Answer:
[tex]m_{KMnO_4}=90.9gKMnO_4[/tex]
Explanation:
Hello there!
In this case, according to the given chemical equation for the reaction for the production of potassium permanganate, we can see a 2:2 mole ratio of this product to the starting manganese (II) oxide, which means, we can calculate the theoretical yield of the former via stoichiometry:
[tex]m_{KMnO_4}=50.0gMnO_2*\frac{1molMnO_2}{86.94gMnO_2}*\frac{2molKMnO_4}{2molMnO_2} *\frac{158.034gKMnO_4}{1molKMnO_4} \\\\m_{KMnO_4}=90.9gKMnO_4[/tex]
Regards!
A 66.4 gram sample of Ba(ClO4)2 3 H2O was heated thoroughly in a porcelain crucible, until its weight remained constant. After heating, how many grams of the anhydrous compound remained?
Answer:
57.2 g
Explanation:
First we convert 66.4 grams of Ba(ClO₄)₂·3H₂O into moles, using its molar mass:
Molar mass of Ba(ClO₄)₂·3H₂O = Molar mass of Ba(ClO₄)₂ + (Molar Mass of H₂O)*3Molar mass of Ba(ClO₄)₂·3H₂O = 390.23 g/mol66.4 g ÷ 390.23 g/mol = 0.170 mol Ba(ClO₄)₂·3H₂O0.170 moles of Ba(ClO₄)₂·3H₂O would produce 0.170 moles of 0.170 moles of Ba(ClO₄)₂. Meaning we now convert 0.170 moles of Ba(ClO₄)₂ into grams, using the molar mass of Ba(ClO₄)₂:
0.170 mol * 336.23 g/mol = 57.2 gThe citric acid cycle has a catabolic role, oxidizing acetate into CO2 and generating energy, and an anabolic role.
a. True
b. False
Use the periodic table to identify the noble gas that would be included in the noble-gas notation for each of the following elements.
Answer:
Explanation:
Use the periodic table to identify the noble gas that would be included in the noble-gas notation for each of the following elements.
Si:
Fr:
Hg:
V:
Answer:
Si: Ne
Fr: Rn
Hg: Xe
V: Ar
Explanation: trust me bro
For the reaction...
N2 + O2 <=> 2NO: AH = +182 kJ mol-1.
If the temperature is increased the equilibrium position will shift
Your answer:
a) to the left
b) to the right
c) to the left and right
d) neither left nor
right
Answer:
B
Explanation:
AH is positive so the forward reaction is endothermic. Thus, increasing temperature would cause equilibrium to shift to the right as endothermic reaction favors higher temperature. This increases the yield of NO.
Sodium reacts vigorously with water while potassium, the next element down in Group I, has a faster reaction rate. What would you expect of the rate of reaction of elements further down in Group I when exposed to water?
a. They would react with the same rate as potassium.
b. They would react much faster than sodium or potasssium.
c. They would react more slowly than potassium.
d. The rate of reaction would be unpredictable
Answer:
c. They would react more slowly than potassium. It's because sodium reacts slowly with water!
Explanation:
i am giving you the answer according to the doubtnut channel!
c. They would react more slowly than potassium.
Because their ability to displace hydrogen will be less.
*
The crystal structure of salt is
quimica espero que si sea corecta
which of these molecules is nonpolar?
Answer:
option b is your right answer
True or false? All producers are at the top of the food web
Answer:
false all producers are at the top of food web
What is the pH of 0.6 M NaOH?
Answer:
pOH = - log[OH-]
[OH-] = 0.6M
[tex]pOH \: = - log(0.6) \\ = 0.2218487496 \\ pH \: + pOH \: = 14 \\ pH \: + 0.221848749 = 14 \\ pH = 14 - 0.221848749 \\ = 13.77815125 \\ 13.8[/tex]
Rank the species below in order of decreasing leaving group capabilities in SN2 reactions (best leaving group to worst).
A. H2N-
B. CH3O-
C. C6H5SO3-
D. H2O
1. Best
2. Worst
Answer:
H2O> C6H5SO3- > CH3O- > H2N-
Explanation:
SN2 reaction is a reaction in which there is a synchronous departure of the leaving group and attachment of the incoming nucleophile in the transition state. Hence, an SN2 reaction passes through a single transition state. It is a bimolecular reaction.
