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drek231 [11]
3 years ago
12

On a temperature versus time graph, how does the temperature at the beginning of a change of state compare with the temperature

at the end of the change?
Chemistry
2 answers:
gogolik [260]3 years ago
6 0

Answer:B

Explanation:

sweet-ann [11.9K]3 years ago
5 0

the answer is ALWAYS THE SAME.

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What is the electron configuration for Bromine? It has 27 electrons. I think the electron configuration for Bromine is : I'm sor
vfiekz [6]
The electron configuration for cobalt is:
[Ar] 3d7 4s2
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A double covalent bond involves two atoms sharing two electrons? <br><br> True or false
jarptica [38.1K]

Answer:

False

Explanation:

A double covalent bond means 2 atoms or elements are sharing <u>4</u><u> </u><u>e</u><u>l</u><u>e</u><u>c</u><u>t</u><u>r</u><u>o</u><u>n</u><u>s</u>.

*single covalent bond shares 2 electrons.

4 0
3 years ago
What is the percent by mass of oxygen in carbon dioxide (CO2)?
krok68 [10]

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I think that it will be

Explanation:

72.71%

8 0
2 years ago
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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
3 years ago
A 2.04 g lead weight, initially at 10.8 oC, is submerged in 7.74 g of water at 52.2 oC in an insulated container. clead = 0.128
QveST [7]

Answer:

Explanation:

Hello, in this case, the lead is catching heat and the water losing it, that's why the heat relation ship is (D is for Δ):

DH_{lead}=-DH_{water}

Now, by stating the heat capacity definition:

m_{Pb}C_{Pb}*(T_{eq}-T_{lead}=-m_{H_2O}C_{H_2O}*(T_{eq}-T_{H_2O})\\

Solving for the equilibrium temperature:

T_{eq}=\frac{m_{Pb}C_{Pb}T_{Pb}+m_{H_2O}C_{H_2O}T_{H_2O}}{m_{Pb}C_{Pb}+m_{H_2O}C_{H_2O}} \\\\T_{eq}=\frac{2.04g*0.128J/(g^oC)*10.8^oC+7.74g*4.18J/(g^oC)*52.2^oC}{2.04g*0.128J/(g^oC)+7.74g*4.18J/(g^oC)} \\\\T_{eq}=51.87^oC

Which is very close to the water's temperature since the lead's both mass and head capacity are lower than those for water.

Best regards.

5 0
2 years ago
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