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3 years ago

A- Gases do not have mass.

Chemistry

1 answer:

finlep [7]3 years ago

7 0

The answer is going to be d. hope that helped

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A fan pulls in air from the surrounding area, which typically has high moisture levels.

When the air passed through the dehumidifier , it touches the cooling coils , which in turn pull moisture from the air by lowering temperature.

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3 years ago

In witch of these examples does chemical energy change to electrical energy

Anuta_ua [19.1K]

D the chemical energy in a batery changes to electrical when its used

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3 years ago

Weather stations send up weather balloons with _____ to transmit temperature, humidity, wind speed, and wind direction measureme

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Technology? GPS? Analyzers? try those

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3 years ago

Consider the second-order reaction:

kirza4 [7]

Answer:

Initial concentration of HI is 5 mol/L.

The concentration of HI after $4.53\times 10^{10} s$ is 0.00345 mol/L.

Explanation:

$2HI(g)\rightarrow H_2(g)+I_2(g)
$

Rate Law: $k[HI]^2
$

Rate constant of the reaction = k = $6.4\times 10^{-9} L/mol s$

Order of the reaction = 2

Initial rate of reaction = $R=1.6\times 10^{-7} Mol/L s$

Initial concentration of HI = $[A_o]$

$1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2$

$[A_o]=5 mol/L$

Final concentration of HI after t = [A]

t = $4.53\times 10^{10} s$

Integrated rate law for second order kinetics is given by:

$\frac{1}{[A]}=kt+\frac{1}{[A_o]}$

$\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}$

$[A]=0.00345 mol/L$

The concentration of HI after $4.53\times 10^{10} s$ is 0.00345 mol/L.

5 0

3 years ago

A 59.1g sample of aluminum is put into a calorimeter (see sketch at right) that contains 250.0g of water. The aluminum sample st

Rainbow [258]

Answer:

The specific heat capacity of aluminum according to this experiment is 0.863 J/g°C

Explanation:

Step 1: Data given

Mass of aluminium = 59.1 grams

Mass of water = 250.0 grams

Initial temperature of aluminium = 91.3 °C

Initial temperature of water = 16.0 °C

Final temperature = 19.5 °C

Pressure remains constant

Specific heat capacity of water = 4.186 J/g°C

Step 2: Calculate specific heat of aluminium

Heat lost = heat gained

Qlost = -Q heat

Q = m*c*ΔT

heat aluminium = - heat water

m(aluminium) * c(aluminium) * ΔT(aluminium) = -m(water) * c(water) * ΔT(water)

⇒m(aluminium) = mass of aluminium = 59.1 grams

⇒c(aluminium) = the specific heat of aluminium = TO BE DETERMINED

⇒ΔT = the change in temperature = T2 -T2 = 19.5 - 91.3 = -71.8 °C

⇒ m(water) = 250.0 grams

⇒c(water) = the specific heat of water = 4.186 J/g°C

⇒ΔT = the change in temperature = T2 -T2 = 19.5 - 16.0 = 3.5 °C

59.1 * c(aluminium) * -71.8 °C = 250.0 * 4.186 J/g°C * 3.5 °C

c(aluminium) = 0.863 J/g°C

The specific heat capacity of aluminum according to this experiment is 0.863 J/g°C

3 0

3 years ago