Answer:
none All have the same kinetic energy
Explanation:
According to the Arrhenius equation, changing which factors will affect the
rate constant?
A. Temperature and the ideal gas constant
B. The activation energy and the constant A
C. The constant A and the temperature
D. Temperature and activation energy
Answer:
e−(Ea/RT): the fraction of the molecules present in a gas which have energies equal to or in excess of activation energy at a particular temperature
Answer:
D. Temperature and activation energy is the correct answer
Explanation:
^_^
1 or 2 topics or two lessons should be explained in an illustrated childrens book minimum of 10 pages must have 3 or more sentences
Answer:
Yes because same topic are long
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
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 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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Question 16(Multiple Choice Worth 5 points)
(04.01 LC) Which statement is true about the total mass of the reactants during a chemical change?
O It is destroyed during chemical reaction.
O It is less than the total mass of the products. O It is equal to the total mass of the products.
O It is greater than the total mass of the products.
Answer:
It is equal to the total mass of the products.
Explanation:
Hope this helps :)
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
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]
an emerald can be described as...
Answer:
green gemstone
Explanation:
hope this helps someone
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.
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.
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
One main difference between the heating of gases on the one hand and solids or liquids on the other is that ___________________. One main difference between the heating of gases on the one hand and solids or liquids on the other is that ___________________. heating of gases depends not only on the temperature difference, but also on the process as well as the amount of gas present. heating of gases depends on temperature difference as well as the amount of gas present. specific heat is not defined for gases. heat cannot be exchanged with gases.
Answer:
heating of gases depends not only on the temperature difference, but also on the process as well as the amount of gas present.
Explanation:
The work done when a gas is heated does not only depends on the initial and final states of the gas but also on the process used to achieve the change of state of the gas.
Several processes can be applied in changing the state of a gas such as; adiabatic process, isobaric process, isochoric process and isothermal process.
Hence, the heating of a gas, depends not only on the temperature difference, as well as the amount of gas present according to the ideal gas laws but also on the process used to achieve the change of state.
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.
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]
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.
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
Please help me ASAP I’ll mark Brainly
Answer:
1. Vacuole
2. chloroplast
3. Nucleus
4. Plasma membrane - cell membrane
5. Vacuole (same as #1 ?) could be vesicle
Explanation:
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
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.
The data shows the number of years that 30 employees worked for an insurance company before retirement. is the population mean for the number of years worked, and % of the employees worked for the company for at least 10 years. (Round off your answers to the nearest integer.)
Answer:
14
73%
Explanation:
The mean Number of years worked :
. (sum of service years) / employees in the
(8+13+15+3+13+28+4+12+4+26+29+3+10+3+17+13+15+15+23+13+12+1+14+14+17+16+7+27+18+24) /
(417 / 30)
= 13.9 years
= 14 years
The percentage of employees who have worked for atleast 10 years :
Number of employees with service years ≥ 10 years = 22 employees
Total number of employees
Percentage (%) = (22 / 30= * 100% = 0.7333 * 100% = 73.33% = 73%
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
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
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.
Explain why caffeine can be extracted from the tea leaves into hot water and how you extracted the aqueous solution and isolated the crude caffeine.
Answer:
The hot water dissolves the flavor and color components.
Explanation:
Caffeine can be extracted from the tea leaves into hot water because the hot water dissolves the flavor and color components away from the solid vegetable. This is an example of a solid-liquid extraction. We can extracted the aqueous solution and isolated the crude caffeine by converting the components of caffeine into their calcium salts which are insoluble in water. Then the caffeine can be extracted from the water by using methylene chloride.
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.
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 sample of copper was heated at 275.1 C and placed into 272 g of water at 21.0 C. The temperature of the water rose at 29.7 C. How many grams of copper were in the sample
Answer:
104.8 g
Explanation:
From the question given above, the following data were obtained:
Initial temperature of copper (T꜀) = 275.1 °C
Mass of water (Mᵥᵥ) = 272 g
Initial temperature of water (Tᵥᵥ) = 21 °C
Equilibrium temperature (Tₑ) = 29.7 °C
Mass of copper (M꜀) =?
NOTE:
Specific heat capacity of copper (C꜀) = 0.385 J/gºC
Specific heat capacity of water (Cᵥᵥ) = 4.184 J/gºC
Finally, we shall determine the mass of the copper in the sample. This can be obtained as follow:
Heat loss by copper = Heat gained by water
M꜀C꜀(T꜀ – Tₑ) = MᵥᵥCᵥᵥ(Tₑ – Tᵥᵥ)
M꜀ × 0.385 (275.1 – 29.7) = 272 × 4.184(29.7 – 21)
M꜀ × 0.385 × 245.4 = 1138.048 × 8.7
M꜀ × 94.479 = 9901.0176
Divide both side by 94.479
M꜀ = 9901.0176 / 94.479
M꜀ = 104.8 g
Thus, the mass of the copper in the sample is 104.8 g
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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