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stiv31 [10]
3 years ago
12

What's the difference between groups and periods in a periodic table?

Physics
2 answers:
Leya [2.2K]3 years ago
5 0
The periodic table is going from left to right. It also has a special name for its vertical columns. Each column is called a group. Each horizontal row is called a period.
Bad White [126]3 years ago
4 0

On the periodic table, there are both groups and periods.

Periods go horizontal on the periodic table are there are 7 groups in all. Groups are very important because they tell you the number of electrons shells also called orbitals.

On the other hand, groups are the vertical columns on the period table. There are 18 groups and the elements in each group correspond to the number of valence electrons.

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Which two statements describe force?
Len [333]

Answer:

I think its C and D

Explanation:Force is like a energy that moves stuff. :)

6 0
2 years ago
From a branch 35 m high, a 0.75 kg bird dives into a small fish tank containing
Bad White [126]

Answer:

ΔT = 1.22*10^-3 °C

Explanation:

First, you calculate the potential energy of the bird when it is at 35 m high. The potential energy is also the mechanical energy of the bird in this case.

U=mgh

m: mass of the bird = 0.75kg

g: gravitational constant = 9.8m/s^2

h: height = 35m

U=(0.75kg)(9.8m/s^2)(35m)=257.25\ J

All this energy is given to the water. You use the following formula in order to calculate the change in temperature:

Q=mc\Delta T

m: mass of the water = 50kg

c: specific heat of water = 4186 J/kg°C

Q is equal to U (potential energy of the bird) because the bird gives all its energy to water. By doing ΔT the subject of the formula you obtain:

\Delta T=\frac{Q}{mc}=\frac{257.25J}{(50kg)(4186J/kg°C)}=1.22*10^{-3}\ \°C

hence, the maximum rise in temperature is 0.00122 °C

7 0
3 years ago
A cube icebox of side 3cm has a thickness of 5.0cm. If 4.0 kg of ice is put in the box estimate the amount of ice remaining afte
qaws [65]

Answer:

The amount of solid ice remaining after 6 hours is approximately 3.68664 kg

Explanation:

The given parameters are;

The side length of the cube box, s = 3(0) cm = 0.3 m

The thickness of the cube box, d = 5.0 cm = 0.05 m  

The mass of ice in the box, m = 4.0 kg

The outside temperature of the cube box, T₁ = 45°C

The temperature of the melting ice inside the box, T₂ = 0°C

The latent heat of fusion of ice, L_f = 3.35 × 10⁵ J/K/hr/kg

The surface area of the box, A = 6·s² 6 × (0.3 m)² = 0.54 m²

The coefficient of thermal conductivity, K = 0.01 J/s·m⁻¹·K⁻¹

For thermal equilibrium, we have;

The heat supplied by the surrounding = The heat gained by the ice

The  heat supplied by the surrounding, Q = K·A·ΔT·t/d

Where;

ΔT = T₁ - T₂ =  45° C - 0° C = 45° C

ΔT = 45° C

Q = K·A·ΔT·t/d = 0.01 × 0.54 × 45 × 6× 60×60/0.05 = 104976

∴ The  heat supplied by the surrounding, Q = 104976 J

The heat gained by the ice = L_f × m_{melted \ ice} =3.35 × 10⁵ J/kg × m_{melted \ ice}

Therefore, from Q =  L_f × m_{melted \ ice}, we have;

Q = 104976 J =  L_f × m_{melted \ ice} = 3.35 × 10⁵ J/kg × m_{melted \ ice}

104976 J = 3.35 × 10⁵ J/kg × m_{melted \ ice}

m_{melted \ ice} = 104976 J/(3.35 × 10⁵ J/kg) ≈ 0.31336 kg

The mass of melted ice, m_{melted \ ice} ≈ 0.31336 kg

∴ The amount of solid ice remaining after 6 hours, m_{ice} = m - m_{melted \ ice}

Which gives;

m_{ice} = m - m_{melted \ ice} = 4.0 kg - 0.31336 kg ≈ 3.68664 kg

The amount of solid ice remaining after 6 hours, m_{ice} ≈ 3.68664 kg.

8 0
3 years ago
Consider an insulated tank with a volume V = 2 L is separated into two equal-volume parts by a thin wall. On the left is an idea
steposvetlana [31]

Answer

given,

V = 2 L

the left is an ideal gas at  P = 100 k Pa and T = 500 K

mass is constant

 m_1 = m_2

\dfrac{P_1V_1}{RT_1} = \dfrac{P_2V_2}{RT_2}

Pressure is same because it's not changing due to process

\dfrac{V}{500} = \dfrac{2 V}{T_2}

T_2 = 1000\ K

\Delta S_{univ} = \Delta S_{sys} + (\Delta S)_{surr}

\Delta S_{univ} =m(C_v ln (\dfrac{T_2}{T_1}))+ R ln (\dfrac{V_2}{V_1})

m = \dfrac{P_1V_1}{RT_1}

m = \dfrac{100 \times 10^3 \times 2 \times 10^{-3}}{287\times 500}

m = 1.39 x 10⁻³ Kg

\Delta S_{univ} =1.39\times 10^{-3}(0.718 ln\ 2+ 0.287 ln (2)

\Delta S_{univ} =0.968\times 10^{-3}\ kJ/K

5 0
3 years ago
Animals that can hear ultrasonic sounds are the what
bulgar [2K]
Here are the different animals that can hear ultrasonic sounds: dogs<span>, </span>cats, porpoises, rodents, whales,dolphins<span>, seals, </span>sea lions<span> and </span>bats<span>.</span>
4 0
3 years ago
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