Answer:
oshfjidgshsjdh
Explanation:
918474828
Calculate the moment of inertia of a CH³⁵CL₃ molecule around a rotational axis that contains the C-H bond. The C-Cl bond length is 177pm and the HCCl angle is 107⁰f
Answer:
The correct answer is "[tex]4.991\times 10^{-45} \ kg.m^2[/tex]".
Explanation:
According to the question,
[tex]R_{C-Cl} = 177 \ pm[/tex]
or,
[tex]=1.77\times 10^{-10} \ m[/tex]
[tex]\alpha = 107^{\circ}[/tex]
[tex]m_{Cl}=34.97 \ m.u[/tex]
or,
[tex]=34.97\times 1.66\times 10^{-27}[/tex]
[tex]=5.807\times 10^{-26} \ kg[/tex]
The moment of inertia around the rotational axis will be:
⇒ [tex]I=3\times m_{Cl}\times (R_{C-Cl})^2 \ Sin^2 \alpha[/tex]
By putting the values, we get
[tex]=3\times 5.807\times 10^{-26}\times (1.77\times 10^{-10})^2 \ Sin^2 (107)[/tex]
[tex]=3\times 5.807\times 10^{-26}\times (1.77\times 10^{-10})^2\times 0.91452[/tex]
[tex]=4.991\times 10^{-45} \ kg.m^2[/tex]
What is bond energy
Which substrate is used in the last step of glycolysis
Pyruvate Kinase
Pyruvate Kinase performs a substrate level phosphorylation on ADP to generate an ATP and pyruvate, the final product of glycolysis.
PK dificiency is transmitted in an autosomal recessive disorder in which both alleles must contain the mutated gene, PK-LR.
Hope it helps you! ＼(^ᴥ^)／
Please help me ASAP I’ll mark Brainly
Answer:
cell
chloroplast and cell wall
nucleus
life processes
cell membrane
shape and size
vacuole
Hope it helps
If 11g of a gas occupies 5.6dm'3at s.t.p., calculate it's vapour density (1.0mol of a gas occupies 22.4dm'3at s.t.p.).
Answer:
[tex]{ \boxed{ \bf{vapour \: density = 2 \times molecular \: mass}}} \\{ \tt{ PV= (\frac{m}{ m_{r}}) RT}} \\ { \tt{3 \times 5.6 = \frac{11}{m _{r}} \times 0.0831 \times 273}} \\ { \tt{m _{r} = 14.85 \: g}} \\ \\ { \bf{vapour \: density = 2 \times m _{r}}} \\ = 2 \times 14.85 \\ = 29.7 \: { \tt{g {dm}^{ - 3} }}[/tex]
Question 1 Points 3 23 and Louis immerses his left hand in a beaker containing cold water and immerses his right hand in a beaker containing warm water. Then, he immerses both his hands on a beaker containing water at room temperature. Which of the following statements are true? 1. The hand that was in hot water would feel cold. 2. The hand that was in cold water would feel hot. 3. His two hands will feel the same hotness. Que O2 and 3 0 1 and 2 o 1 and 3 1.2, and 3
Answer:look down below
Explanation:
The statements that are true about hands that are immersed in the water are:
1. The hand that was in hot water would feel cold.
2. The hand that was in cold water would feel hot.
The correct option is B 1. and 2.
What is temperature?
Temperature is the measurement of the hotness or coldness of any object. It is measured in Celsius or kelvin. Our body has nerves that feel the different temperatures of any object. The high temperature is called hot and the low temperature is called cold.
When Louis put his hand in the warm water and one hand in the cold water. He feels the temperature of both glasses of water. Then he put both hands in the normal water.
So the hand that is warm would feel the water as cold and the hand with cold water would feel the water as hot.
Thus, the correct option is B. 1. and 2.
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the ability of organism to sense changes in its body is an example of
Answer:
the ability of organism to sense changes in its body is an example of responsiveness.Hope it is helpful to you
A sample of gas is placed into an enclosed cylinder and fitted with a movable piston. Calculate the work (in joules) done by the gas if it expands from 5.33 L to 11.05 L against a pressure of 1.50 atm.
Explanation:
here is the answer. Feel free to ask for more chem help
Based on the equations below, which metal is the least active? Pb(NO3)2(aq) + Ni (s) --> Ni(NO3)2 (aq)+ Pb(s) Pb(NO3)2(aq) + Ag(s) --> No reaction Cu(
Answer:
Ni
Explanation:
An active metal is a highly reactive metal. Active metals are found high up in the activity series.
