown identity and concentration. Your only clue to its identity is that it was found between bottles of silver fluoride and sodium fluoride, so it is likely an aqueous solution of one of those two compounds. You will need to develop a procedure to determine the following: a) The identity of the unknown solution b) The concentration of the unknown solution
Answer:
Explanation:
Given that for Part A,
the arrangement can be either AgF or NaF.
On the off chance that the arrangement is AgF ,
For this situation AgF conc. can be controlled by titration with standard choride arrangement with potassium chromate as marker.
Get ready 1 M NaCl standard arrangement by dissolving 58.5 gm of NaCl in one liter refined water. Take 25 mL of 1 N NaCl arrangement in a funnel shaped carafe include hardly any drops of potassium chromate marker.
Take the example in the burette and titrate until a perpetual ruddy earthy colored hasten is seen.
[AgF responds with NaCl as follows
AgF + NaCl \small \rightarrow AgCl ( white precipitate)+ NaF
For whatever length of time that CaCl is available in the arrangement Ag structures AgCl hasten as above. When all the chloride particles expelled this way, the overabundance silver structures Ag2CrO4 ( rosy earthy colored encourage ) with the marker.
The dissolvability of AgCl is a lot of lower than the solvency of Ag2CrO4. Consequently the later won't hasten until all chloride particles exhausted.]
Let V2 be the volume of test utilized. At that point grouping of AgF in the example is given by
C2 = V1C1/V2 = 25 x 1/V2.
Rehash the titration by making legitimate weakening if any required.
On the off chance that the arrangement is NaF (Indirect strategy)
Here the Fluoride particle in a referred to volume of test is hastened as leadchlorofluoride with NaCl and PbNO3.
NaF + NaCl + Pb(NO3)2 \small \rightarrow PbClF (hasten) + 2 NaNO3
The above hasten is separated and washed and the washed encourage is taken in a measuring utencil and re-broke up in nitric corrosive to deliver the chloride particles.
PbClF + 2 HNO3 \small \rightarrow Pb(NO3)2 + HCl + HF
Presently include a known abundance of Silver nitrate answer for the above blend to encourage the chloride as silver chloride. Channel wash and evacuate the AgCl encourage.. The Filtrate contains the overabundance silver nitrate is to be dictated by titration with Standard chloride arrangement as in the past technique.
From the first amount of silver nitrate taken and the overabundance silver nitrate decided from the titration, we can decide the amount of silver nitrate responded.
The moles of silver nitrate responded = the moles of Cl delivered = the moles of F in the example.
In this way the convergence of sodium fluoride in the obscure example might be resolved.
Consider the two gaseous equilibria: The values of the equilibrium constants K 1 and K 2 are related by
The question is missing parts. The complete question is as follows.
Consider the two gaseous equilibria involving SO2 and the corresponding equilibrium constants at 298K:
[tex]SO_{2}_{(g)} + \frac{1}{2}O_{2}[/tex] ⇔ [tex]SO_{3}_{(g)}[/tex]; [tex]K_{1}[/tex]
[tex]2SO_{3}_{(g)}[/tex] ⇔ [tex]2SO_{2}_{(g)}+O_{2}_{(g)}; K_{2}[/tex]
The values of the equilibrium constants are related by:
a) [tex]K_{1}[/tex] = [tex]K_{2}[/tex]
b) [tex]K_{2} = K_{1}^{2}[/tex]
c) [tex]K_{2} = \frac{1}{K_{1}^{2}}[/tex]
d) [tex]K_{2}=\frac{1}{K_{1}}[/tex]
Answer: c) [tex]K_{2} = \frac{1}{K_{1}^{2}}[/tex]
Explanation: Equilibrium constant is a value in which the rate of the reaction going towards the right is the same rate as the reaction going towards the left. It is represented by letter K and is calculated as:
[tex]K=\frac{[products]^{n}}{[reagents]^{m}}[/tex]
The concentration of each product divided by the concentration of each reagent. The indices, m and n, represent the coefficient of each product and each reagent.
The equilibrium constants of each reaction are:
[tex]SO_{2}_{(g)} + \frac{1}{2}O_{2}[/tex] ⇔ [tex]SO_{3}_{(g)}[/tex]
[tex]K_{1}=\frac{[SO_{3}]}{[SO_{2}][O_{2}]^{1/2}}[/tex]
[tex]2SO_{3}_{(g)}[/tex] ⇔ [tex]2SO_{2}_{(g)}+O_{2}_{(g)}[/tex]
[tex]K_{2}=\frac{[SO_{2}]^{2}[O_{2}]}{[SO_{3}]^{2}}[/tex]
Now, analysing each constant, it is easy to see that [tex]K_{1}[/tex] is the inverse of [tex]K_{2}[/tex].
