very high concentration of hydroxide ions
slippery feeling
very low concentration of hydronium ions
Answer : The correct option is, sour taste
Explanation :
Acid : It is a substance that has ability of donating proton or hydrogen ion or hydronium ion, .
The properties of an acid are :
Base : It is a substance that has ability of donating hydroxide ion, .
The properties of an base are :
As per question, the pH is 5 that means the solution is acidic in nature and follow the properties of an acid. So, the pH = 5 has the sour taste characteristic. While the other options are the properties of a base.
Hence, the correct option is, sour taste
If you have to prepare the reactants by dissolving 1.00 g of BaCl₂ and 1.00 g of NaOH in water,
(a) What is the limiting reactant?
(b) How many grams of Ba(OH)₂ are produced?
(c) If your experiment produced 0.700 g of Ba(OH)₂, what is the percent yield of Ba(OH)₂?
(d) Based on this percent yield, how much limiting reactant should be used to achieve the target Ba(OH)₂ theoretical yield?
Answer:
The answer to your question is:
a) BaCl2
b) 0.8208 g
c) yield = 85.3 %
d)
Explanation:
BaCl₂(aq) + NaOH(aq) ----> Ba(OH)₂(s) + 2NaCl(aq)
Data
a) 1 g of BaCl₂
1 g of NaOH
MW BaCl2 = 137 + (35.5x2) = 208 g
MW NaOH = 23 + 16 + 1 = 40 g
208 g of BaCl2 ------------- 1 mol
1 g of BaCl2 ------------- x
x = ( 1 x 1) / 208 = 0.0048 mol of BaCl2
40 g of NaOH ------------ 1 mol
1 g of NaOH ------------ x
x = (1 x 1) / 40
x = 0.025 mol of NaOH
The ratio BaCl2 to NaOH is 1:1 (in the equation)
But experimentally we have 0.0048 : 0.025, so the limiting reactant is BaCl2, because is in lower concentration.
b)
1 mol of BaCl2 -------------- 1 mol of Ba(OH)2
0.0048 mol --------------- x
x = (0.0048 x 1) / 1
x = 0.0048 mol of Ba(OH)2
MW Ba(OH)2 = 137 + 32 + 2 = 171 g
171 g of Ba(OH)2 -------------------- 1 mol
x -------------------- 0.0048 mol
x = (0.0048 x 171) / 1
x = 0.8208 g
c)Data
Ba(OH)2 = 0.700 g
% yield = 0.700 / 0.8208 x 100
% yield = 85.3
d)
Sorry, i don't understand this question
2) empirical formulas
3) molecular structures
4) physical properties
Answer:
Name: Aluminum Sulfite
Explanation:
Systematic nomenclature: dialuminium tris [trioxosulfate (IV)]
Oxoanions: anions derived from removing H + from an oxo acid with a vulgar name:
(SO3)3^-2 (sulfite)
Metal cation according to valence number:
Al (III) or Al+3 (as a cation) (Aluminum)
There are different types of salts, in this case we have a simple salt and the nomenclature for binary substances is used when only one class of cation and one class of anion are present
So:
Al2(SO3)3: Aluminum Sulfite
Ions are charged particles that form when an atom (or group of atoms) receives or loses an electron, and ionic compounds are composed of these charged particles. An anion is a negatively charged ion, whereas a cation is a positively charged ion. Here Al₂(SO₃)₃ is aluminium sulfite.
Ionic compounds are defined as substances that are joined by ionic bonds. In order to reach their closest arrangement as a noble gas, elements can either gain or lose electrons. For the completion of the octet, ions are formed (either by gaining or losing electrons), which aids in their stabilization.
The name of the compound Al₂(SO₃)₃ is aluminium sulfite.
To know more about ionic compounds, visit;
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Answer:
Ka = 1.5 -10^-5
Explanation:
Step 1: Data given
Molarity of the solution = 0.25 M
pH = 2.71
Step 2 The equation
C3H7COOH + NaOH ⇆ C3H7COONa + H2O
Step 3: Calculate pKa
pH = (pKa-log[acid])/2
2pH = pKa -log[acid]
pKa = 2pH +log[acid]
⇒with pH = 2.71
⇒with log[acid] = -0.60
pKa=2*2.71 -0.60
pKa = 5.42 - 0.60
pKa = 4.82
Step 4: Calculate Ka
pKa = -log(Ka)
Ka = 10^-4.82
Ka = 1.5 -10^-5
2g of Copper 2 ions
Explanation:
96.485 columbs=1 faraday will
deposit 64/2g= 32 g cu ion
therfore it will require
96,485 ×2/32 =? coulombs or 1/16 of
Faraday= 1 / 16 mole of electrons .