Active metals react with other metals that are lower than them in the activity thereby displacing the lower metals from a solution of their salts. This is what may have happened in the other two reactions.
Ni is the most active metal listed in the question since it can react a compounds with Pb(NO3)2(aq) to liberate Pb metal.
an emerald can be described as...
Answer:
green gemstone
Explanation:
hope this helps someone
When electrons in a molecule are not found between a pair of atoms but move throughout the molecule, this is called Group of answer choices
Answer:
delocalised electrons
Explanation:
they are called delocalised electrons because that can move freely in the molecule
What is the percent yield of the reaction below if 84.0 grams of Al2O3(s) is recovered from a reaction whose theoretical yield of Al2O3(s) is 104 grams?
4 Al(s) + 3 O2(g) → 2 Al2O3(s)
Answer:
80.8%
Explanation:
Let's consider the following balanced equation.
4 Al(s) + 3 O₂(g) → 2 Al₂O₃(s)
The mass obtained of Al₂O₃ (experimental yield) is 84.0 g. The theoretical yield of Al₂O₃ is 104 g. We can calculate the percent yield of Al₂O₃ using the following expression.
%yield = (experimental yield / theoretical yield) × 100%
%yield = (84.0 g / 104 g) × 100% = 80.8%
Answer:
Percent Yield = 80.8%
Explanation:
We can find the percent yield of a reaction using the equation:
Percent yield = Actual yield (g) / Theoretical Yield (g) * 100
Where Actual yield is the amount of product produced (84.0g)
And theoretical yield is the mass produced assuming a 100% of product (104.0g)
Replacing the computed values:
Percent yield = 84.0g / 104.0g * 100
Percent Yield = 80.8%
I have an unknown volume of gas held at a temperature of 115 K in a container with a pressure of 60atm. If by increasing the temperature to 225 K and decreasing the pressure to 30. atm causes the volume of the gas to be 29 liters, how many liters of gas did I start with?
SHOW YOUR WORK
Explanation:
here is the answer to your question.
Plz help me ASAP in my final project I am ready to pay 20$
Answer:
what do you need help with
c) Solar energy is the source of all forms of energy.give reasons
Answer:
All energy is made by the sun because without the sun there would be no humans to produce other energy
Explanation:
We use many different forms of energy here on earth, but here’s the thing: almost all of them originate with the sun, not just light and heat (thermal) energy! The law of conservation of energy says that energy can’t be created or destroyed, but can change its form. And that’s what happens with energy from the sun—it changes into lots of different forms:
Plants convert light energy from the sun into chemical energy (food) by the process of photosynthesis. Animals eat plants and use that same chemical energy for all their activities.
Heat energy from the sun causes changing weather patterns that produce wind. Wind turbines then convert wind power into electrical energy.
Hydroelectricity is electrical energy produced from moving water, and water flows because heat energy from the sun causes evaporation that keeps water moving through the water cycle.
Right now, much human activity uses energy from fossil fuels such as coal, oil, and natural gas. These energy sources are created over very long periods of time from decayed and fossilized living matter (animals and plants), and the energy in that living matter originally came from the sun through photosynthesis.
solar panel shows what is the ultimate source of energy
if a bottle of vinegar has 4.78g of acetic acid (CH3COOH) per 100.0 g of solution (mixed with water, what is the molarity of the vinegar? Density of the solution is 1.00g/mL.
Answer:
0.796 M
Explanation:
Step 1: Given data
Gravimetric concentration (Cg): 4.78 g%g
Density of the solution (ρ): 1.00 g/mL
Step 2: Calculate the volumetric concentration of the solution (Cv)
We will use the following expression.
Cv = Cg × ρ
Cv = 4.78 g%g × 1.00 g/mL = 4.78 g%mL
Step 3: Calculate the molarity of the solution (M)
The volumetric concentration is 4.78 g%mL, that is, there are 4.78 g of acetic acid per 100 mL of solution. We can calculate the molarity using the following expression.
M = mass solute / molar mass solute × liters of solution
M = 4.78 g / 60.05 g/mol × 0.1 L = 0.796 M
4019K Express your answer as a nuclear equation.
