Answer:
There is 17,114825 g of powdered drink mix needed
Explanation:
<u>Step 1 :</u> Calculate moles
As given, the concentration of the drink is 0.5 M, this means 0.5 mol / L
Since the volume is 100mL, we have to convert the concentration,
⇒0.5 / 1 = x /0.1 ⇒ 0.5* 0.1 = x = 0.05 M
This means there is 0.05 mol per 100mL
e
<u>Step 2 </u>: calculate mass of the powdered drink
here we use the formula n (mole) = m(mass) / M (Molar mass)
⇒ since powdered drink mix is usually made of sucrose (C12H22O11) and has a molar mass of 342.2965 g/mol.
0.05 mol = mass / 342.2965 g/mol
To find the mass, we isolate it ⇒0.05 mol * 342.2965 g/mol = 17,114825g
There is 17,114825 g of powdered drink mix needed
Answer is: the hydronium ion concentratio is 1.71×10⁻⁷ mol/dm³ and pH<6.76.
The Kw (the ionization constant of water) at 40°C is 2.94×10⁻¹⁴ mol²/dm⁶ or 2.94×10⁻¹⁴ M².
Kw = [H₃O⁺] · [OH⁻].
[H₃O⁺] = [OH⁻] = x.
Kw = x².
x = √Kw.
x = √2.94×10⁻¹⁴ M².
x = [H₃O⁺] = 1.71×10⁻⁷ M; concentration of hydronium ion.
pH = -log[H₃O⁺].
pH = -log(1.71×10⁻⁷ M).
pH = 6.76.
pH (potential of hydrogen) is a numeric scale used to specify the acidity or basicity an aqueous solution.
Gain or lose.
The exchange of electrons in chemical bonding seeks to fulfill the octet rule. There are some exceptions, such as with hydrogen and helium, whose valence shells have a capacity of two electrons.
Answer: I believe the answer you are looking for is 0.02180
44. (a) N2O3 (b) SF4 (c) AlCl3 (d) Li2CO3
46. H Br
δ+ δ−
48. The metallic potassium atoms lose one electron and form +1 cations,
and the nonmetallic fluorine atoms gain one electron and form –1 anions.
K → K+
+ e–
19p/19e–
19p/18e–
F + e–
→ F–
9p/9e–
9p/10e–
The ionic bonds are the attractions between K+
cations and F–
anions.
50. See Figure 3.6.
52. (a) covalent…nonmetal-nonmetal (b) ionic…metal-nonmetal
54. (a) all nonmetallic atoms - molecular (b) metal-nonmetal - ionic
56. (a) 7 (b) 4
58. Each of the following answers is based on the assumption that nonmetallic
atoms tend to form covalent bonds in order to get an octet (8) of
electrons around each atom, like the very stable noble gases (other than
helium). Covalent bonds (represented by lines in Lewis structures) and lone
pairs each contribute two electrons to the octet.
(a) oxygen, O
If oxygen atoms form two covalent bonds, they will have an octet of electrons
around them. Water is an example:
H O H
(b) fluorine, F
If fluorine atoms form one covalent bond, they will have an octet of electrons
around them. Hydrogen fluoride, HF, is an example:
H F
(c) carbon, C
If carbon atoms form four covalent bonds, they will have an octet of electrons
around them. Methane, CH4, is an example:
H H
H
H
C
(d) phosphorus, P
If phosphorus atoms form three covalent bonds, they will have an octet
