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2 years ago

What is the purpose of zigzag line on a periodic table ?

Physics

1 answer:

LekaFEV [45]2 years ago

4 0

The zigzag line on the periodic table separates the metals from the nonmetals.

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The number of wavelengths that pass a point each per second is:

Lesechka [4]

Explanation:

d frequency I learned it last year

3 0

2 years ago

A tank is filled with an ideal gas at 400 K and pressure of 1.00 atm.

bekas [8.4K]

To find the temperature it is necessary to use the expression and concepts related to the ideal gas law.

Mathematically it can be defined as

$PV=nRT$

Where

P = Pressure

V = Volume

n = Number of moles

R = Gas constant

T = Temperature

When the number of moles and volume is constant then the expression can be written as

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

Or in practical terms for this exercise depending on the final temperature:

$T_2 = \frac{P_2T_1}{P_1}$

Our values are given as

$T_1 = 400K\\P_1 = 1atm\\P_2 = 2atm$

Replacing

$T_2 = \frac{(2)(400)}{1}\\T_2 = 800K$

Therefore the final temperature of the gas is 800K

6 0

2 years ago

Which best describes a value for density

scoray [572]

Density can be any number, as long as it has the right units.

A unit of density has to be (a unit of mass) divided by (a unit of volume).
The most common one is gram/cm^3.

8 0

2 years ago

Scientific theories are deductive in nature.?

dolphi86 [110]

Answer:

deductive reasoning usually follows steps .

That is, how we predict what the observations should be if the theory were correct

8 0

2 years ago

"What is the magnifying power of an astronomical telescope using a reflecting mirror whose radius of curvature is 5.9 m and an e

notka56 [123]

Answer:

The Magnifying power of a telescope is $M = 109.26$

Explanation:

Radius of curvature R = 5.9 m = 590 cm

focal length of objective $f_{objective}$ = $\frac{R}{2}$

⇒ $f_{objective}$ = $\frac{590}{2}$

⇒ $f_{objective}$ = 295 cm

Focal length of eyepiece $f_{eyepiece}$ = 2.7 cm

Magnifying power of a telescope is given by,

$M = \frac{f_{objective} }{f_{eyepiece} }$

$M = \frac{295}{2.7}$

$M = 109.26$

therefore the Magnifying power of a telescope is $M = 109.26$

4 0

2 years ago