Answer:
[tex]_{19} ^{40} K[/tex] → [tex]_{20} ^{40} Ca[/tex] + [tex]_{-1} ^{0} e[/tex] + ¯v
Explanation:
[tex]_{19} ^{40} K[/tex] is an unstable potassium isotope which has the capacity to undergo beta decay. During the decay process, it transforms to calcium atom with a release of beta particle and an antineutrino.
The required nuclear equation is:
[tex]_{19} ^{40} K[/tex] → [tex]_{20} ^{40} Ca[/tex] + [tex]_{-1} ^{0} e[/tex] + ¯v
The calcium formed due to this process is a stable atom.
In the reactionpyruvate lactatethat is catalyzed by the enzyme lactate dehydrogenase, the compound on which the enzyme works, pyruvate, is called the _______.
Answer:
Substrate
Explanation:
In biochemical sciences, a substrate is a substance that is acted upon by an enzyme to yield a product. Enzymes are known for catalyzing biochemical reactions. The substances that are usually worked with during this catalytic process are termed as SUBSTRATES.
Substrates, which are usually changed during the process, binds to the active site on the enzyme and form an enzyme-substrate complex.
According to this question, pyruvate is converted to lactate in a reaction that is catalyzed by the enzyme lactate dehydrogenase. This means that the compound on which the enzyme works, pyruvate, is called the SUBSTRATE.
A
(c) 2 C(s) + MnO2(s)
Mn(s) + 2 CO(g)
O combination reaction
O decomposition reaction
O combustion reaction
O single-displacement reaction
Answer: The reaction, [tex]2C(s) + MnO_{2}(s) \rightarrow Mn(s) + 2CO(g)[/tex] is a single-displacement reaction.
Explanation:
A chemical reaction in which one element of a compound is replaced by another element participating in the reaction.
For example, [tex]2C(s) + MnO_{2}(s) \rightarrow Mn(s) + 2CO(g)[/tex]
Here, the element manganese is replaced by carbon atom. As only one element gets replaced so, it is a single-displacement reaction.
Thus, we can conclude that [tex]2C(s) + MnO_{2}(s) \rightarrow Mn(s) + 2CO(g)[/tex] is a single-displacement reaction.
I need help solving this!
For the reaction C + 2H2 → CH4, how many moles of hydrogen are needed to make 146.6 grams of methane, CH4 ?
Round your answer to the nearest tenth. If you answer is a whole number like 4, report the answer as 4.0
Use the following molar masses. If you do not use these masses, the computer will mark your answer incorrect.:
Element
Molar Mass
Hydrogen
1
Carbon
12
Answer: Moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, [tex]CH_{4}[/tex].
Explanation:
Given: Mass of methane = 146.6 g
As moles is the mass of a substance divided by its molar mass. So, moles of methane (molar mass = 16.04 g/mol) are calculated as follows.
[tex]Moles = \frac{mass}{molar mass}\\= \frac{146.6 g}{16.04 g/mol}\\= 9.14 mol[/tex]
The given reaction equation is as follows.
[tex]C + 2H_{2} \rightarrow CH_{4}[/tex]
This shows that 2 moles of hydrogen gives 1 mole of methane. Hence, moles of hydrogen required to form 9.14 moles of methane is as follows.
[tex]Moles of H_{2} = \frac{9.14}{2}\\= 4.57 mol[/tex]
Thus, we can conclude that moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, [tex]CH_{4}[/tex].
Chloride ion is a strong nucleophile and bromide is a good leaving group. The major product of treating (S)-2-bromobutane with NaCl in CH3C(O)CH3 (acetone) is _________. (1S,2R)-1-chloro-2-bromobutane cis-2-butene (1R,2S)-1-chloro-2-bromobutane (S)-2-chlorobutane trans-2-butene (R)-2-chlorobutane
Answer:
Chloride ion is a strong nucleophile and bromide is a good leaving group. The major product of treating (S)-2-bromobutane with NaCl in CH3C(O)CH3 (acetone) is _________. (1S,2R)-1-chloro-2-bromobutane cis-2-butene (1R,2S)-1-chloro-2-bromobutane (S)-2-chlorobutane trans-2-butene (R)-2-chlorobutane
Explanation:
The reaction of (S)-2-bromobutane with NaCl in CH3C(O)CH3 (acetone) forms the following product:
The answer is (R)-2-chlorobutane.
The reaction take splace through [tex]S_{N} _2[/tex] mechansim and inversion in configuration happens.
