compare the mass and volume of each object what is true of the mass and volume of all the floating objects

Reaction A has a high activation energy, whereas reacton B has a low activation energy. Which of the statements about reaction A

Alenkinab [10]

Answer:

The correct answer is:

Reaction B is more likely to occur at all than reaction A.

Explanation:

The activation energy in chemistry is the smallest amount of energy required to cause chemical or nuclear reaction in the reactants in chemical or nuclear systems. The activation energy is denoted by $E_{a}$ , and it is measured in Joules (J), KiloJoules (KJ) or Kilocalories per mole (Kcal/mol)

The activation energy can be thought of simply as the minimum amount of energy required to overcome a barrier that prevents a reaction from occurring, hence, from our question, if Reaction A has a high activation energy, it means that the barrier to be overcome before a reaction will occur is large, meaning that the reaction system is more stable and the reaction is less likely to occur than Reaction B which has a low activation energy, meaning that just a relatively small amount of energy, when applied to the reaction system, will initiate a reaction, making it more likely to occur than reaction A.

You should also note that catalysts are substances that are capable of reducing the activation energy of a system, but remains unchanged at the end of the system.

6 0

2 years ago

The first five ionization energies of an element are as follows (in kJ/mol): 577.9, 1820, 2750, 11600, 14800. Which of the follo

Ne4ueva [31]

Answer:option d==> Si.

Explanation:

The energy required to remove electron from a gaseous atom or ion is what is called an ionization energy. As we remove electrons continually in a gaseous atom or ion, the ionization energy increases which are know as the first ionization energy, the second ionization energy, third ionization energy and so on.

Looking at the electronic configuration of Silicon, Si; Ne 3s2 3p2. We can can see that the first four ionization energies are from the removal of the 3p2 and 3s2 electrons and the fifth ionization energy, which is the highest ionization energy of 14800 kJ/mol is the the electron removed from the core shell.

4 0

2 years ago

Calculate ΔH and ΔStot when two copper blocks, each of mass 10.0 kg, one at 100°C and the other at 0°C, are placed in contact in

ira [324]

Explanation:

The given data is as follows.

m = 10.0 kg = 10,000 g (as 1 kg = 1000 g)

Initial temp. of block 1, $T_{1} = 100^{o}C$ = (100 + 273) K = 373 K

Initial temp. of block 2, $T_{2} = 0^{o}C$ = (0 + 273) K = 273 K

So, heat released by block 1 = heat gained by block 2

$mC \Delta T = mC \times \Delta T$

$10000 g \times 0.385 \times (T_{f} - 100)^{o}C = 10000 g \times 0.385 \times (0 - T_{f})^{o}C$

$T_{f} - 100^{o}C = 0^{o}C - T_{f}$

$2T_{f} = 100^{o}C$

$T_{f} = 50^{o}C$

Convert temperature into kelvin as (50 + 273) K = 323 K.

Also, we know that the relation between enthalpy and temperature change is as follows.

$\Delta H = mC \Delta T$

= $10000 g \times 0.385 J/K g \times 323 K$

= 1243550 J

or, = 1243.5 kJ

Now, calculate entropy change for block 1 as follows.

$\Delta S_{1} = mC ln \frac{T_{f}}{T_{i}}$

= $10000 g \times 0.385 J/K g \times ln \frac{323}{373}$

= $10000 g \times 0.385 J/K g \times -0.143$

= -554.12 J/K

Now, entropy change for block 2 is as follows.

$\Delta S_{2} = mC ln \frac{T_{f}}{T_{i}}$

= $10000 g \times 0.385 J/K g \times ln \frac{323}{273}$

= $10000 g \times 0.385 J/K g \times 0.168$

= 647.49 J/K

Hence, total entropy will be sum of entropy change of both the blocks.

$\Delta S_{total} = \Delta S_{1} + \Delta S_{2}$

= -554.12 J/K + 647.49 J/K

= 93.37 J/K

Thus, we can conclude that for the given reaction $\Delta H$ is 1243.5 kJ and $\Delta S_{total}$ is 93.37 J/K.

6 0

2 years ago

If you compared 1 m solutions, was a 1 m nacl solution more or less hypertonic than a 1 m sucrose solution? what is your evidenc

igor_vitrenko [27]

When you say the solution is hypertonic, it means that the solution has a higher osmotic pressure. The formula for this is:

P = iMRT,
for strong electrolytes, i = number of ions.
for nonelectrolytes, i = 1

1. The P for sucrose solution which is a nonelectrolyte (assuming room temp):
P = (1)(1m)(8.314 J/mol-K)(298 K)
P = 2477.572 Pa

The P for NaCl solution, which is a strong electrolyte:
P = (2)(1 m)(8.314)(298 K)
P = 4955.144 Pa

<em>So, that means that NaCl is more hypertonic than the sucrose solution.</em>

2. For the second question, the P for the combination of 1 m glucose (nonelectrolyte) and 1 m sucrose is:
P = (1)(1 m)(8.314)(298 K) + (1)(1)(8.314)(298 K) = 4955.144 Pa
<em>In this case, the osmotic pressures are now equal. It is not hypertonic, but isotonic.</em>

7 0

2 years ago

A 45 mL sample of nitrogen gas is cooled from 135ºC to 15C in a container that can contract or expand at constant pressure. Wha

Vanyuwa [196]

Answer:

V₂ =31.8 mL

Explanation:

Given data:

Initial volume of gas = 45 mL

Initial temperature = 135°C (135+273 =408 K)

Final temperature = 15°C (15+273 =288 K)

Final volume of gas = ?

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 45 mL × 288 K / 408 k

V₂ = 12960 mL.K / 408 K

V₂ =31.8 mL

8 0

2 years ago