2. 3H2 + N2 means you have 6 Hs and 2 Ns. NH3 has one N and 3 Hs, so you need 2 NH3s in order to have the 2 and 6 of each that you need on both sides of the reaction.
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:
oook
hi I am so sorry sorry sorry sorry I don't no answer
Explanation:
but you follow me and give me brainliest ok by by by by
Answer:
Explanation:
A. Linear Relationships
x 2 4 6 8 10 12
y 4.4 8.8 13.2 17.6 22 26.4
(you can plot this by using the (x,y) coordinates for each pair above.)
B. y=2.2x
mass is 2.2 times larger than the volume.
C. The mass is 2.2 times the volume.
Answer:
1.30464 grams of glucose was present in 100.0 mL of final solution.
Explanation:
Moles of glucose = 
Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)
Molarity of the solution = 
A 30.0 mL sample of above glucose solution was diluted to 0.500 L:
Molarity of the solution before dilution = 
Volume of the solution taken = 
Molarity of the solution after dilution = 
Volume of the solution after dilution= 
Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:
Volume of solution = 100.0 mL = 0.1 L
Moles of glucose = 
Mass of 0.007248 moles of glucose :
0.007248 mol × 180 g/mol = 1.30464 grams
1.30464 grams of glucose was present in 100.0 mL of final solution.
