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:
616,0 ng is the right answer.
Explanation:
You should know that 1 mole = 1 .10^9 nanomoles
Get the rule of three.
1 .10^9 nanomoles ...................... 56.0 gr
11 nanomoles .....................
(11 x 56) / 1 .10^9 nanomoles = 6.16 x 10^-7 gr
Let's convert
6.16 x 10^-7 gr x 1 .10^9 = 616 ngr
Answer: If you think about it, B. would be the most reasonable answer with the given factors.
Neon I think. Go to the periodic table and see which one is the 11th
Answer:
d
Explanation:
pv=nrt
2.5×1.01×10^5×8×10^-3=3×8.31×T
T=
