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skad [1K]
3 years ago
13

0.15 grams of Mg is combined with HCl to

Chemistry
1 answer:
gladu [14]3 years ago
3 0

Answer:

A. 1.01

Explanation:

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Checking if my answer is correct: I got LARQY, but I'm pretty sure it's wrong. Can somebody please help?
babunello [35]

Answer:

you are right

Explanation:

and i do not need to explain it because you did

8 0
3 years ago
Five kg of carbon dioxide (CO2) gas undergoes a process in a well-insulated piston-cylinder assembly from 2 bar, 280 K to 20 bar
kaheart [24]

This question is incomplete, the complete question is;

Five kg of carbon dioxide (CO2) gas undergoes a process in a well-insulated piston-cylinder assembly from 2 bar, 280 K to 20 bar, 520 K. If the carbon dioxide behaves as an ideal gas, determine the amount of entropy produced, in kJ/K. Assuming;

a) constant specific heats Cp = 0.939 kJ/Kg K

b) variable specific heats

Answer:

a) the amount of entropy produced is 0.731599 kJ/K

b) the amount of entropy produced is 0.69845 kJ/K

Explanation:

Given the data in the question;

5 kg of carbon dioxide (CO₂) gas undergoes a process in a well-insulated piston-cylinder assembly.

m = 5 kg

Molar mass M = 44.01 g/mol

P₁ = 2 bar, P₂ = 20

T₁ = 280 K, P₂ = 520 K

Since its insulated { q = 0 } ( kinetic and potential energy effects = 0 )

Now,

a) the amount of entropy produced, in kJ/K, Assuming constant specific heats with Cp = 0.939 kJ/Kg K

S_{Generation = m × ((Cp × In( T₂/T₁) - R × In( P₂/p₁ ))

we substitute

S_{Generation = 5 × (( 0.939  × In( 520/280) - 0.1889 × In( 20/2 ))

= 5 × ( 0.5812778 - 0.434958 )

= 5 × 0.1463198

= 0.731599 kJ/K

Therefore, the amount of entropy produced is 0.731599 kJ/K

b) the amount of entropy produced, in kJ/K, Assuming variable specific heats.

Now, from  Table A-23: Ideal Gas Properties of Selected Gases;

T₁,T₂ : s₁⁰ = 211.376 kJ/kmol-K, s₂⁰ = 236.575 kJ/kmol-K

now, s₁ = s₁⁰ / M and s₂ = s₂⁰ / M

we substitute

s₁ = s₁⁰ / M = 211.376 / 44.01  = 4.8029 kJ/kg

s₂ = s₂⁰ / M = 236.575 / 44.01 = 5.37548 kJ/kg

S_{Generation = m × (( s₂ - s₁ ) - R × In( p₂ / p₁ ))

we substitute

S_{Generation = 5 × (( 5.37548 - 4.8029  ) - 0.1880 × In( 20 / 2 ))

= 5 × ( 0.57258 - 0.432885997 )

= 5 × 0.13969

= 0.69845 kJ/K

Therefore, the amount of entropy produced is 0.69845 kJ/K

5 0
3 years ago
Sound wave a has lower frequency than sounds wave b but both waves have the same amplitude . What must also be true of these two
Nady [450]

Explanation:

Higher the frequency smaller will be the wavelength. Higher frequency have shorter wavelength and lower frequency waves have larger wavelength. Also, Beats are formed by the superposition of two waves with slightly different frequencies but with similar amplitudes. In time, waves switch between constructive interference and disruptive interference, giving the resultant wave a time-varying amplitude.

8 0
3 years ago
Read 2 more answers
Can you help me find the role for each organism
Rainbow [258]
The first one is Producer and second level consumer
the second one is: decomposer and producer
the third one is: carnivore and  second level consumer
the fourth one is: herbivore and first level consumer
7 0
3 years ago
1.2 kg of aluminum at 20oC is added to 1.5 kg of water at 80oC. After the system reaches thermal equilibrium, what is its final
Lubov Fominskaja [6]

Answer:

The final temperature is 71.19 °C

Explanation:

Step 1: Data given

Mass of aluminium = 1.2 kg = 1200 grams

Temperature of aluminium = 20.0 °C

Specific heat of aluminium = 0.900 J/g°C

Mass of water = 1.5 kg = 1500 grams

Temperature of water = 80.0 °C

Specific heat of water = 4.184 J/g°C

Step 2: Calculate the final temperature

heat gained = heat lost

Q(aluminium) = - Q(water)

Q = m*c*ΔT

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

⇒ with mass of aluminium = 1200 grams

⇒ with specific heat of aluminium = 0.900 J/g°C

⇒ with ΔT = The change in temperature = T2 - T1 = T2 - 20.0 °C

⇒ with mass of water = 1500 grams

⇒ with specific heat of water = 4.184 J/g°C

⇒ with ΔT = The change in temperature = T2 - T1 = T2 - 80.0°C

1200 * 0.900 * (T2-20.0°C) = -1500 * 4.184 * (T2 - 80.0°C)

1080 * (T2 - 20.0°C) = -6276 * (T2 - 80.0°C)

1080 T2 - 21600 = -6276T2 + 502080

7356T2 = 523680

T2 = 71.19 °C

The final temperature is 71.19 °C

8 0
3 years ago
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