here's your answer..
Each of the following models represents particles present before and after a
chemical reaction which correctly shows conservation of mass during a
reaction?
Answer:
D
Explanation:
According to the law of conservation of mass, matter can neither be created nor destroyed. The implication of this is that the total mass of reactants should be equal to the total mass of products.
There should be equal masses of reactants and products on both sides of the reaction equation. This most important condition is only satisfied by option D.
in some applications nickel-cadmium batteries have been replaced by nickel-zinc batteries a single nickel-cadmium cell has a voltage of 1.30 V. Based on the idfference in the standard reduction potentials of CD2 and ZN2_, what votlage would you estimate a nickel-zinc a battery would produce
If the Bunsen burner gave a luminous flame and some soot was deposited on
the tube, what effect would this situation have on the calculated % of oxygen ?
Explanation:
If bunsen burner gave a luminous flame then there will deposition of soot at the bottom of the test tube which is actually pure carbon.
This deposition of soot actually depicts that there is incomplete combustion reaction that has taken place.
Also, the deposition of soot will provide a limited supply of oxygen to the reaction that has been calculated.
Therefore, in order to avoid any formation of soot it is advisable to adjust the burner flame till it produces a blue flame.
What is Hall-Heroult process?
Answer:
... Sorry I don't know the answer
What is the relationship between the magnitude of Δ (crystal-field splitting energy for an octahedral crystal field) and the energy of the d-d transition for a d1 complex?
a. The energy of the d-d transition is four times as big as Δ.
b. Δ is twice as big as the energy of the d-d transition.
c. Δ is four times as big as the energy of the d-d transition.
d. Δ is equal to the energy of the d-d transition.
e. The energy of the d-d transition is twice as big as Δ.
Answer:
B
Explanation:
The law of partial pressures was developed by ___________.
Answer:
John Dalton
Explanation:
) Dinitrogen Tetroxide partially decomposes according to the following equilibrium: N2O4 (g) 2NO2 (g) A 1.00-L flask is charged with 0.400 mol of N2O4. At equilibrium at 373 K, 0.0055 mol of N2O4 remains. Keq for this reaction is __________.
Answer: The value of [tex]K_{eq}[/tex] for this reaction is 1.578.
Explanation:
Given: Initial moles of [tex]N_{2}O_{4}[/tex] = 0.4 mol
Volume = 1.00 L
Therefore, initial concentration of [tex]N_{2}O_{4}[/tex] is calculated as follows.
[tex]Concentration = \frac{moles}{volume}\\= \frac{0.4}{1.0 L} mol\\= 0.4 M[/tex]
Now, ICE table for the given reaction equation is as follows.
[tex]N_{2}O_{4}(g) \rightleftharpoons 2NO_{2}[/tex]
Initial: 0.4 0
Change: -x +2x
Equib: 0.4 - x = 0.0055 2x
Hence, the value of x is calculated as follows.
0.4 - x = 0.0055
x = 0.4 - 0.0055
= 0.3945
Now, the [tex][NO_{2}][/tex] is calculated as follows.
2x = [tex][NO_{2}][/tex] = [tex]2 \times 0.3945 = 0.789[/tex]
Therefore, [tex]K_{eq}[/tex] for the given reaction is calculated as follows.
[tex]K_{eq} = \frac{[NO_{2}]^{2}}{[N_{2}O_{4}]}\\= \frac{(0.789)^{2}}{(0.3945)}\\= 1.578[/tex]
Thus, we can conclude that [tex]K_{eq}[/tex] for this reaction is 1.578.
The electrolysis of molten AlCl3 for 2.50 hr with an electrical current of 15.0 A produces ________ g of aluminum metal. Group of answer choices
Answer:
The correct answer is 12.58 grams.
Explanation:
Based on the given information, the electrolysis equation will be,
Al³⁺ + 3e⁻ ⇔ Al
1 mol of Al needs 3 moles of electron, and the value for 1 mole of electron is 96485 C.
Thus, 1 mole of Al needs 3 × 96485 C = 289455 C
Now the amount of charge passed is,
T = 2.5 hours
= 2.5 × 3600 s = 9 × 10³ s
Q = Current × Time
= 15A × 9 × 10³ s
= 13.5 × 10⁴ C
The moles of Al plated will be,
= 13.5 × 10⁴ / 289455
= 0.4664 mol
The molecular mass of Al is 26.98 grams per mole
Now the mass of Al will be,
= Number of moles × Molecular mass
= 0.4664 × 26.98
= 12.58 grams
Bryce and his lab partner come up with an idea they think will save time: We just used the fast titrations. You can stop the video when the solution turns pink. We know that when the solution turns pink, the titration is complete, so we just read the volume from the burette as soon as it turns pink. This is faster than going through all the shorter videos and works just as well. Do you agree with Bryce
Answer:
Yes.
Explanation:
Yes, I agree with Bryce and his lab partner because titration is completed when the solution changes its colour. Add chemical from the burette until the solution change its colour so then calculate the amount of chemical is used from the burette and the time at which the titration is completed. Always be careful for calculating the titration in order to get accurate data of the solution.
