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Oxana [17]
2 years ago
8

A laboratory procedure calls for making 590.0 mL of a 1.1 M KNO3 solution. How much KNO3 in grams is needed?

Chemistry
1 answer:
AveGali [126]2 years ago
8 0
590 mL = 590 cm³= 0,59 dm³

C = n/V
n = 1,1M × 0,59 dm³
n = 0,649 mol
_____________________________

M KNO₃ = 39g+14g+16g×3 = 101 g/mol

1 mole -------- 101g
0,649 --------- X
X = 101×0,649
X = 65,549g KNO₃

:)
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Dry trees, shrubs, and other vegetation as well as lightening strikes.
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How could you safely determine if a base is stronger than an acid? Compare the taste of the base and the acid. Use a conductivit
Gnoma [55]
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when a solution of sodium chloride is added to a solution of copper(ii) nitrate, no precipitate is observed. Write the molexular
zhuklara [117]

Explanation:

1.

Cu(NO3)2 + 2NaCl(aq) --> CuCl2(aq) + 2NaNO3(aq)

2.

Cu(NO3)2 + 2NaOH(aq) --> Cu(OH)2(s) + 2NaNO3(aq)

A light blue precipitate of Cu(OH)2 is formed and NaNO3 in solution.

3.

Cu(NO3)2(aq) --> Cu2+(aq) + 2NO3^-2(aq)

2NaOH(aq) --> 2Na+(aq) + 2OH-(aq)

Cu2+(aq) + 2OH-(aq) --> Cu(OH)2(aq)

2Na+(aq) + 2NO3^-2(aq) --> 2NaNO3(aq)

4.

The reaction in both Questions 1 and 2 is called Double displacement reaction. A double-replacement reaction exchanges the cations and/or or the anions of two ionic compounds. A precipitation reaction is a double-replacement reaction in which one product is a solid precipitate (precipitated) while the other in solution.

Since the cation and anions in Qustion 1 were exchanged, the same was done for Question 2, hence the identity of the precipitate in Question 2 was got.

6 0
3 years ago
If the pressure in the room is 759.2 torr and the vapor pressure of water is 23.8 torr, what is the pressure of hydrogen gas in
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The partial pressure of Hydrogen gas can directly be calculated by simply taking the difference of the overall pressure and the vapour pressure of water. That is:

 

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6 0
3 years ago
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If we start with 8000 atoms of radium-226 how much would remain after 3200 years?
olya-2409 [2.1K]
<h3>Answer:</h3>

2000 atoms

<h3>Explanation:</h3>

We are given the following;

Initial number of atoms of radium-226 as 8000 atoms

Time taken for the decay 3200 years

We are required to determine the number of atoms that will remain after 3200 years.

We need to know the half life of Radium

  • Half life is the time taken by a radio active material to decay by half of its initial amount.
  • Half life of Radium-226 is 1600 years
  • Therefore, using the formula;

Remaining amount = Original amount × 0.5^n

where n is the number of half lives

n = 3200 years ÷ 1600 years

 = 2

Therefore;

Remaining amount = 8000 atoms × 0.5^2

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Thus, the number of radium-226 that will remain after 3200 years is 2000 atoms.

6 0
3 years ago
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