Classify each of the following as either macroscopic, microscopic or particulate:
a. a red blood cell.
b. a sugar molecule.
c. baking powder.
Answer:
Classify each of the following as either macroscopic, microscopic or particulate:
a. a red blood cell.
b. a sugar molecule.
c. baking powder.
Explanation:
a. A red blood cell is a microscopic particle.
It can be viewed under a microscope.
b. A sugar molecule is also a microscopic substance.
It can be viewed under a microscope.
c. Baking powder is macroscopic substance.
Predict the products of below reaction, and whether the solution at equilibrium will be acidic, basic, or neutral.
N2O5 + 3H2O → __________
Answer: The product of the given reaction is [tex]HNO_{3}[/tex] and the solution at equilibrium will be acidic.
Explanation:
When two or more chemical substances react together then it forms new substances and these new substances are called products.
For example, [tex]3N_{2}O_{5} + 3H_{2}O \rightarrow 6HNO_{3}[/tex]
This shows that nitric acid [tex](HNO_{3})[/tex] is the product formed and it is an acidic substance.
Hence, the solution at equilibrium will be acidic in nature.
Thus, we can conclude that the product of the given reaction is [tex]HNO_{3}[/tex] and the solution at equilibrium will be acidic.
The mass of a single tantalum atom is 3.01×10-22 grams. How many tantalum atoms would there be in 37.1 milligrams of tantalum?
Answer: There are [tex]1.23 \times 10^{22}[/tex] atoms present in 37.1 mg of tantalum.
Explanation:
Given: Mass of single tantalum atom = [tex]3.01 \times 10^{-22} g[/tex]
Mass of tantalum atoms = 37.1 mg (1 mg = 0.001 g) = 0.0371 g
Therefore, number of tantalum atoms present in 0.0371 grams is calculated as follows.
[tex]No. of atoms = \frac{0.0371 g}{3.01 \times 10^{-22}}\\= 1.23 \times 10^{20}[/tex]
Thus, we can conclude that there are [tex]1.23 \times 10^{22}[/tex] atoms present in 37.1 mg of tantalum.
There are [tex]1.23\times 10^{20}[/tex] atoms of tantalum in 37.1 mg of tantalum.
Explanation:
Given:
Mass of single atom of tantalum =[tex]3.01\times 10^{-22} g[/tex]
To find:
The number of atoms of tantalum in 37.1 milligrams.
Solution:
Mass of tantalum = 37.1 mg
[tex]1 mg = 0.001 g\\37.1 mg=37.1\times 0.001 g=0.0371 g[/tex]
The number of atoms in 0.0371 grams of tantalum = N
Mass of a single atom of tantalum = [tex]3.01\times 10^{-22} g[/tex]
Then a mass of N atoms of tantalum will be:
[tex]0.0371 g=N\times 3.01\times 10^{-22} g\\N=\frac{0.0371 g}{ 3.01\times 10^{-22} g}\\=1.23\times 10^{20}[/tex]
There are [tex]1.23\times 10^{20}[/tex] atoms of tantalum in 37.1 mg of tantalum.
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A buffer is prepared containing 0.75 M NH3 and 0.20 M NH4 . Calculate the pH of the buffer using the Kb for NH3. g
Answer:
pH=8.676
Explanation:
Given:
0.75 M [tex]NH_{3}[/tex]
0.20 M [tex]NH_{4}[/tex]
The objective is to calculate the pH of the buffer using the kb for [tex]NH_3[/tex]
Formula used:
[tex]pOH=pka+log\frac{[salt]}{[base]}\\[/tex]
pH=14-pOH
Solution:
On substituting salt=0.75 and base=0.20 in the formula
[tex]pOH=-log(1.77*10^{-5})+log\frac{0.75}{0.20}\\ =4.75+0.5740\\ =5.324[/tex]
pH=14-pOH
On substituting the pOH value in the above expression,
pH=14-5.324
Therefore,
pH=8.676
A frozen TV dinner contains 21 g of protein, 59 g of carbohydrate, and 18 g of fat. What is the total number of kilojoules (kJ) of potential energy within this TV
dinner? The accepted values for potential energy are 17 kJ per gram of protein, 17 kJ per gramof carbohydrate, and 38 kJ per gram of fat.
Round your answer to the nearest tens place and with the appropriate units.
Answer:
2040 kJ
Explanation:
Step 1: Calculate the energy provided by 21 g of protein
17 kJ are provided per gram of protein.