PLEASE HELP ME!!!!
TRUE or FALSE: When sperm and egg cells combine in fertilization, the
offspring ends up with the same number of chromosomes as their
parents.
Answer: False
Explanation:
Hope this help
Answer:
True.
Explanation:
Every child will contain the same number of chromosomes as the parents (otherwise they wouldn't be considered the same species). Additionally, animals can only mate with a species containing the same number of chromosomes as themselves. This means if the offspring of the parents had a different number fo chromosomes the offspring would be unable to mate with animals of it's own species.
What is the mass of 25.5 L of water assuming it is at 4 degrees Celsius?
Answer:
25.5kg or about 56.2lbs
Explanation:
At 4°C pure water has a mass of about 1 kg/liter.
So, if there was 25.5 L of water it would be 25.5kg or about 56.2lbs.
Choose the group that corresponds to each element.
Alkali Metal
Alkali earth metal
Halogen
Noble gas
The group corresponding to alkali metal, alkali earth metal, halogen, and noble gas is IA,IIA, 17, 18 group respectively.
Alkali metals have one outermost electron in their valence shell. They are kept under oil and are in the group IA of the periodic table.Alkaline earth metals have two outermost electrons in their valence shell. Examples are beryllium, magnesium, etc.Halogens have seven electrons in their outermost shells. They are in the periodic table in group 17 naming fluorine to iodine.Noble gas is the one that has fulfilled electronic configuration and doesn't react with any other compound.This is present in group 18 of periodic table.Learn more about alkali metals at:
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If the specific heat capacity of copper is 387 J/kg/°C, then how much energy is needed to raise the temperature of 400 g of copper from 30°C to 55°C?
Answer:
Explanation:
mass = 400 grams * [1 kg/1000 grams] = 0.400 kg
c = 387 Joules / (oC * kg)
Δt = 55 - 30 = 25 oC
E = m*c * Δt
E = 0.4 * 387 * 25
E = 3870 Joules
Which of these is an ion with a charge of 1+?
As the diagram model the processo se reproduction lomon) formand change formation Sexual reproduction results in a child that is different from its parents because
A. the child receives all of is genetic information from its mother
B. the child produces its own genetic information once it is formed
C. the child receives half of is genetic information from each parent
D. the child receives all of it's genetic information from it's father there
Answer:
The answer is C. the child receives half of is genetic information from each parent
Explanation:
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i. Complete the chemical reaction to show this. Make sure your equation is balanced. (2 points)
C6H12O6 + 02 →
This is the reaction of respiration. Hence, the complete reaction is:
C₆H₁₂O₆ + O₂ → 6CO₂ + 6H₂O
What is respiration?Respiration is defined as a metabolic process wherein the living cells of an organism obtain energy (in the form of ATP) by taking in oxygen and releasing carbon dioxide from the oxidation of complex organic compounds. There are two forms of respiration: aerobic and anaerobic.
Aerobic respiration occurs in the presence of oxygen, while anaerobic respiration occurs in the absence of oxygen.
Because it generates the energy required for body function, respiration is significant. It gives the cells oxygen and releases harmful carbon dioxide. Hence, the reaction of cellular respiration is:
C₆H₁₂O₆ + O₂ → 6CO₂ + 6H₂O
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4. Are there any solutions that have the measure of -3
? Briefly explain why.
H
CH
ll
C C
C001
CH H
2)
C C
1
NO ,
2
BY
Answer:
hakdog
Explanation:
pangitag lainbbbsjsbshsisbsjsbshxx
If the copper ion just GAINED electrons,
which process did it go through?
Reduction or
Oxidation
A 0.200 M solution of a week acid, HA, is 9.4% ionized. The molar concentration of H+ is 0.0188 M. the Acid-dissociation constant, Ka, for HA is...?
We are given:
Initial concentration of HA: 0.200 Molar
The acid is 9.4% ionized
Dissociation constant (α) = (Percent Ionized) / 100 = 0.094
Molar concentration of H+ = 0.0188
Let's Chill! (making the ICE box):
Reaction: HA ⇄ H⁺ + A⁻
Initial: 0.200M - -
Equilibrium: 0.200(1-α) 0.200α 0.200α
while we're here, let's confirm the given equilibrium concentration of H⁺ ions
from the table here, we can see that the equilibrium concentration of H⁺ ions is 0.200α, we know that α = 0.094
[H⁺] = 0.200α = 0.200 * 0.094 = 0.0188 M
which means that we're on the right track
We're basically scientists at this point (finding the dissociation constant):
Acid dissociation is nothing but the equilibrium constant, but for the dissociation of Acids
From the reaction above, we can write the equation of the acid dissociation constant:
Ka = [H⁺][A⁻] / [HA]
now, let's take the values from the 'equilibrium' row of the ice box the plug those in this equation
Ka = (0.200α)(0.200α) / [0.200(1-α)]
Ka = (0.200α)²/[0.200(1-α)]
plugging the value of α
Ka = (0.200*0.094)² / [0.200(0.906)]
Ka = (0.0188)² / 0.1812
Ka = 1.95 * 10⁻³
Calculate the volume of 0.07216 M AgNO3 needed to react exactly with 0.3572 g of pure Na2CO3 to produce solid Ag2CO3.
