Answer:
3.33 M
Explanation:
It seems your question is incomplete, however, that same fragment has been found somewhere else in the web:
" <em>A chemist prepares a solution of silver nitrate (AgNO3) by measuring out 85.g of silver nitrate into a 150.mL volumetric flask and filling the flask to the mark with water.</em>
<em>Calculate the concentration in mol/L of the chemist's silver nitrate solution. Be sure your answer has the correct number of significant digits.</em> "
In this case, first we <u>calculate the moles of AgNO₃</u>, using its molecular weight:
- 85.0 g AgNO₃ ÷ 169.87 g/mol = 0.500 mol AgNO₃
Then we<u> convert the 150 mL of the volumetric flask into L</u>:
Finally we <u>divide the moles by the volume</u>:
- 0.500 mol AgNO₃ / 0.150 L = 3.33 M
Answer:
A decrease in [H3O+] and an increase in pH (option a)
Explanation:
Equilibrium of water is shown in this equation
2H₂O ⇄ H₃O⁺ + OH⁻
When you add NaOH, you are modifying [OH⁻]
NaOH → Na⁺ + OH⁻
In equilibrium of water, the [OH⁻] increases
2H₂O ⇄ ↓ H₃O⁺ + OH⁻ ↑
As the [OH⁻] increases, by Le Chatellier, the equilibrium tends to decrease [H₃O⁺].
If the [OH⁻] is higher, pH is also high so the solution of water and sodium hydroxide would be totally basic.
Answer:
0.209M
Explanation:
M1V1=M2V2
(28.5 mL)(0.183M)=(25.0mL)(M)
M2= 0.209M
*Text me at 561-400-5105 for private tutoring if interested: I can do homework, labs, and other assignments :)
Answer:
The differemt isotopes that differ in atomic mass
Explanation:
Salt=compound, soda=liquid solution, aluminum foil=element, milk=colloid,
steel=solid solution