If you doubled the first reaction, it will have the same coefficients of the second reaction. Since coefficients are "transformed" in power for the constant, the relationship is:
[tex]K_{2}=\frac{1}{K_{1}^{2}}[/tex]
What is the mass (g) of a 17.4 mL piece of material, if the density is 3.4 g/mL?
Answer:
59.16 g
Explanation:
Mass = Density × Volume
= 3.4 g/mL × 17.4 mL
= 59.16 g or
59.2 g (rounded to three significant figures) or
59 g (rounded to two significant figures)
Hope that helps.
The mass of a 17.4 mL piece of material, if the density is 3.4 g/mL is 59.16 grams.
What is density?Density is defined as the degree to which a material is packed together.
It can also be defined as a material is defined as its total mass (m) divided by its total volume (V).
Density is an important topic because it tells us which compounds will float and which will sink when placed in a liquid. Usually substances float as long as their density is less than that of the liquid in which they are immersed.
Density can be calculated in kilogram per cubic meter or gram per cubic meter.
Density can be expressed as
Density = mass / volume
Given Volume = 17.4 ml
Density = 3.4 g / ml
Mass = density x volume
Mass = 3.4 x 17.4
Mass = 59.16 grams
Thus, the mass of a 17.4 mL piece of material, if the density is 3.4 g/mL is 59.16 grams.
To learn more about density, refer to the link below:
https://brainly.com/question/15164682
#SPJ5
what is 8.000 mol / 2.00 L
Answer:
[tex]M=4.00\frac{mol}{L}=4M[/tex]
Explanation:
Hello.
In this case, we should remember that a molar concentration is defined in terms of the moles and volume in liters as shown below:
[tex]M=\frac{mol}{L}[/tex]
Thus, for the given information we obtain:
[tex]M=\frac{8.000mol}{2.00L}\\ \\M=4.00\frac{mol}{L}=4M[/tex]
Best regards.
Is HCL a pure substance? Look at me already asking chemistry questions! //!roll eyes emoji/2#/
Answer:
yes it is as it has a chemical formula which cannot be changed
Perform the following mathematical operations and express the correct answer to the proper number of significant figures or decimal places as required.
7.71 + 4.31 + 1.7 + 4.00141=
7.71 x 4.31 x 1.7 x 4.00141=
Answer: 7.71 + 4.31 + 1.7 + 4.00141= 17.72141
7.71 x 4.31 x 1.7 x 4.00141= 226.04433255
Explanation:
(i) 7.71 + 4.31 + 1.7 + 4.00141
= 7.71000+4.31000+1.70000+ 4.00141 [make like decimals]
= 17.72141 [By adding the corresponding places ]
(ii) 7.71 x 4.31 x 1.7 x 4.00141 =( 7.71 x 4.31) x 1.7 x 4.00141
= 33.2301 x 1.7 x 4.00141
= (33.2301 x 1.7) x 4.00141
= 56.49117 x 4.00141
= ( 56.49117 x 4.00141 )
= 226.04433255
Hence, 7.71 + 4.31 + 1.7 + 4.00141= 17.72141
7.71 x 4.31 x 1.7 x 4.00141= 226.04433255
Which of the changes are chemical changes?
Sugar is dissolved in water.
A warm soda bottle fizzes when opened.
The copper roof turns green over time.
Baking soda is dissolved in vinegar and bubbles form.
A fog appears on the windshield in cold weather.
Answer:
two of them are chemical changes
Explanation:
1.) the copper roof turns green over time: it forms a new substance which is the green deposit known as verdigris .
2.) baking soda is dissolved in vinegar and bubbles form: it forms a new substance which is carbon dioxide gas.
I hope it helps:)
In your own words, describe the concept of specific heat capacity and the effects that it has on temperature changes. Give an example of specific heat capacity that you see in your everyday life. Why did you select this example
Answer:
The specific heat capacity is the heat that a body or a system needs to administer so that it can increase its internal temperature.
Explanation:
The calorific capacity is measured in several units, it varies a lot between products, reactants or the same systems since each one is independent in its composition and this conditions it.
