Answer:
Elements can be separated into smaller atoms
Answer:
The answer is 1.06g.
Explanation:
Analysis of question:
1. Identify the information in the question given.
- volume of HCl is 2 dm3
- pH of HCl is 2.0
2. What the question want?
- mass of Na2CO3 is ?(unknown)
- 3. Do calculation.
- 1st-Write a balanced chemical equation:
Na2CO3 + 2HCl (arrow) 2NaCl + H20 + CO2
- 2nd-Determine the molarity of HCl with the value of 2.0.
pH= -log[H+]
2.0= -log[H+]
log[H+]= -2.0
[H+]= 10 to the power of negative 2(10-2)
=0.01 mol dm-3
molarity of HCl is 0.01 mol dm-3
- 3rd-Find the number of moles of HCl
n=MV
=0.01 mol dm-3 × 2 dm3
=0.02 mol of HCl
- 4th-Find the second mol of it.
Based on the chemical equation,
2.0 mol of HCl reacts with 1.0 mol of Na2CO3
0.02 mol of HCl reacts with 0.01 mol of Na2CO3
<u>N</u><u>a</u>2CO3>a=<u>1</u><u> </u>mol
<u>2</u><u>H</u>Cl>b=<u>2</u><u> </u>mol
mass= number of mole × molar mass
g=0.01 × [2(23)+ 12+ 3(16)]
g=0.01 × 106
# =1.06 g.
<h3><u>Answer</u>;</h3>
pH = 4.38
<h3><u>Explanation;</u></h3>
Consider that ;
[H3O+] = 10-pH mol/L
Therefore; the pH is the -log of the [hydronium ion].
pH = - log [H3O+]
Thus;
pH = - log (4.2 × 10^–5)
= 4.38
5 times dilution
0.780M x 1/5 = 0.156M
Hope this help.
Answer:inform the lab instructor and get instructions
Explanation:
If you come across a chemical in the laboratory which has been wrongly labelled, do not be quick to dilute it or take any further action. The laboratory instructor who may have prepared the reagent himself or has better knowledge about the reagent should be contacted immediately so that he/she can give you instructions about what to do with the wrongly labelled reagent.
