In North America, where is population density the highest ?

Which of the following examples from the story "The Dentist" uses direct characterization to tell you how Lemon

padilas [110]

Answer: Don't know sorry

Explanation: And I-oop SkSk

8 0

3 years ago

BRAINLIESTTT ASAP!!! PLEASE HELP ME :)

Mrac [35]

Answer:

Energy level (n = 4), d subshell (l = 2)

Step-by-step explanation:

The number in 4d tells you the energy level of the subshell: n = 4.

The letter d gives you the secondary quantum number l.

Depending on its value, the letter gives you the shape of the orbital.

l = 0 corresponds to an s orbital (spherical)

l = 1 corresponds to a p orbital (dumb-bell shaped)

l = 2 corresponds to a d orbital (four four-leaf clovers + one that looks like a dumbbell with a doughnut around its middle)

5 0

3 years ago

Analysis of a 10.15 g sample of a compound known to contain only phosphorus and oxygen indicates a phosphorus content of 4.433 g

Anna007 [38]

find mass of oxygen by subtracting mass of phosphorus from the total mass

divide the masses by the molar mass to get moles

divide moles by the smallest amount of moles

multiply by 2 to get a nice number

P4O5

5 0

3 years ago

Balance each skeleton reaction, use Appendix D to calculate E°cell, and state whether the reaction is spontaneous:(b) Hg₂²⁺(aq)

Olegator [25]

The given reaction is not spontaneous.

We must recognize changes in oxidation states that take place across elements in order to balance these equations. To accomplish this, keep in mind following guidelines:

A neutral element on its own has an oxidation number of zero.For a neutral molecule, the total number of oxidations must be zero.The net charge of an ion is equal to the sum of its oxidation numbers.In a compound: hydrogen prefers +1, oxygen prefers -2, fluorine prefers -1.In a compound with no oxygen present the other halogens will also prefer -1.

One of the mercury atoms is oxidized from +1 to +2 in the simple aqueous ion, for a loss of 1 electron.

Oxidation half-reaction:

$0.5 Hg^{2+} _{2} (aq)$ → $Hg^{2+} (aq) + 1e^-$

$E^o _{ox} = - 0.92 V$

The other mercury is reduced from +1 to zero in mercury metal, for a gain of 1 electron.

Reduction half-reaction:

$0.5 Hg^{2+} _{2} (aq) + 1 e^-$ → $Hg(l)$

$E^o _{red} = 0.85V$

This is a disproportionation redox reaction !

Net reaction:

$Hg^{2+} _{2} (aq)$ → $Hg^{2+} (aq) + Hg (l)$

$E^o _{cell} = 0.85 - 0.92 = -0.07V$

The cell potential is negative so this reaction is NOT spontaneous.

To learn more about the non spontaneous reaction please click on the link brainly.com/question/20358734

#SPJ4

5 0

2 years ago

Part A: Three gases (8.00 g of methane, CH_4, 18.0g of ethane, C_2H_6, and an unknown amount of propane, C_3H_8) were added to t

myrzilka [38]

Explanation:

Part A:

Total pressure of the mixture = P = 5.40 atm

Volume of the container = V = 10.0 L

Temperature of the mixture = T = 23°C = 296.15 K

Total number of moles of gases = n

PV = nRT (ideal gas equation)

$n=\frac{PV}{RT}=\frac{5.40 atm\times 10.0 L}{0.0821 atm L/mol K\times 296.15 K}=2.22 mol$

Moles of methane gas = $n_1=\frac{8.00 g}{16 g/mol}=0.5 mol$

Moles of ethane gas = $n_2=\frac{18.0 g}{30 g/mol}=0.6 mol$

Moles of propane gas = $n_3$

$n=n_1+n_2+n_3$

$2.22=0.5 mol +0.6 mol+ n_3$

$n_3= 2.22 mol - 0.5 mol -0.6 mol= 1.12 mol$

Mole fraction of methane = $\chi_1=\frac{n_1}{n_1+n_2+n_3}=\frac{n_1}{n}$

$\chi_1=\frac{0.5 mol}{2.22 mol}=0.2252$

Similarly, mole fraction of ethane and propane :

$\chi_2=\frac{n_2}{n}=\frac{0.6 mol}{2.22 mol}=0.2703$

$\chi_3=\frac{n_3}{n}=\frac{1.12 mol}{2.22 mol}=0.5045$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

Partial pressure of methane gas:

$p_1=P\times \chi_1=5.40 atm\times 0.2252=1.22 atm$

Partial pressure of ethane gas:

$p_2=P\times \chi_2=5.40 atm\times 0.2703=1.46 atm$

Partial pressure of propane gas:

$p_3=P\times \chi_3=5.40 atm\times 0.5045=2.72 atm$

Part B:

Suppose in 100 grams mixture of nitrogen and oxygen gas.

Percentage of nitrogen = 37.8 %

Mass of nitrogen in 100 g mixture = 37.8 g

Mass of oxygen gas = 100 g - 37.8 g = 62.2 g

Moles of nitrogen gas = $n_1=\frac{37.8 g g}{28g/mol}=1.35 mol$

Moles of oxygen gas = $n_2=\frac{62.2 g}{32 g/mol}=1.94 mol$

Mole fraction of nitrogen= $\chi_1=\frac{n_1}{n_1+n_2}$

$\chi_1=\frac{1.35 mol}{1.35 mol+1.94 mol}=0.4103$

Similarly, mole fraction of oxygen

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{1.94 mol}{1.35 mol+1.94 mol}=0.5897$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

The total pressure is 405 mmHg.

P = 405 mmHg

Partial pressure of nitrogen gas:

$p_1=P\times \chi_1=405 mmHg\times 0.4103 =166.17 mmHg$

Partial pressure of oxygen gas:

$p_2=P\times \chi_2=405 mmHg\times 0.5897=238.83 mmHg$

3 0

3 years ago