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
Answer:
B 24 m west I belive that it is the answer
Answer: 287.8 cm3
Explanation:
Given that:
Initial volume of gas V1 = 350 cm3
Initial pressure of gas P1 = 740 mmHg
New volume V2 = ?
New pressure P2 = 900 mmHg
Since, pressure and volume are involved while temperature is constant, apply the formula for Boyle's law
P1V1 = P2V2
740 mmHg x 350 cm3 = 900mmHg x V2
V2 = (740 mmHg x 350 cm3) /900mmHg
V2 = 259000 mmHg cm3 / 900mmHg
V2 = 287.8 cm3
Thus, the gas will occupy 287.8 cubic centimeters at the new pressure.
Answer:
<h2>B i have to search it like 23 hours to find it but Thank you for sharing your question :)</h2>
Answer:
18.5 years
Explanation:
(6 x 10^23)/(10^15 x 1.03 x 3600 x 24 x 365)
Hope this helps!
