<span>100 g of KClO3 @ 122.55 g/mol = 0.816 moles of KClO3
by the reaction
2 KClO3 --> 2 KCl & 3 O2
0.816 moles of KClO3 @ 3 moles O2 / 2 moles KClO3 = 1.224 moles of O2 can be made
using molar mass
1.224 moles of O2 @ 32.0 g/mol =
39.2 grams of O2 can be made</span>
Answer:
The pH of the solution is 10,46.
Explanation:
The pH gives us an idea of the acidity or basicity of a solution. More precisely, it indicates the concentration of H30 + ions present in said solution. The pH scale ranges from 0 to 14: from 0 to 7 corresponds to acid solutions, 7 neutral solutions and between 7 and 14 basic solutions. It is calculated as:
pH = -log (H30 +)
pH= -log (3.5*10^-11)
<em>pH= 10,46</em>
Answer:
Electron-pair geometry: tetrahedral
Molecular geometry: trigonal pyramidal
Hybridization: sp³
sp³ - 4 p
Explanation:
There is some info missing. I think this is the original question.
<em>For NBr₃, What are its electron-pair and molecular geometries? What is the hybridization of the nitrogen atom? What orbitals on N and Br overlap to form bonds between these elements?</em>
<em>The N-Br bonds are formed by the overlap of the ___ hybrid orbitals on nitrogen with ___ orbitals on Br.</em>
<em />
Nitrogen is a central atom surrounded by 4 electron domains. According to VESPR, the corresponding electron-pair geometry is tetrahedral.
Of these 4 electron domains, 3 represent covalent bonds with Br and 1 lone pair. According to VESPR, the corresponding molecular geometry is trigonal pyramidal.
In the nitrogen atom, 1 s orbital and 3 p orbitals hybridize to form 4 sp³ orbitals for each of the electron domains.
The N-Br bonds are formed by the overlap of the sp³ hybrid orbitals on nitrogen with 4p orbitals on Br.
Answer:
E = 1053.365 J
Explanation:
∴ Q = 440μC * ( 1 E-6 C / μC ) = 4.4 E-4 C
∴ C = k*εo*A / d
∴ k mica = 5.4
∴ εo = 8.8542 E-12 C² / N.m²
∴ A = 6.2cm * 6.2 cm = 38.44 cm² * ( m/100cm)² = 3.844 E-3 m²
∴ d = 2mm * ( m / 1000mm ) = 2 E-3 m
⇒ C = (( 5.4 ) * (8.8542 E-12 ) * ( 3.844 E-3 )) / 2 E-3
⇒ C = 9.1896 E-11 C²/N.m
⇒ E = ( 4.4 E-4 )² / (2*9.1896 E-11)
⇒ E = 1053.365 N.m = 1053.365 J
Answer: see below
<u>Explanation:</u>
When multiplying or dividing, use the smallest number of significant figures (sf) of the digits provided.
37(a) 24 × 3.26 = 78.24 <em>2 sf × 3 sf </em>--> round to 2 sf's
≈ 78
(b) 120 × 0.10 = 12 <em>2 sf × 2 sf </em>--> round to 2 sf's
= 12
(c) 1.23 × 2.0 = 2.46 <em>3 sf × 2 sf </em>--> round to 2 sf's
= 2.5
(d) 53 × 1.53 = 81.09 <em>2 sf × 3 sf </em>--> round to 2 sf's
= 81
38(a) 4.84 ÷ 2.4 = 2.0167 <em>3 sf × 2 sf </em>--> round to 2 sf's
= 2.0
(b) 60.2 ÷ 20.1 = 2.995 <em>3 sf × 3 sf </em>--> round to 3 sf's
= 3.00
(c) 102.4 ÷ 51.2 = 2 <em>4 sf × 3 sf </em>--> round to 3 sf's
= 2.00
(d) 168 ÷ 58 = 2.896 <em>3 sf × 2 sf </em>--> round to 2 sf's
= 2.9