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Elza [17]
3 years ago
5

There are two different compounds of sulfur and fluorine.

Chemistry
1 answer:
olga nikolaevna [1]3 years ago
5 0
What's your question......................
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A weighed piece of magnesium ribbon is added to a dried crucible, which is reweighed and heated in air to form the compound MgO.
olga nikolaevna [1]

Answer:

One of the errors for low percentage of magnesium could be because not all the magnesium may have reacted.

Explanation:

During the heating process, if the magnesium have not reacted completely, it can lead to low percentage of magnesium in the oxide formed. The product may still look a bit greyish rather than whitish after the heating process.

4 0
3 years ago
The class of organic compound below contains a carbonyl group as a part of its structure?
kolbaska11 [484]

Aldehydes and Ketones. Aldehydes and ketones are classes of organic compounds that contain a carbonyl (C=O) group.

that's the correct answer

can I get brainilest

5 0
2 years ago
Which atom would form a cation?<br> F<br> Na<br> N<br> P
sineoko [7]

Answer:

I believe Na

Explanation:

Copper, because it is the only metal out of all of them.

Therefore copper is the only element that can loose electrons to have a positive charge, it is the most likely to become a cation.

5 0
3 years ago
A sample of an unknown metal has a mass of 58.932g. it has been heated to 101.00 degrees C, then dropped quickly into 45.20 mL o
yaroslaw [1]
<h3>Answer:</h3>

0.111 J/g°C

<h3>Explanation:</h3>

We are given;

  • Mass of the unknown metal sample as 58.932 g
  • Initial temperature of the metal sample as 101°C
  • Final temperature of metal is 23.68 °C
  • Volume of pure water = 45.2 mL

But, density of pure water = 1 g/mL

  • Therefore; mass of pure water is 45.2 g
  • Initial temperature of water = 21°C
  • Final temperature of water is 23.68 °C
  • Specific heat capacity of water = 4.184 J/g°C

We are required to determine the specific heat of the metal;

<h3>Step 1: Calculate the amount of heat gained by pure water</h3>

Q = m × c × ΔT

For water, ΔT = 23.68 °C - 21° C

                       = 2.68 °C

Thus;

Q = 45.2 g × 4.184 J/g°C × 2.68°C

    = 506.833 Joules

<h3>Step 2: Heat released by the unknown metal sample</h3>

We know that, Q =  m × c × ΔT

For the unknown metal, ΔT = 101° C - 23.68 °C

                                              = 77.32°C

Assuming the specific heat capacity of the unknown metal is c

Then;

Q = 58.932 g × c × 77.32°C

   = 4556.62c Joules

<h3>Step 3: Calculate the specific heat capacity of the unknown metal sample</h3>
  • We know that, the heat released by the unknown metal sample is equal to the heat gained by the water.
  • Therefore;

4556.62c Joules = 506.833 Joules

c = 506.833 ÷4556.62

  = 0.111 J/g°C

Thus, the specific heat capacity of the unknown metal is 0.111 J/g°C

8 0
3 years ago
A representative particle of carbon dioxide is
KonstantinChe [14]
Carbon atom with iron and helium
6 0
2 years ago
Read 2 more answers
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