Answer:
0.00676 M
Explanation:
A chemist prepares a solution of calcium bromide by weighing out 0.607g of calcium bromide into a 450ml volumetric flask and filling the flask to the mark with water. Calculate the concentration in mol/L of the chemist's calcium bromide solution. Be sure your answer has the correct number of significant digits.
Step 1: Given data
Mass of calcium bromide (solute): 0.607 g
Volume of solution: 450 mL
Step 2: Calculate the moles corresponding to 0.607 g of calcium bromide
The molar mass of CaBr₂ is 199.89 g/mol.
0.607 g × 1 mol/199.89 g = 0.00304 mol
Step 3: Convert the volume of solution to liters
We will use the conversion factor 1 L = 1000 mL.
450 mL × 1 L/1000 mL = 0.450 L
Step 4: Calculate the molar concentration of calcium bromide
The molarity of the solution is:
M = moles of solute / liters of solution
M = 0.00304 mol / 0.450 L
M = 0.00676 M
Answer:
No.
Explanation:
During chemical reaction, atomes cannot be created or destroyed, they can only react together to form <em>E</em><em>l</em><em>e</em><em>m</em><em>e</em><em>n</em><em>t</em><em> </em>or <em>C</em><em>o</em><em>m</em><em>p</em><em>o</em><em>u</em><em>n</em><em>d</em><em> </em>at the <em>P</em><em>r</em><em>o</em><em>d</em><em>u</em><em>c</em><em>t</em><em> </em>side.
The universe<span> today is observed to contain one helium atom for every ten or eleven atoms of </span><span>hydrogen</span>
Solution :
It is given that :
Weight of the antacid tablet = 5.4630 g
4.3620 gram of antacid is crushed and is added to the stomach acid of 200 mL and is reacted.
25 mL of the stomach acid that is partially neutralized required 13.6 mL of NaOH to be titrated for a red end point.
27.7 mL of 
solution is equivalent to 
of the original stomach acid. Therefore, 13.6 mL of NaOH will take x 
= 12.27 ml of the original stomach acid.