As for its mathematical calculation, it is the quotient, that is, the division between the dose of energy transferred to a body and the change in temperature that it experiences.
The solubility of silver sulfate (Ag2SO4), in moles per liter, can be expressed in terms of the resulting ion concentrations. Which relationship is correct?
Answer:
Solubility = [SO₄²⁻]
Explanation:
Solubility of Ag₂SO₄ can be understood seen its Ksp equilibrium:
Ag₂SO₄(s) ⇄ SO₄²⁻ + 2Ag⁺
Where 1 mole of silver sulfate dissolves producing 1 mole of sulfate ion and 2 moles of silver ion.
Solubility is defined as: "the amount of solid that can be dissolved per liter of solution".
In the dissolution process, the moles of Ag₂SO₄ dissolved are equal to moles of SO₄²⁻.
That means:
Solubility = [SO₄²⁻]Rhodium has an atomic radius of 0.1345 nm and density of 12.41 gm/cm3 . Determine whether it has an FCC or BCC crystal structure.
Answer:
FCC.
Explanation:
Hello,
In this case, since the density is defined as:
[tex]\rho =\frac{n*M}{Vc*N_A}[/tex]
Whereas n accounts for the number of atoms per units cell (2 for BCC and 4 for FCC), M the atomic mass of the element, Vc the volume of the cell and NA the Avogadro's number. Thus, for both BCC and FCC, the volume of the cell is:
[tex]Vc_{BCC}=(\frac{4r}{\sqrt{3} } )^3=(\frac{4*0.1345x10^{-7}cm}{\sqrt{3} } )^3=2.997x10^{-23}cm^3\\\\Vc_{FCC}=(2\sqrt{2}r)^{3} =(2\sqrt{2} *0.1345x10^{-7}cm)^3=5.506x10^{-23}cm^3[/tex]
Hence, we compute the density for each crystal structure:
[tex]\rho _{BCC}=\frac{n_{BCC}*M}{Vc_{BCC}*N_A}=\frac{2*102.9g/mol}{2.337x10^{-23}cm^3*6.022x10^{23}/mol} =14.62g/cm^3\\\\\rho _{FCC}=\frac{n_{FCC}*M}{Vc_{FCC}*N_A}=\frac{4*102.9g/mol}{5.506x10^{-23}cm^3*6.022x10^{23}/mol} =12.41g/cm^3[/tex]
Therefore, since the density computed as a FCC crystal structure matches with the actual density, we conclude rhodium has a FCC crystal structure.
Regards.
What is the mass of an object if its height is 10 meters above ground and its gravitational potential energy (GPE) is 3920 Joules? Remember Gravity = 9.8 meters per seconds squared
Answer:
40
Explanation:
Your trying to find out the meters so your going to divide 3920J by 10 and 9.8
3920/10/9.8
Which of the following is the most effective buffer system for a pH value of 4.45? a) H3PO4/H2PO4 (Kal for H3PO4 is 6.9 times 10-3) b) H_2CO_3/HCO_3^- (K_a1 for H_2CO_3 is 4.8 times 10^-7) c) HCO_3^-/CO_3^2- (K_a2 for H2CO_3 is 4.8 times 10^-11) d) H_2S/HS^- (K_a1 for H_2S is 8.9 times 10^-8) e) HC_2O_4^-/C_2O_4^2- (K_a2 for H_2C_2O_4 is 5.1 times 10^-5)
Answer:
Option e.
Acid oxalate / oxalate
Explanation:
To determine the most efecttive buffer, we must look the pKas of the system.
If pKa value is similiar to buffer's pH, the variation of pH by the addition of a small amount of strong acid or base is minimal. It would be a good buffer.
a. H₃PO₄ / H₂PO₄⁻ Ka 1 = 6.9×10⁻³
pKa = - log Ka → - log 6.9×10⁻³ = 2.16
b. H₂CO₃ / HCO₃⁻ Ka1 = 4.8×10⁻⁷
pKa = - log Ka → - log 4.8×10⁻⁷ = 6.32
c. HCO₃⁻ / CO₃⁻² Ka2 = 4.8×10⁻¹¹
pKa = - log Ka → - log 4.8×10⁻¹¹ = 10.32
d. H₂S / HS⁻ Ka1 = 8.9×10⁻⁸
pKa = - log Ka → - log 8.9×10⁻⁸ = 7.05
e. HC₂O₄⁻ / C₂O₄⁻² Ka2 = 5.1×10⁻⁵
pKa = - log Ka → - log 5.1×10⁻⁵ = 4.29
pKa which is so close to the ph's value is e, acid oxalate / oxalate
How many signals does the aldehyde (CH3)3CCH2CHO have in 1H NMR and 13C NMR spectra?
a. five 'H signals and six 13C signals.
b. three 'H signals and four 13C signals.
c. five fH signals and four 13C signals.
d. three 'H signals and six 13C signals.
e. Impossible to know without knowing which isomer was under study.
