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stiks02 [169]
2 years ago
8

Why can scratching at the bottom of a flask induce crystallization?

Chemistry
1 answer:
Ksenya-84 [330]2 years ago
7 0

Scratching causes cracks and crevices on the surface of the flask (though microscopically). These will act as favorable sites for nucleation, which leads to the formation of crystals.

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IM ON A TIMER! What are the main types of star clusters? Check all that apply.
jekas [21]

There are globular and open star clusters, but there are no binary, eclipsing, or wobbling ones.


8 0
3 years ago
Read 2 more answers
How does the electron-cloud model describe electrons?
Sav [38]

Answer:

C. An electron has a high probability of being in certain regions.

Explanation:

In the electron cloud model, there are no electron-orbits around the nucleus but a cloud. This cloud has various densities with respect to distance from the nucleus. The most dense region of the cloud (which is the region close to the nucleus) is where electrons has the highest probability of existence.

The model explains that an electron a greater chance of being in the region closer to the nucleus. Thus, an electron has a high probability of being in certain region of the cloud about the central nucleus. And an electrostatic force exists between the nucleus and the electrons.

7 0
2 years ago
How many grans of NaoH Are needed to neutralize grans of H2So4​
user100 [1]

Answer :]

to convert from g NaOH to mol NaOH. = 1.48 g NaOH are needed to neutralize the acid.

5 0
2 years ago
The isomerization of methylisonitrile to acetonitrileCH3NC(g)→CH3CN(g)is first order in CH3NC . The rate constant for the reacti
lisabon 2012 [21]

Answer:

Option E, Half life = 2.96\times 10^3\ s

Explanation:

For a first order reaction, rate constant and half-life is related as:

            t_{1/2}=\frac{0.693}{k}

Where,

t_{1/2} = Half life

k = Rate constant

Rate constant given = 2.34\times 10^{-4}\ s^{-1}

t_{1/2}=\frac{0.693}{k}

=\frac{0.693}{2.34 \times 10^{-4}}=2.96\times 10^3\ s

So, the correct option is option E.

4 0
3 years ago
A mouse is placed in a sealed chamber with air at 769.0 torr. This chamber is equipped with enough solid KOH to absorb any CO2 a
svlad2 [7]

<u>Answer:</u> The amount of oxygen gas consumed by mouse is 0.202 grams.

<u>Explanation:</u>

We are given:

Initial pressure of air = 769.0 torr

Final pressure of air = 717.1 torr

Pressure of oxygen = Pressure decreased = Initial pressure - Final pressure = (769.0 - 717.1) torr = 51.9 torr

To calculate the amount of oxygen gas consumed, we use the equation given by ideal gas which follows:

PV=nRT

where,

P = pressure of the gas = 51.9 torr

V = Volume of the gas = 2.20 L

T = Temperature of the gas = 292 K

R = Gas constant = 62.364\text{ L. Torr }mol^{-1}K^{-1}

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

51.9torr\times 2.20L=n\times 62.364\text{ L. Torr }mol^{-1}K^{-1}\times 292K\\\\n=\frac{51.9\times 2.20}{62.364\times 292}=0.0063mol

To calculate the mass from given number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of oxygen gas = 0.0063 moles

Molar mass of oxygen gas = 32 g/mol

Putting values in above equation, we get:

0.0063mol=\frac{\text{Mass of oxygen gas}}{32g/mol}\\\\\text{Mass of oxygen gas}=(0.0063mol\times 32g/mol)=0.202g

Hence, the amount of oxygen gas consumed by mouse is 0.202 grams.

5 0
3 years ago
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