Answer:
The validation of seafloor spreading in the 1950s and 60s
Explanation:
The theory of seafloor spreading was supported by numerous evidence including thermal probes that showed that heat flow over the mid-ocean ridges measured up to four times those measured in general bottom sediments, which are taken as due to the presence of molten Earth material close to the ridge crest
The ridge crest also show signs unusually seismic wave velocities that are considered to be due to microfracturing and thermal expansion from upwelling magma
 
        
                    
             
        
        
        
Electrons in sigma <span>bonds remain localized between two atoms. Sigma </span><span>bond results from the formation of </span><span>a molecular orbital </span><span>by the end to </span><span>end overlap of atomic </span>orbitals. Electrons<span> in pi</span> bonds can become delocalized between more than two atoms. Pi bonds result from the formation of molecular orbital by side to side overlap of atomic orbitals. 
 
<span> </span>
 
        
             
        
        
        
1)
<span>m(NaCl) = 1.95 g
V(H2O) = 250mL
M(NaCl) = </span><span>58.5 g/mole
Since waters density value is 1g/mL, it can be assumed that volume and mass of water are same values:
</span>V(H2O) = 250ml = 250g = 0.25 kg<span>
</span><span>molality of NaCl:
</span><span>
n(NaCl)=m/M=1.95/58.5= 0.033 mole
</span>molality b(NaCl)=n(NaCl) / V (H2O)= 0.033/0.25 = 0.132 mol/kg
<span>
milimolality of NaOH = 0.132/0,001 = 132 mmole/kg
</span>
milliosmolality of NaOH = milimolality x N of ions formed in dissociation
Since NaCl dissociates into 2 ions in solution:
<span>                                        
</span>milliosmolality of NaOH = 132 x 2 = 264  osmol<span>es/kg
</span>
2)
m(gl) = 9 g
V(H2O) = 250mL
M(NaCl) = 180 g/mole
Since waters density value is 1g/mL, it can be assumed that volume and mass of water are same values:
V(H2O) = 250ml = 250g = 0.25 kg
molality of glucose:
n(gl)=m/M=9/180= 0.05 mole
molality b(gl)=n(gl) / V (H2O)= 0.05/0.25 = 0.2 mol/kg
milimolality of glucose = 0.132/0,001 = 200 mmole/kg
milliosmolality of glucose = milimolality x N of ions formed in dissociation
Since glucose does not dissociate, milimolality and milliosmolality are same:
                                        
milliosmolality of glucose = 200 osmoles/kg
3)
The osmosis represents the diffusion of solvent molecules through a semi-permeable membrane that allows passage solvent molecules but does not to the dissolved substance molecule. The osmosis occurs when the concentrations of the solution on both sides of the membrane are different. Since the semi-permeable membrane only permeates the solvent molecules, but not the particles of the dissolved substance, it occurs the solvent diffusion through the membrane, i.e. the solvent molecules pass through the membrane to equalize the concentration on both sides of the membrane. Solvents molecules move from the middle with a lower concentration in the middle with a higher concentration of dissolved substances.
In our case, osmosis will occur because the concentration of NaCl solution and the concentration of glucose solution do not have same values. Osmosis will occur in the direction of glucose solution because it has a lower concentration.
        
             
        
        
        
Answer:
d) Cr⁺³
Explanation:
Consideremos un ion que contiene 24 protones, 28 neutrones y 21 electrones.
Para encontrar el simbolo del elemento, tenemos que considerar el número atómico (Z) que es igual al número de protones. Con esta información, buscamos en la tabla periódica el  elemento con Z=24 es el Cromo.
La carga total está dada por la diferencia entre protones y electrones. Los ´protones tienen carga +1 y los electrones carga -1. Luego, este ion tiene carga:
24 protones + 21 electrones = 24 . (1) + 21 . (-1) = +3
El simbolo del ion es Cr⁺³.