Answer:
B- three 1H signals and four 13C signals
Explanation:
Protons in similar environment have the same chemical shift; deshielding by electro-negative elements is also a factor that plays in signal emission in spectroscopy.
All C- atoms in methyl-3 group would count as a single peak, as well as the hydrogen atoms present there on the 13C and 1H spectra respectively.
Let me know if you have any further questions.
72. A piece of unknown metal with mass 30 g is heated to 1100 C and dropped into 100.0 g of water at 200 C. The final temperature of the system is 25 degree Celsius. What is the specific heat of the metal
Answer:
THE SPECIFIC HEAT OF THE METAL IS 28.69 J/g C
Explanation:
Mass of unknown metal = 30 g
Mass of water = 100 g
Initial Temperature of metal = 110 C
Initial temperature of water = 200 C
Final temperature of the system = 25 C
Specific heat of water = 4200 J/ g C
Specific heat of metal = unknown
Heat lost = Heat gained\
M C Q = M C Q
30 * C * ( 25 - 110) = 100 * 4200 * (25 -200)
C = 100 * 4.18 * -175 / 30 * - 85
(the minus sign cancels out and we have):
C = 100 * 4.18 * 175 / 30 * 85
C = 73150 / 2550
C = 28.69 J/ g C
The specific heat of the unknown metal is hence 28.69 J/ g C
Please Help Me What is chromatography? Question 2 options: A) A picture taken with a chrome camera. B) The stationary phase. C) The separation of substances in a mixture by differences in their attraction to a substance over which they are passed.
Explanation:
Chromatography = It is a method in which a drop of mixture is spot is spotted at one of the end of the paper. After some time whenever it gets dried, it is then dipped into solvent without toushing the spot. As a result the mixtures gets separated.
Hence, the correct option is (c) "The separation of substances in a mixture by differences in their attraction to a substance over which they are passed"
Significant Figures
Tell me what's wrong. When adding and subtracting, you line up decimal and look for the least decimal places.
0.87
4.061
+10.4
--------------
15.331
My answer is 15 but teacher said I'm wrong
Thank You
Based on the bond energies for the reaction below, what is the enthalpy of the reaction?HC≡CH (g) + 5/2 O₂ (g) → 2 CO₂ (g) + H₂O (g)
Answer:
1219.5 kj/mol
Explanation:
To reach this result, you must use the formula:
ΔHºrxn = Σn * (BE reactant) - Σn * (BE product)
ΔHºrxn = [1 * (BE C = C) + 2 * (BE C-H) + 5/2 * (BE O = O)] - [4 * (BE C = O) + 2 * (BE O-H).
The BE values are:
BE C = C: 839 kj / mol
BE C-H: 413 Kj / mol
BE O = O: 495 kj / mol
BE C = O = 799 Kj / mol
BE O-H = 463 kj / mol
Now you must replace the values in the above equation, the result of which will be:
ΔHºrxn = [1 * 839 + 2 * (413) + 5/2 * (495)] - [4 * (799) + 2 * (463) = 1219.5 kj/mol
Based on the bond energies, the reaction
HC≡CH(g) + 5/2 O₂(g) → 2 CO₂(g) + H₂O(g) has a standard enthalpy of reaction of - 1222 kJ/mol.
The given reaction is: HC≡CH(g) + 5/2 O₂(g) → 2 CO₂(g) + H₂O(g)
The bond energies (E) of the reactions can help to calculate the standard enthalpy of the reaction (ΔH°rxn) using the following expression.
