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.
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Answer:
When the two gases are mixed, the ammonium chloride precipitates in the tube walls.
Explanation:
This is the reaction:
HCl (g) + NH₃(g) → NH₄Cl (s) ↓
As the product formed is solid at room temperature, a suspension is first formed in the internal air of the tube that appears as a cloud. Afterwards it finally precipitates into the walls forming a white layer
A radioactive element has an unstable nucleus that emits particles in the form of alpha, beta, or gamma radiation. A stable element has a nucleus that does not emit such particles
Answer:
15.35 g of (NH₄)₃PO₄
Explanation:
First we need to look at the chemical reaction:
3 NH₃ + H₃PO₄ → (NH₄)₃PO₄
Now we calculate the number of moles of ammonia (NH₃):
number of moles = mass / molecular wight
number of moles = 5.24 / 17 = 0.308 moles of NH₃
Now from the chemical reaction we devise the following reasoning:
if 3 moles of NH₃ are produce 1 mole of (NH₄)₃PO₄
then 0.308 moles of NH₃ are produce X moles of (NH₄)₃PO₄
X = (0.308 × 1) / 3 = 0.103 moles of (NH₄)₃PO₄
mass = number of moles × molecular wight
mass = 0.103 × 149 = 15.35 g of (NH₄)₃PO₄
