Arrange these elements based on their atomic radii. ga, f, s, as

4. Positive electric charges are always attracted to-____ charges.

kvv77 [185]

Answer: Negatively

Explanation:

7 0

2 years ago

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An atom of helium has a radius of 31. pm and the average orbital speed of the electrons in it is about 4.4x 10 m/s Calculate the

otez555 [7]

The question is incomplete, here is the complete question:

An atom of helium has a radius of 31. pm and the average orbital speed of the electrons in it is about $4.4\times 10^6m/s$ . Calculate the least possible uncertainty in a measurement of the speed of an electron in an atom of helium. Write your answer as a percentage of the average speed, and round it to 2 significant digits

<u>Answer:</u> The percentage of average speed is 41 %

<u>Explanation:</u>

We are given:

Radius of helium atom = 31 pm = $31\times 10^{-12}m$ (Conversion factor: $1m=10^{12}pm$ )

So, diameter of helium atom = $(2\times r)=(2\times 31\times 10^{-12})=64\times 10^{-12}m$

The diameter of the atom will be equal to the uncertainty in position.

The equation representing Heisenberg's uncertainty principle follows:

$\Delta x.\Delta p=\frac{h}{2\pi}$

where,

$\Delta x$ = uncertainty in position = d = $64\times 10^{-12}m$

$\Delta p$ = uncertainty in momentum = $m\Delta v$

m = mass of electron = $9.1095\times 10^{-31}kg$

h = Planck's constant = $6.627\times 10^{-34}kgm^2/s^2$

Putting values in above equation, we get:

$64\times 10^{-12}m\times 9.1095\times 10^{-31}kg\times \Delta v=\frac{6.627\times 10^{-34}kgm^2/s}{2\times 3.14}\\\\\Delta v=\frac{6.627\times 10^{-34}kgm^2/s^2}{2\times 3.14\times 64\times 10^{-12}m\times 9.1095\times 10^{-31}kg}=1.81\times 10^6m/s$

To calculate the percentage of average speed, we use the equation:

$\text{Percentage of the average speed}=\frac{\text{Uncertainty in velocity}}{\text{Average orbital speed}}\times 100$

We are given:

Average orbital speed = $4.4\times 10^6m/s$

Putting values in above equation, we get:

$\text{Percentage of the average speed}=\frac{1.81\times 10^6m/s}{4.4\times 10^6m/s}\times 100\\\\\text{Percentage of the average speed}=41.\%$

Hence, the percentage of average speed is 41 %

3 0

4 years ago

Which of these molecules is an important source of ENERGY for most organisms?

Nitella [24]

Answer:

Glucose

Explanation:

Glucose molecule is the primary source of energy for most organisms where processes such as glycolysis starts and cellular respiration follows. Bodies of organisms can get energy from carbohydrates, proteins and fats with carbohydrates being the main component for energy where they are converted to glucose which is a form of sugar that reacts to affect blood sugar levels within an hour or two hours after eating.

The second answer option, table salt (NaCl) is incorrect because sodium added in food is vital for functioning of nerves and muscles. Additionally, it helps in fluid regulation and control of blood pressure/volume in the body.

The third answer , hydrochloric acid is not correct because presence of hydrochloric acid in the body helps in breaking down, digestion and absorption of nutrients.

The correct answer choices are : A and D.

6 0

3 years ago

Which question can be answered using the scientific method ?

anyanavicka [17]

The correct answer is D because the the rest will have different answers depending on the person you ask. D can be proven with facts by research

8 0

3 years ago

Which of the following best represents potential energy being converted to kinetic energy? HELP

TiliK225 [7]

Answer:

B) A drawn bow is released, causing an arrow to fly across the field

Explanation:

Potential energy can be thought of as "potential" to do something. For eg, putting a ball on top of hill causes the ball to have the potential to roll down the hill if released. Here the ball converts the potential energy into kinetic energy (energy of motion) to roll down.

Similarly, the bow has been stretched (potential to fly if released), and when its released, it converts the potential energy into kinetic energy.

7 0

2 years ago