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lions [1.4K]
4 years ago
7

What is the particle of an atom that moves rapidly in a cloud like region around the nucleus

Chemistry
1 answer:
lesya [120]4 years ago
4 0
Electron is the particle
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B. How do you think the physical properties (strength, flexibility, and viscosity) of the polymer would change if more borate ha
Talja [164]

It became thicker and its viscosity decreased and cannot flow as easily as before.

You ignite a chemical reaction by adding the borax solution to the glue mixture.

In a chemical reaction, the molecules of glue and borax combine to form a flexible, springy new substance. With rubber's vulcanization serving as a model, chemical cross-linking has been extensively employed to change the physical properties of polymeric materials.

Chemical links between polymer chains provide a substance with a more solid structure and perhaps a better-defined shape. It thickened and lost viscosity, making it more difficult to flow than it once could.

Learn more about the chemical reaction  here brainly.com/question/16714866

#SPJ4.

3 0
1 year ago
A 5 gram round ball has a density of 1.25 grams/milliliter. What is the volume of the round ball?
Arlecino [84]

Explanation:

m=5g

density=1.25g/ml

density=m/v

v=m/density=5/1.25

4 0
4 years ago
Please help me if your good with science:).
stellarik [79]
D an average train goes 300 km/h or 190-220 mph
5 0
3 years ago
Assuming that the experiments performed in the absence of inhibitors were conducted by adding 5 μl of a 2 mg/ml enzyme stock sol
prohojiy [21]

Hey there!:

From the given data ;

Reaction  volume = 1 mL , enzyme content = 10 ug ( 5 ug in 2 mg/mL )

Enzyme mol Wt = 45,000 , therefore [E]t is 10 ug/mL , this need to be express as "M" So:

[E]t in molar  = g/L * mol/g

[E]t  = 0.01 g/L * 1 / 45,000

[E]t = 2.22*10⁻⁷

Vmax = 0.758 umole/min/ per mL

= 758 mmole/L/min

=758000 mole/L/min => 758000 M

Therefore :

Kcat = Vmax/ [E]t

Kcat = 758000 / 2.2*10⁻⁷ M

Kcat = 3.41441 *10¹² / min

Kcat = 3.41441*10¹² / 60 per sec

Kcat = 5.7*10¹⁰ s⁻¹

Hence   kcat of   xyzase is  5.7*10¹⁰ s⁻¹


Hope that helps!



4 0
3 years ago
What pressure will be exerted by 25 g of CO2 at temperature of 25°C and a volume of .50 L?
Grace [21]

Answer:

P = 27.9 atm

Explanation:

Given data:

Mass of CO₂ = 25 g

Temperature = 25°C (25+273.15 K = 298.15 K)

Volume of gas = 0.50 L

Pressure of gas = ?

Solution:

Firs of all we will calculate the number of moles of gas,

Number of moles = mass/molar mass

Number of moles = 25 g/ 44 g/mol

Number of moles = 0.57 mol

Pressure of gas :

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

P × 0.50 L = 0.57 mol × 0.0821 atm.L/ mol.K  × 298.15 K

P = 13.95 atm.L/ 0.50 L

P = 27.9 atm

4 0
3 years ago
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