Answer:
93.4 mL
Explanation:
Let's state the reaction:
2AgNO₃ + Na₂CO₃ → Ag₂CO₃ + 2NaNO₃
We determine the moles of sodium carbonate:
0.3572 g . 1mol / 105.98g = 3.37×10⁻³ moles
Ratio is 1:2. We say:
1 mol of sodium carbonate react to 2 moles of silver nitrate
Then, our 3.37×10⁻³ moles of carbonate may react to: 3.37×10⁻³ . 2
= 6.74×10⁻³ moles
If we convert to mmoles → 6.74×10⁻³ mol . 1000 mmol / mol = 6.74 mmol
Molarity is mol/L but we can use mmol /mL
6.74 mol / volume in mL = 0.07216 M
6.74 mol / 0.07216 M = volume in mL → 93.4 mL
Please type out all of your calculations for this dilution equation: In your vitamin C experiment, it calls for a 5% concentration of iodine. However, your 7 fluid oz. tincture of iodine contains 70% iodine. How would you dilute this
Answer:
The original7 fluid oz. 70% tincture of iodine is diluted to 98 fluid oz. to obtain a 5% solution.
Explanation:
Dilution is a technique employed in experimental sciences such as chemistry and biochemistry as well as in medicine to obtain a less concentrated solution from a more concentrated one.
The dilution formula is the most important formula required in dilution. The dilution formula is given as: C1V1 = C2V2
Where C1 is the initial concentration of the stock solution;
V1 is the volume of the stock solution required;
C2 is the final concentration of the diluted solution to be prepared;
V2 is the final volume of the diluted solution.
Using the dilution formula to determine the answer to the above question:
C1 = 70% = 0.7; V1 = 7 fluid oz.; C2 = 5% = 0.05; V2 = ?
To determine V2, it is made subject of the dilution formula:
V2 = C1V1/C2
V2 = (0.7 × 7) / 0.05
V2 = 98 fluid oz.
Therefore, the original7 fluid oz. 70% tincture of iodine is diluted to 98 fluid oz. to obtain a 5% solution.
(1.21 x 10^-3 + 1.3 x 10^-3) x 6.453 x 10^2 =
Answer:
1.61
Explanation:
I need help with my chemistry
Answer:
Single displacement
Explanation:
Congratulations you have worked hard and now you are done with the year! I am so proud of you!
Answer:
lololol
Explanation:
how many liters of hydrogen gas I needed to react with CS2 to produce 2.5 L of CH4 At STP
Answer:
9.8 L
Explanation:
The reaction that takes place is:
4H₂(g) + CS₂(g) → CH₄(g)+ 2H₂S(g)At STP, 1 mol of any gas occupies 22.4 L.
We calculate how many moles are there in 2.5 L of CH₄ at STP:
2.5 L ÷ 22.4 L/mol = 0.11 mol CH₄Then we convert CH₄ moles into H₂ moles, using the stoichiometric coefficients of the reaction:
0.11 mol CH₄ * [tex]\frac{4molH_2}{1molCH_4}[/tex] = 0.44 mol H₂Finally we calculate the volume that 0.44 moles of H₂ would occupy at STP:
0.44 mol * 22.4 L/mol = 9.8 LWhich atom has the smallest atomic radius between cesium, potassium, rubidium, and francium?
10. What is the molarity of a solution containing 20 moles of NaCl dissolved in 10 liters of water?
0.5 mol/L
2 mol/L
5 mol/L
10 mol/L
Answer:
2 mol/L
Explanation:
M=Mol of solute/L. of solution
M=20mol/10L of H2O= 2mol/L
Answer:
2mol/L
Explanation:
got it correct on my quiz
compare and contrast the three types oflevers.
Answer:
The difference between the three classes depends on where the force is, where the fulcrum is and where the load is. In a first class lever, the fulcrum is located between the input force and output force. In a second class lever, the output force is between the fulcrum and the input force. write the class of lever.
Explanation:
What caused carbon dioxide to decrease in the air (abiotic matter) of the biodome?
Answer:
As organisms release energy during cellular respiration, carbon dioxide is produced from the carbon in energy storage molecules. ... Carbon dioxide in the biodome decreased because decomposers decreased which means there was a decrease in cellular respiration overall.
Explanation:
Carbon dioxide is "a colorless, odorless gas that is present in the atmosphere and is formed when any fuel containing carbon is burned".
What is atmosphere?Atmosphere is "a mixture of gases that includes about 78% nitrogen and 21% oxygen at Earth's surface".
What is fuel?Fuel is "a substance which upon combustion produces a usable amount of energy".
When the population of decomposer decreased, the amount of carbon dioxide in the biodome decreased drastically. The carbon dioxide that used to be in the air (abiotic matter) is now in another part of the ecosystem and the carbon remained constant.
Hence, Carbon dioxide decreased in the air (abiotic matter) of the biodome.
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