A good leaving groups must be stable on its own. Usually, weak bases and neutral species are good leaving groups.
The order decreasing leaving group capabilities in SN2 reactions for the species listed in the question is;
H2O>C6H5SO3-> CH3O-> H2N-
Hence H2O is the best leaving group while H2N- is the worst leaving group.
The ranking from the best-leaving group to worst leaving group in SN2 reactions would be D. H₂O (water), C. C6H5SO3- (phenylsulfonate), B. CH₃O- (methoxide), A. H₂N- (amide).
In SN₂ reactions, leaving group capabilities are determined by the ability of the group to accept electrons and leave the molecule. The better the leaving group, the easier it is for it to leave the molecule. Based on this,
Water is a weak base and a good leaving group in SN2 reactions.
Phenylsulfonate is also a good leaving group, but it is slightly weaker than water.
Methoxide is a weaker base and leaving group compared to water and phenylsulfonate.
Amide is the weakest base and leaving group among the provided species.
To learn more about the SN₂ reactions, follow the link:
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all first level consumers are carnivore: True or false?
Drag the tiles to the correct boxes. Not all tiles will be used.
Match each Sl unit to the quantity it measures.
Answer:
1) Length - Meter
2) Mass - Pound
3) Time - Minute
Please Mark This As Brainliest!
When using vacuum filtration to separate a dissolved solid from an undissolved solid, what techniques should you use to ensure a quantitative separation
Answer: See explanation
Explanation:
Vacuum filtration is referred to as a fast filtration technique that is used in the separation of solids from liquids. It is also used to collect a desired solid. It basically uses a side-arm flask and a Buchner funnel.
Based on the question, the techniques that should be used to ensure a quantitative separation goes thus:
• Wet the filter paper before the mixture is poured into the filter funnel.
• Then, carefully rinse the flask with a little amount of water into the filter funnel.
• After that, the solid on the filter paper should be washed the with a small amount of water.
• Finally, Dry the solid on the filter paper when the separation is done.
A sample of Kr gas is observed to effuse through a porous barrier in 3.79 minutes. Under the same conditions, the same number of moles of an unknown gas requires 2.66 minutes to effuse through the same barrier. The molar mass of the unknown gas is________ g/mol.
Answer:
[tex]M'=41.3g/mol[/tex]
Explanation:
From the question we are told that:
Time Through first barrier [tex]T_1=3.79min[/tex]
Time Through second barrier [tex]T_2=2.66min[/tex]
Generally the Grahams Law equation for time of diffusion is mathematically given by
[tex]T=K\sqrt{M}[/tex]
Where
M=Mass Number of Kr
[tex]M=83.8[/tex]
Therefore
[tex]{T}{\sqrt{M}={T}{\sqrt{M'}[/tex]
[tex]{3.79}{\sqrt{83.8}={2.66}{\sqrt{M'}[/tex]
[tex]M'=0.645[/tex]
[tex]M'=41.3g/mol[/tex]
Calculate the number of moles
309 grams of (SF)4 = how many moles of (SF)4
Answer:
2.85 mol
Explanation:
Step 1: Given data
Mass of sulfur tetrafluoride (SF₄): 309 g
Step 2: Calculate the number of moles corresponding to 309 g of sulfur tetrafluoride
To convert mass to moles we need a conversion factor: the molar mass. The molar mass of SF₄ is 108.07 g/mol.
309 g × 1 mol/108.07 g = 2.85 mol
If 4.00 moles of O2 occupies a volume of 5.0 L at a particular temperature and pressure, what volume will 3.00 moles of oxygen gas occupy under the same condition?
Answer: Volume occupied by 3.00 moles of oxygen gas under the same condition is 3.75 L.
Explanation:
Given: [tex]n_{1}[/tex] = 4.00 moles, [tex]V_{1}[/tex] = 5.0 L
[tex]n_{2}[/tex] = 3.00 moles, [tex]V_{2}[/tex] = ?
Formula used is as follows.
[tex]\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}[/tex]
Substitute the values into above formula as follows.
[tex]\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}\\\frac{5.0 L}{4.00 mol} = \frac{V_{2}}{3.00 mol}\\V_{2} = 3.75 L[/tex]
Thus, we can conclude that volume occupied by 3.00 moles of oxygen gas under the same condition is 3.75 L.
