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Goshia [24]
2 years ago
9

Help me please Will give brainliest if correct (No links or spam)

Chemistry
2 answers:
Paraphin [41]2 years ago
7 0

Answer:

Dip one end of the blue litmus paper into the solution take it out if it turns red then its acidic.

silver

Hydrogen gas

carbon dioxide

beryllium

magnesium chloride

Sorry if any are wrong i tried

PtichkaEL [24]2 years ago
6 0

Answer:

Dip one end of the blue litmus paper into the solution take it out if it turns red then its acidic.

silver

Hydrogen gas

carbon dioxide

beryllium

magnesium chloride

Sorry if any are wrong i tried

Explanation:

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Why all samples of a given substance have the same intensive properties?
Karolina [17]

Answer:

Every sample of a given substance has identical intensive properties because every sample has the same composition

4 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
SI unit for measuring distance is
MrMuchimi
The SI unit for measuring distance is the meter.
I attached a table of SI measurements for you :)
Hope this helped!

6 0
3 years ago
If you add salt continuously to a glass of water, eventually some salt will remain at the bottom. why?
GREYUIT [131]
Http://www.lcmrschooldistrict.com/demers/cbphysicalscience/Chp%208-3%20Solubility%20and%20Concentrat...

This website helped me understand the concept.
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3 years ago
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Express 1123 pg in nanograms.
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