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lina2011 [118]
3 years ago
8

Consider the development of atomic theory. The last piece of the atomic puzzle was the discovery of

Chemistry
1 answer:
Ray Of Light [21]3 years ago
7 0

Answer:

The answer will be B

Explanation:

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If a sample of NH3 gas at 689,1 mmHg has a volume of 607 2 mL and the presure is
never [62]

Answer:

The answer to your question is V2 = 746.1 ml

Explanation:

Data

Pressure 1 = P1 = 689.1 mmHg

Volume 1 = V1 = 607.2 ml

Pressure 2 = P2 = 560.8 mmHg

Volume 2 = V2 = ?

Process

To solve this problem use Boyle's law

              P1V1 = P2V2

-Solve for V2

              V2 = P1V1/P2

-Substitution

               V2 = (689.1 x 607.2) / 560.8

-Simplification

                V2 = 746.1 ml

8 0
3 years ago
Choose the aqueous solution below with the lowest freezing point. These are all solutions of nonvolatile solutes and you should
atroni [7]

Answer:

(NH_4)_3PO_4 0.075 m solution has the lowest freezing point.

Explanation:

Depression in freezing point is given by:

\Delta T_f=T-T_f

\Delta T_f=K_f\times m

\Delta T_f=iK_f\times \frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in Kg}}

where,

\Delta T_f =Depression in freezing point

K_f = Freezing point constant of solvent

1 - van't Hoff factor

m = molality

According question, molality of all the solutions are same and are in prepared with same solvent. So, values of molality and K_f will remain the same and will not effect the freezing point of the solution.

The lowering in freezing point will now depend upon van't Hoff factors of the solutions. Higher the value of van'Hoff factor more will be the lowering in freezing point of the solution.

The van't Hoff factor of KNO_2 solution = i_1=2

The van't Hoff factor of LiCN solution = i_2=2

The van't Hoff factor of (NH_4)_3PO_4 solution = i_3=4

The van't Hoff factor of NaI solution = i_4=2

The van't Hoff factor of NaBrO_3 solution = i_5=2

The solution of ammonium phosphate has the highest values of van't Hoff factor which will result in maximum lowering of the freezing point of the solution.

Hence,(NH_4)_3PO_4 0.075 m solution has the lowest freezing point.

4 0
3 years ago
Read 2 more answers
The total number of each element is not changed during a chemical reaction
Paha777 [63]

false all element when they are combined eachother thier valance number will change

the combination of oxygen and aluminum gives us Al2O3 see even the numbers are changing thank you if you have a comment you can write

3 0
3 years ago
Assessment
MakcuM [25]

i didn’t mean to put this opps

4 0
3 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
3 years ago
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