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ollegr [7]
3 years ago
6

Which of the following are features of the electron cloud model of an atom?

Chemistry
1 answer:
marta [7]3 years ago
3 0
THE ANSWER IS OPTION.A
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A sample of sulfur hexafluoride gas occupies a volume of 5.10 L at 198 ºC. Assuming that the pressure remains constant, what tem
Ludmilka [50]

Answer:

When the volume will be reduced to 2.50 L, the temperature will be reduced to a temperature of 230.9K

Explanation:

Step 1: Data given

A sample of sulfur hexafluoride gas occupies a volume of 5.10 L

Temperature = 198 °C = 471 K

The volume will be reduced to 2.50 L

Step 2 Calculate the new temperature via Charles' law

V1/T2 = V2/T2

⇒with V1 = the initial volume of sulfur hexafluoride gas = 5.10 L

⇒with T1 = the initial temperature of sulfur hexafluoride gas = 471 K

⇒with V2 = the reduced volume of the gas = 2.50 L

⇒with T2 = the new temperature = TO BE DETERMINED

5.10 L / 471 K = 2.50 L / T2

T2 = 2.50 L / (5.10 L / 471 K)

T2 = 230.9 K = -42.1

When the volume will be reduced to 2.50 L, the temperature will be reduced to a temperature of 230.9K

8 0
3 years ago
Which of the following is the correct scientific notation for 0.000056?
svet-max [94.6K]
Hi,

0.000056 = 5.6 \times  {10}^{ - 5}

Hope this helps.
r3t40
7 0
3 years ago
When using 100ml or 50ml graduated cylinder to what decimal place can your volume be estimated?<br>​
Olegator [25]

Answer:

I know that the 100-mL graduated cylinders are always read to 1 decimal place.

I think for 50 mL graduated cylinders, it lets you measure volumes up to 50.0 mL to the nearest 0.1 or 0.2 mL, depending on your exact cylinder.

3 0
3 years ago
which tool was mostly likely used in a procedure if the lab report shows that approximately 300 ml of water was used
DiKsa [7]

it is most likely a beaker, beacuse 300ml is quite a large volume. Otherwise, it would be a measuring cylinder or pippette

7 0
3 years ago
he mineral rhodochrosite [manganese(II) carbonate, MnCO3] is a commercially important source of manganese. Write a half-reaction
valina [46]

Answer:

MnCO_{3}+2H_{2}O-2e^{-}\rightarrow MnO_{2}+HCO_{3}^{-}+3H^{+}

Explanation:

Half reaction:MnCO_{3}\rightarrow MnO_{2}

(1)CO_{3} balance: MnCO_{3}\rightarrow MnO_{2}+HCO_{3}^{-}

(2)H and O balance in acidic medium:MnCO_{3}+2H_{2}O\rightarrow MnO_{2}+HCO_{3}^{-}+3H^{+}

(3) charge balance:MnCO_{3}+2H_{2}O-2e^{-}\rightarrow MnO_{2}+HCO_{3}^{-}+3H^{+}

Hence balanced half-reaction:MnCO_{3}+2H_{2}O-2e^{-}\rightarrow MnO_{2}+HCO_{3}^{-}+3H^{+}

7 0
3 years ago
Read 2 more answers
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