Answer:
The difference in the electronegativities of chlorine and boron is 3.0 - 2.0 = 1.0 ; the difference in between chlorine and carbon is 3.0 = 2.5 = 0.5 . Consequently, the B-Cl bond is more polar ; the chlorine atom asrries the partial negative charge because it has higher electronegativity .
Explanation:
hope it helps!
C. quadruples the rate
<h3>Further explanation</h3>
Given
The rate law :
R=k[A]²
Required
The rate
Solution
There are several factors that influence reaction kinetics :
- 1. Concentration
- 2. Surface area
- 3. Temperature
- 4. Catalyst
- 5. Pressure
- 6. Stirring
The rate is proportional to the concentration.
If the concentration increased, the reaction rate will increase
The reaction is second-order overall(The exponent is 2)
The concentration of A is doubled, the reaction rate will increase :
r = k[A]² ⇒ r= k[2A]²⇒r=4k[A]²
<em>The reaction rate will quadruple.</em>
Ca(OH)2(aq) + 2HCl(aq)------> CaCl2(aq) + 2H2O(l) ΔH-?
CaO(s) + 2HCl(aq)-----> CaCl2(aq) + H2O(l), Δ<span>H = -186 kJ
</span>
CaO(s) + H2O(l) -----> Ca(OH)2(s), Δ<span>H = -65.1 kJ
</span>
1) Ca(OH)2 should be reactant, so
CaO(s) + H2O(l) -----> Ca(OH)2(s)
we are going to take as
Ca(OH)2(s)---->CaO(s) + H2O(l), and ΔH = 65.1 kJ
2) Add 2 following equations
Ca(OH)2(s)---->CaO(s) + H2O(l), and ΔH = 65.1 kJ
<span><u>CaO(s) + 2HCl(aq)-----> CaCl2(aq) + H2O(l), and ΔH = -186 kJ</u>
</span>Ca(OH)2(s)+CaO(s) + 2HCl(aq)--->CaO(s) + H2O(l)+CaCl2(aq) + H2O(l)
Ca(OH)2(s)+ 2HCl(aq)---> H2O(l)+CaCl2(aq) + H2O(l)
By addig these 2 equation, we got the equation that we are needed,
so to find enthalpy of the reaction, we need to add enthalpies of reactions we added.
ΔH=65.1 - 186 ≈ -121 kJ
Ionization energy increases from left to right because the left wants to lose elctrons and the right wants to gain electron
As you go a group it is easier lose lose because the electrons are farther away from the nucleus and there is less attraction from the positive charges.
It should be 3p3. the p level can hold 6 electrons
Rubidium group 1, 1 valence electrons very reactive
Mg2,2 very reactive
Al 3, 3 reactive
Answer:
Cloruro de sodio y fluoruro de sodio.
Dióxido de carbono y monóxido de hidrógeno.
Explicación:
El cloruro de sodio y el fluoruro de sodio son los compuestos que tienen enlaces iónicos. Estos compuestos iónicos se utilizan para diferentes actividades de nuestra vida diaria. El cloruro de sodio se usa para cocinar y el fluoruro de sodio se usa en la pasta de dientes para limpiar nuestros dientes. El dióxido de carbono y el monóxido de hidrógeno son compuestos que tienen enlaces covalentes. El dióxido de carbono se usa en refrescos / refrescos y algunos otros líquidos que se pueden usar en la vida diaria. El monóxido de hidrógeno es el agua pura que bebemos todos los días en nuestra vida diaria y es muy importante para nuestra supervivencia.
