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mote1985 [20]
3 years ago
10

A hot metal plate at 150°C has been placed in air at room temperature. Which event would most likely take place over the next fe

w minutes?
Molecules in both the metal and the surrounding air will start moving at lower speeds.
Molecules in both the metal and the surrounding air will start moving at higher speeds.
The air molecules that are surrounding the metal will slow down, and the molecules in the metal will speed up.
The air molecules that are surrounding the metal will speed up, and the molecules in the metal will slow down.
Chemistry
1 answer:
Oksi-84 [34.3K]3 years ago
8 0

Answer:

d. The air molecules that are surrounding the metal will speed up, and the molecules in the metal will slow down.

Explanation:

hopes this helps  

                                                      sorry if it doesn't

:)

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The gas described in parts a and b has a mass of 1.66 g. the sample is most likely which monatomic gas?
nordsb [41]
Mass of the gas m = 1.66 
The calculated temperature T = 273 + 20 = 293
 We have to calculate molar mass to determine the gas
 Molar Mass = mRT / PV
 M = (1.66 x 8.314 x 293) / (101.3 x 1000 x 0.001)
 M = 4043.76 / 101.3 = 39.92 g/mol
 So this gas has to be Argon Ar based on the molar mass.

7 0
3 years ago
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Sean loves to go camping! When he goes camping, he uses a propane stove to cook all his meals. Propane's chemical formula is C3H
kotykmax [81]

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O + heat

Explanation:

When Sean cook meals with propane he use a combustion reaction:

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O + heat

When propane (C₃H₈) is reacted with oxygen (O₂) produces carbon dioxide (CO₂), water (H₂O) and heat.

Learn more about:

combustion reaction

brainly.com/question/13824679

#learnwithBrainly

8 0
3 years ago
How many internal parts or “organelles” are there in a plant cell?<br> please i need a answer!
Ghella [55]

Answer: it has 2 parts

Explanation:

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3 0
3 years ago
A student places a 100.0°C piece of metal that weighs 85.5 g into 122 mL of 16.0°C water. If the final temperature is 20.2°C, wh
Musya8 [376]

Answer:

The specific heat of the metal is 0.314 J/g°C

Explanation:

Step 1: data given

Temperature of the piece of metal = 100.0 °C

Mass of the metal = 85.5 grams

Volume of water = 122 mL = 122 grams

Temperature of water = 16.0 °C

The final temperature of water = 20.2 °C

The specific heat of water = 4.184 J/g°C

Step 2: Calculate the specific heat of metal

Heat gained= heat lost

Qgained = - Qlost

Qwater = -Qmetal

Q = m*c* ΔT

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

⇒m(metal) = mass of metal = 85.5 grams

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

⇒ΔT(metal) = the change of temperature of metal = T2 - T1 = 20.2 - 100 °C =  -79.8 °C

⇒m(water) = the mass of water = 122 grams

⇒c(water) = the specific heat of water = 4.184 J/g°C

⇒ΔT(water) = the change of temperature of metal = T2 - T1 = 20.2 - 16.0 °C =  4.2 °C

85.5 *c(metal) * -79.8 = -122 * 4.184 * 4.2

c(metal) * (-6822.9) = -2143.9

c(metal) = 0.314 J/g°C

The specific heat of the metal is 0.314 J/g°C

7 0
3 years ago
Just need help with these two questions
kvasek [131]
B/A hope it help thanks
8 0
3 years ago
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