It can be easily judged that only the first reaction is an acid base reaction among the three given in the question. So, we can avoid the other two reactions given in the question. Now let us focus and write down the balanced chemical equation of <span>P-Toluidine + HCl.
</span><span>C7H9N + HCl = C7H10N (+) + Cl (-)
</span>
I hope the answer has come to your help.
Answer:
moles H₂O = 10
Explanation:
The mass of Na₂CO₃⋅xH₂O is 3.837 g and the mass of Na₂CO₃ is 1.42g
Therefore the mass of xH₂O is 3.837 - 1.42 = 2.417 g
The molar mass of Na₂CO₃ is 106 g/mol and for H₂O is 18 g/mol
The moles of Na₂CO₃ and H₂O in the sample are:
Na₂CO₃ = 1.42 / 106 = 0.01340 moles
H₂O = 2.417 / 18 = 0.1343
Now using rule of three :
1 mole of Na₂CO₃ has x moles of H₂O
0.01340 moles of Na₂CO₃ has 0.1343 moles of H₂O
x = 1 * 0.1343 / 0.01340 = 10
In this item, we are simply to find the ions that may bond and are able to form a formula unit. We are also instructed to give out their name. There are numerous possible combinations of ions to form a compound. Some answers are given in the list below.
1. Na⁺ , Cl⁻ , NaCl ---> sodium chloride (this is most commonly known as table salt)
2. C⁴⁺ , O²⁻ , CO₂ ---> carbon dioxide
3. Al³+ , Cl⁻ , AlCl₃ ----> aluminum chloride
4. Ca²⁺ , Cl⁻ , CaCl₂ ---> calcium chloride
5. Li⁺ , Br⁻ , LiBr ---> lithium bromide
6. Mg³⁺ , O²⁻ , Mg₂O₃ ----> magnesium oxide
7. K⁺ , I⁻ , KI ---> potassium iodide
8. H⁺ , Cl⁻ , HCl --> hydrogen chloride
9. H⁺ , Br⁻ , HBr ----> hydrogen bromide
10. Na⁺ , Br⁻ , NaBr ---> sodium bromide
Answer:
1.7 × 10 ^42
Explanation:
Using Nernst equation
E°cell = RT/nF Inq
at equilibrium
Q=K
E°cell = 0.0257 /n Ink= 0.0592/n log K
Fe2+(aq)+2e−→Fe(s) E∘= −0.45 V
Ag+aq)+e−→Ag(s) E∘= 0.80 V
Fe(s)+2Ag+(aq)→Fe2+(aq)+2Ag(s)
balance the reaction
Fe → Fe²⁺ + 2e⁻ reversing for oxidation E° = 0.45 v
2 Ag⁺ +2e⁻ → 2Ag
n = 2 moles and K = equilibrium constant
E° cell = 0.80 + 0.45 = 1.25 V
E° cell = (0.0592 / n) log K
substitute the value into the equations and solve for K
(1.25 × 2) / 0.0592 = log K
42.23 = log K
k = 10^ 42.23
K = 1.7 × 10 ^42
<em>Same group element have same</em><em><u> Valence electron</u></em><em> and behave similarly in </em><em><u>Chemistry.</u></em>
<u>Explanation:</u>
For example. First group elements Alkali metals:- H, Li, K, Rb, Cs, Fr
Valance electron will take part in forming a bond with other elements and compound will form. All the above-given elements (H-Fr) have valence electron 1 in outer most 'S' shell. All have electronic configuration S1
Behavior: Since valence electrons are the same so the behavior of all the elements in this group is the same. All are metal (from Li-Fr, except Hydrogen), all are very reactive, does not found in native state in the environment, and all react with water.