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Alchen [17]
3 years ago
8

Which term below represents a pure substance that can be separated chemically, but cannot be separated physically, that is made

up of two or more different atoms?
chemical bondmoleculemixtureelement
Chemistry
2 answers:
GREYUIT [131]3 years ago
7 0

Molecules are made up of many atoms and can only be separated by chemical means

aleksandr82 [10.1K]3 years ago
7 0

I think the answer is, chemical bond

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325

Explanation:

velocity = 3250 Hz x 0.1 m

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What is the percent composition for the compound NaBr? (8C)
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the answer is D

Explanation:

percentage composition=  mole of the substance divided by the total molar mass of the compound multiplied by 100.

3 0
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How many moles of water h2o are present in 75.0 g h2o?
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7 0
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Determine the mass of sodium carbonate required to produce 23.4g of sodium chloride when it reacts with excess hydrochloric acid
Leni [432]
The balanced chemical reaction for the described reaction above is,
                          Na2CO3 + 2HCl ---> 2NaCl + H2CO3
From the reaction, 1 mole of Na2CO3 is needed to produce 2 moles of NaCl. In terms of mass, 106 grams of Na2CO3 are needed to produce 116.9 grams of NaCl. From this,
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8 0
2 years ago
A 85.2 g copper bar was heated to 221.32 degrees Celsius and placed in a coffee cup calorimeter containing 4250 mL of water at 2
Assoli18 [71]

Answer:

The specific heat of copper is 0.385 J/g°C

Explanation:

A 85.2 g copper bar was heated to 221.32 degrees Celsius and placed in a coffee cup calorimeter containing 425 mL of water at 22.55 degrees Celsius. The final temperature of the water was recorded to be 26.15 degrees Celsius. What is the specific heat of the copper?

Step 1: Data given

Mass of copper = 85.2 grams

Temperature of copper = 221.32 °C

Volume of water = 425 mL

Temperature of water = 22.55 °C

Final temperature = 26.15 °C

Specific heat of water = 4.184 J/g°C

Step 2: Calculat the specific heat of copper

Heat lost = heat gained

Q = m*c*ΔT

Qcopper = -Qwater

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

⇒ m(copper) = 85.2 grams

⇒ c(copper) = TO BE DETERMINED

⇒ ΔT(copper) = the change in temeprature = T2 -T1 = 26.15 -221.32 = -195.17 °C

⇒ m(water) = The mass of water = 425 mL * 1g/mL = 425 grams

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

⇒ ΔT(water) = The change of temperature of water = 26.15 - 22.55 = 3.6

85.2 * c(copper) * (-195.17) = -425 * 4.184 * 3.6

c(copper) = 0.385 J/g°C

The specific heat of copper is 0.385 J/g°C

(Note, The original question says the volume of the water is 4250 mL. IF this is not an error, the specific heat of copper is 3.85 J/g°C (10x higher than the normal value).

8 0
2 years ago
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