[tex]\triangle H_ \degrees {rxn} = \sum E_{reactants} - \sum E_{products}[/tex]
The broken bonds are: HC≡CH 1 triple C-C bond 2 single C-H bonds O₂ 5/2 double O-O bonds The formed bonds are: CO₂ 4 double C-O bonds H₂O 2 single O-H bondsSolution:
bond energies in the formula:
[tex]\triangle H_ \degrees {rxn} = 835+ 2(411) + 2.5(494) - 4(799) - 2(459)[/tex]
= 1222 kJ/mol
Thus, Based on the bond energies, the reaction:
HC≡CH(g) + 5/2 O₂(g) → 2 CO₂(g) + H₂O(g) has a standard enthalpy of reaction of -1222 kJ/mol.
Learn more: brainly.com/question/24857760
Speed of light is 3.0x108 m/s. Convert speed of light unit into miles per hour (mi/s). 3.0x108 m/s =
mi/s
Report your answer in the format with correct sig figs:
5x1013
Given:
1 mile = 1.61 km
1 km = 1000 m
Explanation:
1 mile = 1.61 km and 1 km = 1000 m
The speed of light is [tex]3\times 10^8\ m/s[/tex]. We need to convert it into miles per hour.
1 m = (1/1000) km
[tex]3\times 10^8\ m=\dfrac{3\times 10^8}{1000}\\\\=3\times 10^5\ km[/tex]
1 km = (1/1.61) miles
[tex]3\times 10^5\ km =\dfrac{3\times 10^5}{1.61}\\\\=1.86\times 10^5\ \text{miles}[/tex]
So, [tex]v=1.86\times 10^5\ \text{miles/s}[/tex]
Hence, this is the required solution.
Consider the following numbered processes: 1. A → 2B 2. B → C + D 3. E → 2D ΔH for the process A → 2C + E is
Answer:
ΔH = ΔH₁ + ΔH₂ - ΔH₃
Explanation:
Given that:
1. A → 2B
2. B → C + D
3. E → 2D
Assuming from the corresponding ΔH for process 1, 2 and 3 are ΔH₁, ΔH₂, ΔH₃ respectively.
To estimate the ΔH for the process A → 2C + E
We multiply 2 with equation 2 where (B → C + D)
2B → 2C + 2D ⇒ 2ΔH₂
Also, let's switch equation (3), such that we have,
2D → E -ΔH₃
The summation of all the equation result into :
A → 2C + E
where; ΔH = ΔH₁ + ΔH₂ - ΔH₃
Based on Hess law of constant heat summation, the ΔH for the process A → 2C + E is ΔH1 + ΔH2 - ΔH3
What is the enthalpy of the series steps of reaction steps?According to Hess' law of constant heat summation, the heat of any reaction ΔH∘f for a specific reaction is equal to the sum of the heats of reaction for any set of reactions which in sum are equivalent to the overall reaction.
Thus, ΔH for the process A → 2C + E is calculated as follows:
Reaction step 1: A → 2B ; ΔH1Reaction step 2: B → C + D ; ΔH2Reactipn step 3: E → 2D ; ΔH3The A → 2C + E can be obtained from the summation of the processes above:
A ---> 2B
2B ---> 2C + 2D
2D ---> E by reversing reaction step 3
Thus, ΔH = ΔH1 + ΔH2 - ΔH3
Therefore, the ΔH for the process A → 2C + E is ΔH1 + ΔH2 - ΔH3
Learn more about enthalpy and Hess law at: https://brainly.com/question/11628413
How much energy is transferred when 30.0 g of water is cooled from 25.0’C to 12.7’C
Answer:
1549.8 Joules.
Explanation
Energy transferred = mass ( in kg) * specific heat of water * change in temperature in ( celsius).
= 0.03 * 4200 * (25 - 12.7)
= 1549.8 Joules.
According to specific heat capacity, 1549.8 joules of energy is transferred when 30.0 g of water is cooled from 25.0°C to 12.7°C.
What is specific heat capacity?Specific heat capacity is defined as the amount of energy required to raise the temperature of one gram of substance by one degree Celsius. It has units of calories or joules per gram per degree Celsius.
It varies with temperature and is different for each state of matter. Water in the liquid form has the highest specific heat capacity among all common substances .Specific heat capacity of a substance is infinite as it undergoes phase transition ,it is highest for gases and can rise if the gas is allowed to expand.
It is given by the formula ,
Q=mcΔT, substitution in formula gives, Q=30×4.2×12.3=1549.8 joules.
Learn more about specific heat capacity,here:
https://brainly.com/question/29766819
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Which is not a compound? water silicon dioxide oxygen gas carbon dioxide gas
Answer: Oxygen gas
Explanation: Gradpoint