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notsponge [240]
2 years ago
4

. boiling 21.1 g of water at 100°C

Chemistry
2 answers:
mr Goodwill [35]2 years ago
8 0

Answer:

dont understand

Explanation:

Step2247 [10]2 years ago
3 0

How many joules are required to boil 21.1 g of water at 100°C?

Awnser:47,700 J  

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Metals are considered this if they can be made into sheets
Korvikt [17]

They are considered malleable. They can be made into sheets

Happy to help! Please mark me as the brainliest!

6 0
2 years ago
Which runner has greater kinetic energy: a 46-kilogram runner moving at a speed of 8 meters per second or a 92-kilogram runner m
Lena [83]
Runner a has a greater kinetic energy
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2 years ago
If 34.7 g of AgNO₃ react with 28.6 g of H₂SO₄ according to this UNBALANCED equation below, how many grams of Ag₂SO₄ could be for
Luba_88 [7]

The  number  of grams   of Ag2SO4  that could be formed  is   31.8  grams



    <u><em> calculation</em></u>

Balanced   equation is  as below

2 AgNO3 (aq)  + H2SO4(aq)  →  Ag2SO4 (s)   +2 HNO3 (aq)


  • Find  the  moles  of  each reactant by use  of  mole= mass/molar mass  formula

that is  moles of  AgNO3= 34.7 g / 169.87  g/mol= 0.204 moles

             moles of  H2SO4 =  28.6  g/98  g/mol  =0.292  moles

  • use the  mole  ratio to determine the moles of  Ag2SO4

   that is;

  •    the mole ratio of  AgNo3 : Ag2SO4 is  2:1 therefore  the  moles of Ag2SO4=  0.204  x1/2=0.102 moles

  • The moles  ratio of H2SO4  : Ag2SO4  is  1:1  therefore  the moles of Ag2SO4 = 0.292  moles

 

  •      AgNO3  is the limiting reagent therefore  the moles of   Ag2SO4 = 0.102  moles

<h3>     finally  find  the mass  of Ag2SO4  by use of    mass=mole  x molar mass  formula</h3>

that  is  0.102   moles  x  311.8  g/mol= 31.8 grams

3 0
2 years ago
Consider a beaker of water sitting on the pan of an electronic scale that has been tared. A metal weight hanging from a string i
natka813 [3]

Answer:

See explanation below for answers

Explanation:

We know that the balance is tared, so the innitial weight would be zero. Now, let's answer this by parts.

a) mass of displaced water.

In this case all we need to do is to substract the 0.70 with the 0.13 g. so:

mW = 0.70 - 0.13

mW = 0.57 g of water

b) Volume of water.

In this case, we have the density of water, so we use the formula for density and solve for volume:

d = m/V

V = m/d

Replacing:

Vw = 0.57/0.9982

Vw = 0.5710 mL of water

c) volume of the metal weight

In this case the volume would be the volume displaced of water, which would be 0.5710 mL

d) the mass of the metal weight.

In this case, it would be the mass when the metal weight hits the bottom which is 0.70 g

e) density.

using the above formula of density we calculate the density of the metal

d = 0.70 / 0.5710

d = 1.2259 g/mL

4 0
2 years ago
Ancient metalworkers in East Africa knew the secrets of strengthening their iron. They knew that if they melted their iron in a
Ilia_Sergeevich [38]

Answer is: in steel, the solvent is iron and the solute is the carbon.

Solution is homogeneous mixture composed of two or more substances.

In the solution, the solute molecules interact with solvent molecules.

Solvent (usually a liquid, but can also be a solid or a gas) is a substance that dissolves a solute.  Iron is solvent because there is more iron than carbon and they are both solid state.

Solvent is always the same state of matter as the solution.

In ideal solution  the forces of attraction between the solute-solute and the solvent-solvent and solute-solvent molecules are the same.

7 0
2 years ago
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