All categories

3 years ago

Whitch family has full valence energy level

1 answer:

7 0

Group 0 - they are called the NOBLE GASES
(as they're very unreactive) because...

ALL of their atoms are full.
Example : Helium contains 2 electrons so it's outer shell is full

You might be interested in

Assuming the carbon cycle is a closed system, which of the following statements is true?

ella [17]

Please add answer options :)

5 0

3 years ago

Read 2 more answers

A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º. What is the coefficient of

Alex_Xolod [135]

Given :

A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º.

To Find :

The coefficient of static friction between the box and the plane.

Solution :

Vertical component of force :

$mg\ sin\ \theta = 120\times 10 \times sin\ 47^\circ{}=877.62 \ N$

Horizontal component of force(Normal reaction) :

$mg\ cos\ \theta = 120\times 10 \times cos\ 47^\circ{}=818.40 \ N$

Since, box is on the verge of slipping :

$mg\ sin\ \theta= \mu(mg \ cos\ \theta)\\\\\mu = tan \ \theta\\\\\mu = tan\ 47^o\\\\\mu = 1.07$

Therefore, the coefficient of static friction between the box and the plane is 1.07.

Hence, this is the required solution.

7 0

3 years ago

A 23.9 g sample of iridium is heated to 89.7 oC, and then dropped into 20.0 g of water in a coffee-cup calorimeter. The temperat

Nikolay [14]

Answer:

The specific heat capacity of iridium = 0.130 J/g°C

Explanation:

Assuming no heat losses to the environment and to the calorimeter,

Heat lost by the iridium sample = Heat gained by water

Heat lost by the iridium sample = mC ΔT

m = mass of iridium = 23.9 g

C = specific heat capacity of the iridium = ?

ΔT = change in temperature of the iridium = 89.7 - 22.6 = 67.1°C

Heat lost by the iridium sample = (23.9)(C)(67.1) = (1603.69 C) J

Heat gained by water = mC ΔT

m = mass of water = 20.0 g

C = 4.18 J/g°C

ΔT = 22.6 - 20.1 = 2.5°C

Heat gained by water = 20 × 4.18 × 2.5 = 209 J

Heat lost by the iridium sample = Heat gained by water

1603.69C = 209

C = (209/1603.69) = 0.130 J/g°C

7 0

3 years ago

Read 2 more answers

Occasionally, people can survive falling large distances if the surface they land on is soft enough. During a traverse of Eiger'

uranmaximum [27]

Answer:

4611.58 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 32.174 ft/s²

Equation of motion

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 32.174\times 516+0^2}\\\Rightarrow v=182.218\ ft/s$

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-182.218^2}{2\times 3.6}\\\Rightarrow a=-4611.58\ ft/s^2$

Magnitude of acceleration while stopping is 4611.58 ft/s²

8 0

2 years ago

The apparent depth of an object at the bottom of a tank filled with a liquid of refractive index 1.3 is 7.7 cm. What is the actu

Juliette [100K]

$\orange{\underline{\huge{\bold{\textit{\green{\bf{QUESTION}}}}}}}$

The apparent depth of an object at the bottom of a tank filled with a liquid of refractive index 1.3 is 7.7 cm. What is the actual depth of the liquid in the tank?

${\bold{\blue{GIVEN}}}$

REFRACTIVE INDEX = 1.3

APPARENT DEPTH = 7.7 cm

${ \bold{\green{To \: Find}}}$

REAL DEPTH OF THE OBJECT.

${\red{FORMULA \: \: USED }}$

$Reflective \: Index = \frac{Real \: Depth}{Apparent \: Depth }$

$\huge\mathbb{\red A \pink{N}\purple{S} \blue{W} \orange{ER}}$

Refractive Index = 1.3

Apparent Depth = 7.7 cm

Putting the values in the formula:-

$Reflective \: Index = \frac{Real \: Depth}{Apparent \: Depth } \\ \\ 1.3 = \frac{Real \: Depth}{7.7 \: cm} \\ \\ 1.3 \times 7.7 = Real \: Depth \\ \\ 10.01 \: \: cm = Real \: Depth$

3 0

2 years ago