After the student closes the cooler, there NOT any inputs or outputs from this system because ________. a. no energy is being transferred into or out of the cooler b. there is no flow of energy across the system boundary c. there is no matter flowing into or out of the system d. all of the above
Answer:
d. All of the above
Explanation:
d. All of the above
True or False
Helium belongs to Noble Metals
Answer:
This answer is "True"
PLEASE is possible to calculate the theoretical yield without a balanced equation?
No, you absolutely need a balanced equation because you need the coefficients to figure out the molar ratios plus you need the amount of one reactant or product to determine the theoretical yield.
No, it's not possible to calculate the theoretical yield without a balanced equation.
What is the theoretical yield?Theoretical yield is the yield that is calculated by the complete reaction of the limiting reagent. This is called expected or calculated amount of product.
[tex]\rm percent\; yield = \dfrac{(actual\; yield)}{(theoretical\; yield)} \times 100\\\\[/tex]
Thus, No, it's not possible to calculate the theoretical yield without a balanced equation.
Learn more about theoretical yield
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Energy is just like water, cycles through ecosystems.
True or false?
Answer:
true
Explanation:
I think
Which aqueous solution has the highest vapor pressure at 25ºC? 15.0g of sucrose (C12H22O11) in 100.0mL of water
The question is incomplete, the complete question is;
Which aqueous solution has the highest vapor pressure at 25ºC?
15.0g of glucose (C6H12O6) in 100.0mL of water
25.0g of glucose (C6H12O6) in 100.0mL of water
15.0g of sucrose (C12H22O11) in 100.0mL of water
25.0g of sucrose (C12H22O11) in 100.0mL of water
Answer:
15.0g of sucrose (C12H22O11) in 100.0mL of water
Explanation:
The vapor pressure of a substance is a colligative property. Colligative properties are the properties of a substance that depend on the amount of solute present.
Since vapour pressure is a colligative property, we have to look out for the solution that has the lowest number of moles because as more solute is dissolved in the solvent, the vapor pressure of the solvent decreases.
Hence, 15.0g of sucrose (C12H22O11) in 100.0mL of water has the lowest number of moles in solution thus it is expected to exhibit the highest vapour pressure.
H2C=CH2 + H2 решение?
Answer:
[tex]CH _{3}CH _{3}[/tex]
Ethane
Calculate the amount of heat associated with cooling a 350.0 g aluminum bar from 70.0 oC to 25.0 oC. The specific heat of aluminum is 0.897 J/g oC. (–14,127.75 J)
Explanation:
[tex]q = 350 \times0.897 \times (70 - 25) \\ q = 14127.75[/tex]
HELP PLEASE HURRY!!!
You have 12.51 grams of C2H6, how many moles do you have?
Answer:
How many grams C2H6 in 1 mol? The answer is 30.06904. We assume you are converting between grams C2H6 and mole. You can view more details on each measurement unit
In which two ways does a balanced chemical equation model the law of
conservation of mass?
Answer:Matter cannot be created or destroyed in chemical reactions. This is the law of conservation of mass. In every chemical reaction, the same mass of matter must end up in the products as started in the reactants. Balanced chemical equations show that mass is conserved in chemical reactions
Explanation:
sorry it is late but hope it helps!(o゜▽゜)o☆
Which expression gives the standard enthalpy
change of formation for methanol?
Answer:
25
Explanation:
bec
True or false: this reaction is balanced.
2C2H6(g) + 70215)
True
4CO2(g) + 6H20(3)
False
Answer:
false
Explanation:
i did it
for each 20 grams of glucose made by the plant, calculate the mass of water used
Answer: 6
Explanation:
You would need 6
What the correct answer
Answer:
[Ar] 4s²3d³
Explanation:
Vanadium has atomic number of 23. The electronic configuration of vanadium can be written as:
V (23) => 1s² 2s²2p⁶ 3s²3p⁶ 4s²3d³
NOTE: After the 18th electron, 4s will be filled before 3d.
We can also write the electronic configuration of an element in its condensed form by writing the symbol of the noble before the desired element in a squared bracket followed by the remaining electrons to complete the electronic configuration of the element.
The electronic configuration of vanadium in its condensed form is given below:
The noble gas before vanadium is Argon (Ar) with atomic number of 18. Thus, the electronic configuration of vanadium becomes:
V (23) => [Ar] 4s²3d³
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