Soft of easy help plz B C A D

Dr. Sinko places an aluminum block and a copper block in contact, and allows them to achieve thermodynamic equilibrium. Aluminum

Angelina_Jolie [31]

Answer:

The two blocks have the same temperature

Explanation:

Given;

heat capacity of Aluminum = 0.9 J/g

heat capacity of copper = 0.35 J/g

thermal conductivity of copper = 385 W/(K m)

thermal conductivity of Aluminum = 205 W/(K m)

Based on the given data, it can be concluded that;

Aluminum’s higher heat capacity means it does not get as hot when energy is shared between the two blocks, so the copper will be hotter
Copper’s higher thermal conductivity will cause the aluminum to heat up compared to the copper

However, the two metallic blocks (aluminum block and a copper block), were allowed to achieve thermodynamic equilibrium, and as a result they will have the same temperature.

Therefore, the correct statement is "The two blocks have the same temperature".

7 0

3 years ago

The value of delta for the [C_rF_6]^3- complex is 182 kJ/mol. Calculate the expected wavelength of the absorption corresponding

kirza4 [7]

Answer: Yes the absorb in the visible range.

Explanation:

The relationship between wavelength and energy of the wave follows the equation:

$E=\frac{Nhc}{\lambda}$

where,

$E$ = energy of the wave = 182 kJ/mol = 182000 J/mol

N = avogadro's number = $6.023\times 10^{23}$

h = plank constant = $6.6\times 10^{-34}Js^{-1}$

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength of the wave = ?

Putting all the values:

$182000=\frac{6.023\times 10^{23}\times 6.6\times 10^{-34}\times 3\times 10^8m/s}{\lambda}$

$\lambda=0.65\times 10^{-6}m=650nm$ $(1nm=10^{-9}m)$

The wavelength range for visible rays is 400 nm to 750 nm, thus the complex absorb in the visible range.

5 0

3 years ago

Please help me get the answer to bothe a and b

neonofarm [45]

Answer:

If a metal and metal solution react, the more reactive metal will displace the less reactive metal from solution. If the metal in solution you start with is formed from a more reactive metal than the metal to be added, no reaction will occur.

3 0

2 years ago

Boiling occurs when molecules being changed from liquid to gas plz tell me the answer true or false

Valentin [98]

Answer:

Liquid to Gas

Explanation:

The particles need energy to rise and over come the attractions between them as the liquid gets warmer more particles have sufficient, energy to escape from liquid. eventually even particles in the middle of the liquid form bubbles of gas in the liquid At this point the liquid is boiliing and turning into gas.

5 0

3 years ago

1. Calculate the energy change (q) of the surroundings (water) using the enthalpy equation

Rasek [7]

Answer:

Q1: 728.6 J.

Q2:

a) 668.8 J.

b) 0.3495 J/g°C.

Explanation:

Q1: Calculate the energy change (q) of the surroundings (water) using the enthalpy equation:

The amount of heat absorbed by water = Q = m.c.ΔT.

where, m is the mass of water (m = d x V = (1.0 g/mL)(24.9 mL) = 24.9 g).

c is the specific heat capacity of liquid water = 4.18 J/g°C.

ΔT is the temperature difference = (final T - initial T = 32.2°C - 25.2°C = 7.0°C).

∴ The amount of heat absorbed by water = Q = m.c.ΔT = (24.9 g)(4.18 J/g°C)(7.0°C) = 728.6 J.

Q2: Calculate the energy change (q) of the surroundings (water) using the enthalpy equation

qwater = m × c × ΔT.

We can assume that the specific heat capacity of water is 4.18 J / (g × °C) and the density of water is 1.00 g/mL. calculate the specific heat of the metal. Use the data from your experiment for the unknown metal in your calculation.

a) First part: the energy change (q) of the surroundings (water):

The amount of heat absorbed by water = Q = m.c.ΔT.

where, m is the mass of water (m = d x V = (1.0 g/mL)(25 mL) = 25 g).

c is the specific heat capacity of liquid water = 4.18 J/g°C.

ΔT is the temperature difference = (final T - initial T = 31.6°C - 25.2°C = 6.4°C).

∴ The amount of heat absorbed by water = Q = m.c.ΔT = (25 g)(4.18 J/g°C)(6.4°C) = 668.8 J.

b) second part:

Q water = Q unknown metal.

Q unknown metal = - 668.8 J. (negative sign due to the heat is released from the metal to the surrounding water).

Q unknown metal = - 668.8 J = m.c.ΔT.

m = 27.776 g, c = ??? J/g°C, ΔT = (final T - initial T = 31.6°C - 100.5°C = - 68.9°C).

- 668.8 J = m.c.ΔT = (27.776 g)(c)( - 68.9°C) = - 1914 c.

∴ c = (- 668.8)/(- 1914) = 0.3495 J/g°C.

3 0

4 years ago