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alukav5142 [94]
3 years ago
11

Search...

Chemistry
2 answers:
denis23 [38]3 years ago
6 0
Answer: The answer is the second one
Ira Lisetskai [31]3 years ago
6 0
49484477337282828383838292
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How much water must be added to 6.0 M silver nitrate in order to make 500 mL of 1.2 M solution?
algol [13]

The amount of water that must be added to 6.0 M silver nitrate to make 500mL of 1.2 M solution is : 2000 mL

<u>Given data :</u>

Concentration of siilver nitrate ( M₁ ) = 6.0 M

volume of solution ( V₁ ) = 500 mL

Conc of solution ( M₂ )= 1.2 M

<h3>Determine the amount of water that must be added</h3>

we will apply the equation below

M₁V₁ = M₂V₂ ---- ( 1 )

where : V₂ = V₁ + water added  ---- ( 2 )

V₂ ( Final volume ) = ( M₁V₁ ) / M₂

                              = ( 6 * 500 ) / 1.2

                              = 2500 mL

Back to eqaution ( 2 )

2500 mL = 500 mL + added water

therefore ; added water = 2500 - 500

                                        = 2000 mL

Hence we can conclude that The amount of water that must be added to 6.0 M silver nitrate to make 500mL of 1.2 M solution is : 2000 mL.

Learn more about Volume : brainly.com/question/12410983

#SPJ1

7 0
2 years ago
How many kilograms of water must be added to 6.07 grams of oxalic acid (H2C2O4) to make a 0.025 m solution?
uranmaximum [27]
This dilution problem uses the equation
M
a
V
a
=
M
b
V
b

M
a
= 6.77M - the initial molarity (concentration)
V
a
= 15.00 mL - the initial volume
M
b
= 1.50 M - the desired molarity (concentration)
V
b
= (15.00 + x mL) - the volume of the desired solution
(6.77 M) (15.00 mL) = (1.50 M)(15.00 mL + x )
101.55 M mL= 22.5 M mL + 1.50x M
101.55 M mL - 22.5 M mL = 1.50x M
79.05 M mL = 1.50 M
79.05 M mL / 1.50 M = x
52.7 mL = x
59.7 mL needs to be added to the original 15.00 mL solution in order to dilute it from 6.77 M to 1.50 M.
I hope this was helpful.
4 0
2 years ago
Buffer consists of undissociated acid (ha) and the ion made by dissociating the acid (a-). How does this system buffer a solutio
docker41 [41]

In buffer solution there is an equilibrium between the acid  HA and its conjugate base A⁻: HA(aq) ⇌ H⁺(aq) + A⁻(aq).

When acid (H⁺ ions) is added to the buffer solution, the equilibrium is shifted to the left, because conjugate base (A⁻) reacts with hydrogen cations from added acid, according to Le Chatelier's principle: H⁺(aq) + A⁻(aq) ⇄ HA(aq). So, the conjugate base (A⁻) consumes some hydrogen cations and pH is not decreasing (less H⁺ ions, higher pH of solution).

A buffer can be defined as a substance that prevents the pH of a solution from changing by either releasing or absorbing H⁺ in a solution.

Buffer is a solution that can resist pH change upon the addition of an acidic or basic components and it is able to neutralize small amounts of added acid or base, pH of the solution is relatively stable


3 0
3 years ago
5. Write the names of the following ions.
Talja [164]
A.) Phosphate ion or Orthophosphate
d.) Hydroxide
D.) Ammonium
e.) Iron
C.) Nitrate
f.) Sulfur dioxide
5 0
2 years ago
Iodine would have chemical properties most like?
mrs_skeptik [129]
<span>manganese (Mn)
.tellurium (Te)
.chlorine (Cl).
<span>xenon (Xe).</span></span>
6 0
2 years ago
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