How many moles of ag are formed by the complete reaction of 28.3 mol of pb?

The electrons in the outer shell of an atom are called: outer ionized polar valence

Marrrta [24]

Valence.

The electrons in the outer shell of an atom are called valence electrons.

Valence electrons determine whether the an element is ready form compounds. These electrons can be gained, lost, or shared in the formation of compounds.

7 0

3 years ago

Mercury has a mass density of 13.54 g/ml . how many milliliters would 100. grams occupy

frosja888 [35]

7.39 ml For this problem, simply divide the mass of mercury you have by it's density. 100 g / 13.54 g/ml = 7.3855 ml Since we only have 3 significant digits in 100., you need to round the result to 3 significant digits. So 7.3855 ml = 7.39 ml

6 0

3 years ago

for work to be accomplished we must have (2 points) a)an action which causes movement and an opposite reaction. b)an action whic

Tanya [424]

I think it's A, an action which causes movement and an opposite reaction

5 0

3 years ago

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This graph shows two curves pertaining to a hydrogen s orbital.

fgiga [73]

Answer 1) : According to the complete question attached in the answer,

The radial wave function which is denoted by $R_{nl}(r)$ shown with orange color crosses through zero point. Also, At the the radial nodes, which are spherical shells to some radial distance away from the nucleus there no electron are found.

Also, the radial probability distribution curve denoted as $R^{2}_{nl}(r)$ shown in blue color is observed to touch zero, and shows the place of radial node.

Therefore, the total number of nodes will include both the kinds which has radial and angular nodes which will be represented by 'n'.

It is observed that for any atomic orbital, the total number of nodes will be n-1 .

Considering the s orbital of the hydrogen, which has zero angular momentum (l); (l=0), as it has zero angular nodes.

Hence, there will be only radial nodes, which is

(n−1 = total number of radial nodes in s orbitals)

According to the image, there are 4 radial nodes shown, so n = 5 (as n-1 = 4; therefore, n = 5)

This represents the 5s orbital.

Answer 2) The radial nodes are observed in I'm seeing radial nodes at

$1.9a_{0}$ , $6.4a_{0}$ , $13.9a_{0}$ and $27.0a_{0}$ .

where $a_{0}$ represents the hydorgen bohr atomic radius = 0.0529177 nm

Explanation : It is quite easy to observe the given graph and find out the approximate values of the radial nodes, it does not requires any equation to be solved. Equation can be used to find the radial nodes if it was supplied along with the question. Although by mere speculation one can find out the answer.

3 0

3 years ago

Read 2 more answers

Select True or False from the pull down menus for the following statements.

Shkiper50 [21]

Answer:

First question: 1- False, 2- True, 3- False, 4 -Tue, 5- True.

Second question: 12.7 ºC

Explanation:

First question:

1- When a phase change for a pure substance is taking place under constant pressure, the temperature remains constant, and there's no sensitive heat flowing, but there's latent heat flow, which must be added to separate the molecules and to increase the kinetic energy.

2- When observed the heating and the cooling curve, at the phase change, there is no change in temperature, so it must be a horizontal line, which has a slope equal to 0.

3- Heat is the energy that is transferred by the substances or bodies because of a difference in temperature. The temperature is the measure of average kinetic energy in the molecules, so they are different.

4- As explained above, it's true, that's the definition of temperature.

5- Melting and freezing are the opposite processes and they occur at the same temperature. The difference is that for melting, the substance is absorbing heat, and for freezing it is losing heat, but the heat amount is the same for both process and is calculated by Q = ±m*L, where Q is the heat, m the mass, L the heat capacity, and the signal ± indicates if the substance is absorbing (+) or losing (-) heat.

Second question:

For the conservation of energy, the total amount of heat must be 0. The coin is losing heat, so it must be negative. The water is gaining heat, so it must be positive:

Qw - Qc = 0

Q = m*s*ΔT, where Q is the heat, m is the mass, s is the specif heat, and ΔT the temperature variation (final - initial). Qw is from water and Qc for the coin. The specif heat from the water is 4.184 J/gºC. At the thermal equilibrium, the final temperature must be equal for both.

mw*sw*ΔTw = mc*sc*ΔTc, if the coin is pure silver, sc = 0.233 J/gºC

27.0*4.184*(T - 15.5) = 15.5*0.233*(T - 100)

112.968*(T - 15.5) = 3.6115*(T - 100)

112.968T - 1751.004 = 3.6115T - 361.15

109.3565T = 1389.854

T = 1389.854/109.3565

T = 12.7ºC

So, the final water temperature would be 12.7ºC, which is impossible because it needs to increase. So the coin is not silver pure.

7 0

3 years ago