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3 years ago

DDT is currently banned worldwide. a. True b. False

1 answer:

5 0

B. False. It is banned worldwide for agricultural uses. The <span>use of DDT is still authorized in small proportions in countries that need it, with support deployed for the alteration to safer and more effective alternatives.</span>

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The expected o-c-h angle in this molecule is degrees. the expected hybridization at the central carbon is

WINSTONCH [101]

Each neutral carbon atom contains four valence electrons and may form up to four electron domains. Possible hybridizations include

$sp^{3}$ , four electron domains, as in ethane $\text{C}_2\text{H}_6$
$sp^{2}$ , three electron domains, as in ethene $\text{C}_2\text{H}_4$
$sp$ , two electron domains, as in ethyne $\text{C}_2\text{H}_2$

Molecules of each of the three hybridization demonstrate spatial configurations that would maximizes the separation between the electron domains.

Carbon atoms with a $sp^{3}$ hybridization would demonstrate a tetrahedral configuration with a bond angle of approximately $109.5\textdegree{}$
Carbon atoms with a $sp^{2}$ hybridization would demonstrate a triangular planar configuration with a bond angle of $120\textdegree{}$
Carbon atoms with a $sp$ hybridization would demonstrate a linear configuration with a bond angle of $180\textdegree{}$

Bond angles are characteristic of the spatial configuration of electron domains and identifies the hybridization of the central carbon atom.

Note that each hydrogen atom contains only one valence electron and would form only single bonds. It takes two valence electrons for oxygen atoms to achieve an octet such that each oxygen form only two bonds at a single time. Therefore given the fact that the carbon is bonded to both hydrogen and oxygen, only the following hybridizations are possible

$sp^{3}$ in which the oxygen atom forms a carbon-oxygen double bond with the central carbon atom;
$sp^{2}$ in which the oxygen atom forms a single bond with the central carbon atom and with a second atom.

3 0

3 years ago

The concentration of pb2+ in a commercially available standard solution is 1.00 mg/ml. what volume of this solution should be di

Vanyuwa [196]

<span>The concentration of pb2+ = 1.00mg/ml Diluted Solution is 6.0 x 102 ml = 612 ml Volume of the concentration of pb2+ is 0.054 mg/l is v (vL)(1.00mg/ml) = (.612L)(0.054mg/l) Volume = 0.033048L Volume of the concentration of pb2+ is 0.054 mg/l = 33.048 ml.</span>

3 0

3 years ago

Give the Unabbreviated electron configuration for barium

Shalnov [3]

Answer:

Ba: 1s² 2s²2p⁶ 3s²3p⁶ 4s²3d¹⁰4p⁶ 5s²4d¹⁰5p⁶ 6s²

Step-by-step explanation:

Step 1. Locate barium in the Periodic Table.

It's in Period 6, Group 2: Element 56 (highlighted blue in the Periodic Table below).

Step 2. Add 54 electrons to the energy levels

You add then in the order shown in the diagram below.

The complete electron configuration is:

Ba: 1s² 2s²2p⁶ 3s²3p⁶ 4s²3d¹⁰4p⁶ 5s²4d¹⁰5p⁶ 6s²

n = 2 + 8 + 8 + 18 + 18 + 2 = 56

8 0

3 years ago

A sample of gas originally at 25 C and 1.00 atm pressure in a 2.5 L container is allowed to expand until the pressure is 0.85 at

aivan3 [116]

Answer:

Final volumeof the gas = 2.84 L

Explanation:

The formular to be used here is the general gas equation. the formular is being used because it gives the relationship between the three gas parameters (volume, temperature and pressure) mentioned.

The general gas equation is given as;

$\frac{P1V1}{T1} = \frac{P2V2}{T2}$

where;

P1 = initial pressure

V1 = initial volume

T1 = initial temperature

P2 = Final pressure

V2 = Final volume

T2 = Final temperature

From the question,

P1 = 1.00 atm

P2 = 0.85atm

T1 = 25C + 273 = 298K (Converting to kelvin)

T2 = 15C + 273 = 288K (Converting to kelvin)

V1 = 2.5L

V2 = ?

from the equation, making V2 subject of formula we have;

$V2 = \frac{T2P1V1}{T1P2}$

V2 = (1*2.5*288)/(298*0.85) = 2.84 L.

5 0

3 years ago

On a mission to a newly discovered planet, an astronaut finds chlorine abundances of 13.85 % for 35Cl and 86.15 % for 37Cl. What

diamong [38]

Answer:

36.693

Explanation:

From the question given, the following were obtained:

Let 35Cl be isotope A and 37Cl be isotope

For isotope A (35Cl),

Mass number = 34.9700

Abundance = 13.85 %

For isotope B (37Cl),

Mass number = 36.9700

Abundance = 86.15 %

Atomic mass =

[( Mass of AxAbundance A)/100] + [(Mass of BxAbundance B)/100]

[(34.97x13.85)/100] + [(36.97x86.15)/100]

= 36.693

6 0

3 years ago