21 g × 17 kJ/g = 357 kJ
Step 2: Calculate the energy provided by 59 g of carbohydrate
17 kJ are provided per gram of carbohydrate.
59 g × 17 kJ/g = 1003 kJ
Step 3: Calculate the energy provided by 18 g of fat
38 kJ are provided per gram of fat.
18 g × 38 kJ/g = 684 kJ
Step 4: Calculate the total energy provided by the dinner
357 kJ + 1003 kJ + 684 kJ = 2044 kJ ≈ 2040 kJ
The doctor has ordered Claforan 1 g in 100 ml D5W to run IV piggyback for 30 minutes twice daily. The pharmacy sends Claforn 2 g in a powdered form, which when reconstituted has a concentration of 180 mg Claforan per ml. How much Claforn will you add to the bag of D5W
Answer:
0.111 g
Explanation:
1 g = 1000 mg
Doctor ordered the following concentration of Claforan:
C = 1 g/100 mL x 1000 mg/1 g = 10 mg/mL
If we add 2 g iof Claforan, we obtain:
2 g Claforn ---- 180 mg/mL Claforan
To reach a concentration equal to C (10 mg/mL), we need:
10 mg/mL Claforan x 2 g Claforn/(180 mg/mL Claforan) = 0.111 g Claforn
Therefore, we have to add 0,111 g (111 mg) of Claforn to the bag of 100 ml D5W to obtain the ordered concentration of 10 mg/mL Claforan.
A solution of hydrochloric acid had a hydrogen ion concentration of 1.0 mol/dm3
Water was added to hydrochloric acid until the ph increased by 1
What was the hydrogen ion concentration of the hydrochloric acid after had been added?
Answer:
pH = -log[H+]
Where [H+] = Hydrogen ion concentration
In this case,
[H+] = 1 × 10^(-2) = 10^(-2)
log{10^(-2)} = -2
-log{10^(-2)} = -(-2) = 2
pH = -log{10^(-2)} = 2
and hi.!!!
Answer:
0.1
Explanation:
Hydrogen ion concentration can be calculated using the formula [H+] = 10^-pH
pH can be concentrated using ph = -log[H+]
let's calculate the initial pH before anything was added: pH = -log(1) = 0
it increased by 1 so the final pH is 1.
Now we'll find the [H+] of a solution with a pH of 1:
concentration = 10^(-1) = 0.1
A 1.5-m 3 insulated rigid tank contains 2.7 kg of carbon dioxide at 100 kPa. Now paddle-wheel work is done on the system until the pressure in the tank rises to 150 kPa. Determine the entropy change of carbon dioxide during this process. Assume constant specific heats
Answer:
The entropy change of carbon dioxide = 0.719 kJ/k
Explanation:
Given:
1.5 m - 3 insulated rigid tank contains 2.7 kg of carbon dioxide at 100 kPa
The objective is to determine the entropy change of carbon dioxide
Formula used:
ΔS=
Solution:
On considering,
[tex]C_{P} =0.846 kJ/kg K\\C_V=0.657 kJ/kg k\\[/tex]
ΔS=[tex]mc_{v} lu\frac{p_{2} }{P_{1} }[/tex]
On substituting the values,
ΔS=[tex]2.7*0.657lu\frac{150}{100}[/tex]
ΔS=0.719 kJ/k
The entropy change is "0.719 kJ/K".
Given values are:
Mass of tank,
m = 2.7 kgPressure,
P₁ = 100 kPaRised pressure,
P₂ = 150 kPaAssumption of constant specific heat is,
[tex]C_v = 0.657 \ kJ/kgK[/tex]As we know the formula,
→ [tex]\Delta S = mC_v \ ln(\frac{P_2}{P_1} )[/tex]
[tex]= (2.7)(0.657) \ ln (\frac{150}{100} )[/tex]
[tex]= 1.7739\times 0.4055[/tex]
[tex]= 0.7193 \ kJ/K[/tex]
Thus above answer is right.
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howtocalculatethevolumeofcarbondioxideproducedwhen400gofmarblewereats.t.p
Answer:
so 0.15 moles X 22.4 dm3/mole=3.36 dm3. Next we find the moles of hexane combusted, and then the moles of CO2. Finally, we find the volume of CO2 using the fact that at STP, 1 mole of gas = 22.4 dm3.
