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3 years ago

5

How do these two substances compare? Explain.

1 answer:

Advocard [28]3 years ago

8 0

To be honest the answer should be B. Hope this help.

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32 Points!!! Match the definition with correct keyword.

Temka [501]

Answer: Okay so here's the order lol from top to bottom

2, 1, 3, 4, 5

Explanation:

5 0

3 years ago

write a chemical equation for the ionization of water. how do you indicate that this is a reversible action?

LiRa [457]

H₂O + H₂O ⇌ H₃O⁺ + OH⁻

You use two single-barbed arrows pointing in opposite directions (technically rightwards harpoon over leftwards harpoon) <em>to indicate a reversible reactio</em>n.

8 0

4 years ago

Describe how the law of conservation of energy is demonstrated in this laboratory activity.

Crank

What laboratory activity. The law of conservation of energy is that energy remains constant. This relates to physics

8 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

A heat pump is used to heat a house in the winter and to cool it in the summer. During the winter, the outside air serves as a l

elixir [45]

Answer:

$T_{C}$ = -4.2°C

$T_{H}$ = 49.4°C

Explanation:

A Carnot cycle is known as an ideal cycle in thermodynamic. Therefore, in theory, we have:

| $\frac{Q_{C} }{Q_{H} }$ | = $\frac{T_{C} }{T_{H} }$

Similarly,

| $Q_{H}$ | = | $Q_{C}$ | + | $W_{S}$ |

During winter, the value of | $T_{H}$ | = 20°C = 273.15 + 20 = 293.15 K and | $W_{S}$ | = 1.5 kW. Therefore,

| $Q_{H}$ | = 0.75( $T_{H}$ - $T_{C}$ )

Similarly,

| $\frac{W_{S} }{Q_{H} }$ | = 1 - $\frac{T_{C} }{T_{H} }$

1.5/0.75*(293.15- $T_{C}$ ) = 1 - ( $T_{C}$ /293.15

Further simplification,

$T_{C}$ = -4.2°C

During summer, $T_{C}$ = 25°C = 273.15+25 = 298.15 K, and | $W_{S}$ | = 1.5 kW. Therefore,

| $Q_{C}$ | = 0.75( $T_{H}$ - $T_{C}$ )

Similarly,

| $\frac{W_{S} }{Q_{C} }$ | = $\frac{T_{H} }{T_{C} }$ - 1

1.5/0.75*( $T_{H}$ - 298.15) = ( $T_{H}$ /298.15

Further simplification,

$T_{H}$ = 49.4°C

4 0

4 years ago