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4 years ago

How many places are there for electrons in the third shell of an atom?

2 answers:

5 0

Each shell can contain only a fixed number of electrons: The first shell can hold up to two electrons, the second shell can hold up to eight (2 + 6) electrons, the third shell can hold up to18 (2 + 6 + 10) and so on. The general formula is that the nth shell can in principle hold up to 2(n2) electrons.

Alisiya [41]4 years ago

4 0

The answer is 4 places for electrons

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An atom that has 117 protons in its nucleus has not yet been made. Once this atom is made, to which group will element 117 belon

Pavel [41]

In nomine patris, et filii, et spiritus sancti.

3 0

3 years ago

A 0.877 mol sample of N2(g) initially at 298 K and 1.00 atm is held at constant volume while enough heat is applied to raise the

MissTica

Answer : The value of $q\text{ and }\Delta U$ is 286.2 J and 286.2 J respectively.

Explanation : Given,

Moles of sample = 0.877 mol

Change in temperature = 15.7 K

First we have to calculate the heat absorbed by the system.

Formula used :

$q=n\times c_v\times \Delta T$

where,

q = heat absorbed by the system = ?

n = moles of sample = 0.877 mol

$\Delta T$ = Change in temperature = 15.7 K

$c_v$ = heat capacity at constant volume of $N_2$ (diatomic molecule) = $\frac{5}{2}R$

R = gas constant = 8.314 J/mol.K

Now put all the given value in the above formula, we get:

$q=0.877mol\times \frac{5}{2}\times 8.314J/mol.K\times 15.7K$

$q=286.2J$

Now we have to calculate the change in internal energy of the system.

$\Delta U=q+w$

As we know that, work done is zero at constant volume. So,

$\Delta U=q=286.2J$

Therefore, the value of $q\text{ and }\Delta U$ is 286.2 J and 286.2 J respectively.

8 0

4 years ago

Read 2 more answers

Pleaseeee Please help, I will love you forever and ever

matrenka [14]

Answer:

The answer to your question is

Explanation:

Data

mass = 0.5kg

T1 = 35

T2 = ?

Q = - 6.3 x 10⁴ J = - 63000 J

Cp = 4184 J / kg°C

Formula

Q = mCp(T2 - T1)

T2 = T1 + Q/mCp

Substitution

T2 = 35 - 63000/(0.5 x 4184)

T2 = 35 - 63000/2092

T2 = 35 - 30.1

T2 = 4.9 °C

6 0

3 years ago

A uniform stationary ladder of length L and mass M leans against a smooth vertical wall, while its bottom legs rest on a rough h

ikadub [295]

Answer:A uniform ladder of mass and length leans at an angle against a frictionless wall .If the coefficient of static friction between the ladder and the ground is , determine a formula for the minimum angle at which the ladder will not slip.

Explanation:A uniform ladder of mass and length leans at an angle against a frictionless wall .If the coefficient of static friction between the ladder and the ground is , determine a formula for the minimum angle at which the ladder will not slip.

3 0

3 years ago

A spacecraft in the shape of a long cylinder has a length of 100 m, and its mass with occupants is 1 480 kg. It has strayed too

maks197457 [2]

Answer:

2352645198509.9604 m/s²

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

M = Mass of black hole = $90\times 1.989\times 10^{30}\ kg$

$R_{f}$ = 10000+100 m

$R_{sb}$ = Distance between the nose and the center of the black hole = 10000 m

The difference in the gravitational field in this system is given by

$\Delta F=\dfrac{GMm}{R_{f}^2}-\dfrac{GMm}{R_{sb}^2}\\\Rightarrow \Delta g=GM\left(\frac{1}{R_f^2}-\frac{1}{R_{sb}^2}\right)\\\Rightarrow g=6.67\times 10^{-11}\times 90\times 1.989\times 10^{30}\left(\frac{1}{(10000+100)^2}-\frac{1}{10000^2}\right)\\\Rightarrow \Delta g=-2352645198509.9604\ m/s^2$

The acceleration is 2352645198509.9604 m/s²

4 0

3 years ago