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1 year ago

gelified acetate-based water-in-salt electrolyte stabilizing hexacyanoferrate cathode for aqueous potassium-ion batteries

1 answer:

Nookie1986 [14]1 year ago

4 0

In recent years, gelified acetate-based water-in-salt electrolyte stabilising hexacyanoferrate cathode for aqueous potassium-ion batteries has gained popularity.

Here, water activity in "Water-in-salt" electrolytes (WiSE) based on potassium acetate (KAc) is investigated using Raman and classical molecular dynamics (MD), demonstrating the significantly reduced water activity of these WiSEs because water can be efficiently coordinated by both the acetate anion and the potassium cation. It is discovered that the overall molecular arrangement resembles the "sponge-like" shape seen in several ionic liquids. According to comprehensive electrochemical, scanning electronic microscopy (SEM), and X-ray photoelectron spectroscopy (XPS) study, such WiSE can also be compatible with Al current collectors with appropriately adjusted composition. However, due to the limited stability of the cathode material in the alkaline environment, the use of inexpensive potassium manganese hexacyanoferrate (KMHCF) in combination with these electrolytes is hampered by poor cycling performance.

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An empty graduated cylinder weighs 55.26 g. When filled with 48.1 mL of an unknown liquid, it weighs 92.39 g. The density of the

katrin [286]

<h2>Density of the unknown liquid is 771.93 kg/m³</h2>

Explanation:

An empty graduated cylinder weighs 55.26 g

Weight of empty cylinder = 55.26 g = 0.05526 kg

Volume of liquid filled = 48.1 mL = 48.1 x 10⁻⁶ m³

Weight of cylinder plus liquid = 92.39 g = 0.09239 kg

Weight of liquid = 0.09239 - 0.05526

Weight of liquid = 0.03713 kg

We have

Mass = Volume x Density

0.03713 = 48.1 x 10⁻⁶ x Density

Density = 771.93 kg/m³

Density of the unknown liquid is 771.93 kg/m³

5 0

3 years ago

Seeing a police officer ahead and not being sure of the speed limit, a driver slows from 65 mi/hr to 55 mi/hr in 4.0 seconds, tr

Flauer [41]

Answer:

$-1.13 m/s^2$

Explanation:

The acceleration of the car is given by:

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time

Here we have to convert the velocities from mi/hr to m/s. Keeping in mind that

1 mi = 1609 m

1 h = 3600 s

We get:

$u = 65 mi/h \cdot \frac{1609}{3600}=29.1 m/s\\v = 55 mi/h \cdot \frac{1609}{3600}=24.6 m/s$

While the time interval is

t = 4.0 s

Substituting,

$a=\frac{24.6 - 29.1}{4.0}=-1.13 m/s^2$

3 0

3 years ago

A block of mass 3.20 kg is placed against a horizontal spring of constant k = 865 N/m and pushed so the spring compresses by 0.0

Aliun [14]

Answer:

a) The initial elastic potential energy of the block-spring system is 28.113 joules.

b) The final speed of the block is approximately 4.192 meters per second.

Explanation:

a) By applying Hooke's law and definition of work, we define the elastic potential energy ( $U_{g}$ ), measured in joules, by the following formula:

$U_{g} = \frac{1}{2}\cdot k\cdot x^{2}$ (1)

Where:

$k$ - Spring constant, measured in newtons per meter.

$x$ - Deformation of the spring, measured in meters.

If we know that $k = 865\,\frac{N}{m}$ and $x = 0.065\,m$ , then the elastic potential energy is:

$U_{g} = \frac{1}{2}\cdot \left(865\,\frac{N}{m} \right) \cdot (0.065\,m)$

$U_{g} = 28.113\,J$

The initial elastic potential energy of the block-spring system is 28.113 joules.

b) According to the Principle of Energy Conservation, the initial elastic potential energy of the block-spring system becomes into translational kinetic energy, that is:

$U_{g} = \frac{1}{2}\cdot m\cdot v^{2}$ (2)