Answer:
The reaction rate of the both questions remain unchanged.
Explanation:
For question 1: The reaction 1-iodo -2- methylbutane with cyanide ion is an SN2 reaction because the Alkyl halide is a primary alkyl halide. The rate of reaction is dependent on concentration of the nucleophile and the alkyl halide at the same. For the rate of reaction to be affected (increased or decreased), the concentration of nucleophile and the alkyl halide have to be altered.
For question 2: The reaction of 2-iodo -2- methylbutane with ethanol is an SN1 reaction because the Alkyl halide is a tertiary alkyl halide. There are two-step reaction mechanism in this reaction. The first step is the rate determining step which determines the extent of the reaction and hence the rate of reaction. For the rate of reaction to be affected (increased or decreased), the concentration of the Alkyl halide alone will be altered. The rate of reaction is independent of the concentration of the nucleophile.
The correct answers are ,
A) C
B) N
C) Ti
D) Zn
E) Fe
F) Phosphorus
G)Calcium
H) Helium
I) Lead
J) Silver
<h3>How are elements named?</h3>
Elements have been given names based on a variety of factors, <u>including their characteristics</u>, the compound or ore from which they were extracted, the method by which they were found or acquired, mythical characters, locations, and well-known individuals. Some components have <u>names that are descriptive and are based on one of their attributes.</u>
The International Union of Pure and Applied Chemistry chooses the official element names and symbols (IUPAC). However, different nations frequently use similar names and symbols for elements. Official names and symbols for elements are not given until after their discovery has been confirmed. The discoverer may then suggest a name and a symbol.
There are name standards for several element groupings. Names of halogens end in -ine. All noble gas names, save helium, end in -on. The names of most other elements finish with -ium.
To learn more about elements:
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Answer:
2.1 × 10⁻¹ M
2.0 × 10⁻¹ m
Explanation:
Molarity
The molar mass of aniline (solute) is 93.13 g/mol. The moles corresponding to 3.9 g are:
3.9 g × (1 mol/93.13 g) = 0.042 mol
The volume of the solution is 200 mL (0.200 L). The molarity of aniline is:
M = 0.042 mol/0.200 L = 0.21 M = 2.1 × 10⁻¹ M
Molality
The moles of solute are 0.042 mol.
The density of the solvent is 1.05 g/mL. The mass corresponding to 200 mL is:
200 mL × 1.05 g/mL = 210 g = 0.210 kg
The molality of aniline is:
m = 0.042 mol/0.210 kg = 0.20 m = 2.0 × 10⁻¹ m
The volume of oxygen required to burn 12.00 L ethane is calculated as follows
find the moles of C2H6 used
At STP 1 mole is always = 22.4 L, what about 12.00 L
= ( 12.00L x 1 moles) 22.4 L = 0.536 moles
write the reacting equation
2C2H6+ 7O2 = 4CO2 + 6H2O
by use of mole ratio between C2H6 :O2 which is 2:7 the moles of O2
= 0.536 x7/2= 1.876 moles
again at STP 1mole = 22.4 L what about 1.876 moles
= 22.4 L x 1.876 moles/ 1 mole = 42.02 L
Answer:
Option A.
2Na + 2H2O —> 2NaOH + H2
Explanation:
To know which option is correct, we shall do a head count of the number of atoms present on both side to see which of them is balanced. This is illustrated below below:
For Option A:
2Na + 2H2O —> 2NaOH + H2
Reactant >>>>>>> Product
2 Na >>>>>>>>>>> 2 Na
4 H >>>>>>>>>>>> 4 H
2 O >>>>>>>>>>>> 2 O
Thus, the above equation is balanced.
For Option B:
2Na + 2H2O —> NaOH + H2
Reactant >>>>>>> Product
2 Na >>>>>>>>>>> 1 Na
4 H >>>>>>>>>>>> 3 H
2 O >>>>>>>>>>>> 1 O
Thus, the above equation is not balanced.
For Option C:
2Na + H2O —> 2NaOH + H2
Reactant >>>>>>> Product
2 Na >>>>>>>>>>> 2 Na
2 H >>>>>>>>>>>> 4 H
1 O >>>>>>>>>>>> 2 O
Thus, the above equation is not balanced.
For Option D:
Na + 2H2O —> NaOH + 2H2
Reactant >>>>>>> Product
1 Na >>>>>>>>>>> 1 Na
4 H >>>>>>>>>>>> 5 H
2 O >>>>>>>>>>>> 1 O
Thus, the above equation is not balanced.
From the illustrations made above, only option A is balanced.
