Answers
Answer: Option (b) is the correct answer.
Explanation:
Reduction is defined as the process in which there occurs gain of hydrogen. Whereas oxidation is defined as the process in which there occurs loss of hydrogen.
As the given reaction is as follows.
Since, hydrogen is being added in this chemical reaction. It means that reduction is taking place and carbon atom is reduced.
Thus, we can conclude that in the given reaction carbon atoms are reduced.
Related Questions
Round each answer to the number of significant figures in the parentheses.
A cubic box with sides of 20.0 cm contains 2.00 × 1023 molecules of helium with a root-mean-square speed (thermal speed) of 200 m/s. The mass of a helium molecule is 3.40 × 10-27 kg. What is the average pressure exerted by the molecules on the walls of the container? (The Boltzmann constant is 1.38 × 10-23 J/K and the ideal gas constant is R = 8.314 J/mol•K .) (12 pts.)
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Answers
Answer:
The mass per unit volume percentage is equal to 3.43%
Explanation:
given
mass=7 grams
volume = 2 * 102 ml = 204ml
The percentage mass per volume is given by
%(m/v) = w grams of solute * 100 / Volume of solution in ml
=
= 3.43%
Final answer:
The percent mass/volume (% m/v) for a solution containing 7.00 g of dextrose in 2.00×102 mL of solution is calculated as (7.00 g / 2.00×102 mL) * 100 = 3.47%, so the dextrose solution is 3.47% m/v.
Explanation:
The percent mass/volume (% m/v) is a way of expressing the concentration of a solute in a solution. It is calculated as the mass of the solute divided by the volume of the solution, multiplied by 100%. In this case, to calculate the % m/v for a solution that contains 7.00 g of dextrose in 2.00×102 mL of solution, you would use the following equation:
% m/v = (mass of solute / volume of solution) * 100
Substituting the given values into this equation, you get:
% m/v = (7.00 g / 2.00×102 mL) * 100 = 3.47%
Therefore, the dextrose solution is 3.47% m/v.
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b) All bases
c) all solvents
d) nonpolar solvents
Answers
Answer:
d) Non-polar solvents
True
False
Answers
Answers
Answer : The mass of sucrose added to 375 mL of water must be, 10.6 grams.
Explanation :
As we are given that 2.75 m/m percent solution of sucrose. That means, 2.75 grams of sucrose present in 100 grams of solution.
Mass of solution = 100 g
Mass of sucrose = 2.75 g
Mass of water = Mass of solution - Mass of sucrose
Mass of water = 100 g - 2.75 g
Mass of water = 97.25 g
First we have to calculate the mass of water.
Density of water = 1.00 g/mL
Volume of water = 375 mL
Now we have to calculate the mass of sucrose in 375 g of water.
As, 97.25 grams of water contain 2.75 grams of sucrose
So, 375 grams of water contain grams of sucrose
Therefore, the mass of sucrose added to 375 mL of water must be, 10.6 grams.
To make a 2.75% m/m sucrose solution, you need to add approximately 1062 grams of sucrose to 375 mL of water, considering the density of water as 1 g/mL.
To prepare a mass/mass (m/m) percent solution of sucrose, you need to calculate the mass of sucrose (in grams) that needs to be added to 375 mL of water to achieve a 2.75% concentration.
Here's how you can calculate it:
1. Convert the volume of water to grams, considering the density of water:
Density of water ≈ 1 g/mL
Mass of water = Volume of water × Density of water
Mass of water = 375 mL × 1 g/mL = 375 g
2. Determine the desired mass of sucrose as a percentage of the total mass:
Desired m/m percent = 2.75%
3. Calculate the mass of sucrose needed:
Mass of sucrose = (Desired m/m percent / 100) × Total mass
Mass of sucrose = (2.75 / 100) × (375 g + Mass of sucrose)
4. Rearrange the equation to solve for the mass of sucrose:
Mass of sucrose = (2.75 / 100) × (375 g) / (1 - (2.75 / 100))
Now, calculate:
Mass of sucrose = (2.75 / 100) × (375 g) / (1 - 0.0275)
Mass of sucrose ≈ (2.75 / 100) × (375 g) / 0.9725
Mass of sucrose ≈ (2.75 × 375 g) / 100 / 0.9725
Mass of sucrose ≈ (1031.25 g) / 0.9725
Mass of sucrose ≈ 1061.98 g
So, approximately 1062 grams of sucrose must be added to 375 mL of water to prepare a 2.75 m/m percent solution of sucrose.
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Answers
Answer:
Ea =22542.6
Explanation:
The rate constant k is affected by the temperature and this dependence may be represented by the Arrhenius equation:
where the pre-exponential factor A is assumed to be independent of temperature, R is the gas constant, and T the temperature in K. Taking the natural logarithm of this equation gives:
ln k = ln A - Ea/(RT)
or
ln k = -Ea/(RT) + constant
or
ln k = -(Ea/R)(1/T) + constant
These equations indicate that the plot of ln k vs. 1/T is a straight line, with a slope of -Ea/R. These equations provide the basis for the experimental determination of Ea.
now applying the above equation in the problem
we can write that
solve for Ea:
Ea = R[Ln(k2/k1)] / [(1/T1) - (1/T2)]
but k_2 = 2 k_1, hence:
Ea = (8.314 J/moleK)[ln(2)] / [(1/273+45) - (1/273+73)]
Ea =22542.6
Answer:
The activation energy for this reaction is 22.6 kJ/ mol
Explanation:
Step 1: Data given
Rate constant doubles when Temperature goes from 45.0 °C to 73.0 °C
R = 8.314 J/K*mol
Step 2: Calculate the activation energy
Log (k2/k1) = Ea / 2.303R *((1/T1) - (1/T2))
⇒ with k1 = initial rate constant
⇒ with k2 = rate constant after doubled = 2k1
⇒ T1 = initial temperature = 45.0 °C = 318 Kelvin
⇒ T2 = Final temperature = 73.0 °C = 346 Kelvin
log (2) = Ea / (2.303*8.314) *((1/318) - (1/346))
log(2) = Ea / (2.303*8.314) * 0.00025448
Ea = 22649 J/mol = 22.6 kJ/mol
The activation energy for this reaction is 22.6 kJ/ mol
Answers
Answer:
ROYGBIV or Roy G
Explanation:
Thanks for the points!