The elements in Groups 1A(1) and 7A(17) are all quite reactive.
<h3>Major difference between Groups 1A(1) and 7A(17) : </h3>
Group 7's halogens, which are non-metal elements, become less reactive as you move down the group. In contrast to the alkali metals in Group 1 of the periodic table, this trend is the opposite. The most reactive element in Group 7 is fluorine.
Alkali metals are soft and reactive metals. They react vigorously with water and become more reactive. And other hand halogens are reactive non metals.
- Elements of group 1A are known as alkali metals. Elements of this group are lithium, sodium, potassium, rubidium, cesium.
- Reactivity increase down group 1 but decrease up group 7 this is because group 7 elements react by gaining an electron. As one move down the group, the amount of electron shielding increases, meaning that the electron is less attracted to the nucleus.
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Answer:
20L is the new volume
Explanation:
In this case, moles and T° from the gas remain constant. This is the formula we must apply, to solve this:
P₁ . V₁ = P₂ . V₂
5 atm . 10 L = P₂ . 2.5L
P₂ = (5 atm . 10 L) / 2.5L →20L
<span>Answer is: pH of solution of sodium cyanide is 11.3.
Chemical reaction 1: NaCN(aq) → CN</span>⁻(aq)
+ Na⁺<span>(aq).
Chemical reaction 2: CN</span>⁻ +
H₂O(l) ⇄ HCN(aq) + OH⁻<span>(aq).
c(NaCN) = c(CN</span>⁻<span>)
= 0.021 M.
Ka(HCN) = 4.9·10</span>⁻¹⁰<span>.
Kb(CN</span>⁻) = 10⁻¹⁴ ÷
4.9·10⁻¹⁰ = 2.04·10⁻⁵<span>.
Kb = [HCN] · [OH</span>⁻]
/ [CN⁻<span>].
[HCN] · [OH</span>⁻<span>] =
x.
[CN</span>⁻<span>] = 0.021 M - x..
2.04·10</span>⁻⁵<span> = x² / (0.021 M
- x).
Solve quadratic equation: x = [OH</span>⁻<span>] = 0.00198 M.
pOH = -log(0.00198 M) = 2.70.
pH = 14 - 2.70 = 11.3.</span>
Answer:
NaCl.
Explanation:
In the solution, ZnSe ionizes to 
and 
. Following reaction represents the ionization of ZnSe in solution -
⇄ 
As we want to increase the solubility of ZnSe, we must decrease the concentration of dissociated ions so that the reaction continues to forward direction.
If we add NaCl to this solution, then we have 
and 
in the solution which will be formed by the ionization of NaCl.
Now, in the solution will react with two ions to form 
as follows -
⇄
Due to this reaction the concentration of will decrease in the solution and more ZnSe can be soluble in the solution.
Answer:
The unknown temperature is 304.7K
Explanation:
V1 = 100mL = 100*10^-3L
P1 = 99.10kPa = 99.10*10³Pa
V2 = 74.2mL = 74.2*10^-3L
P2 = 133.7kPa = 133.7*10³Pa
T2 = 305K
T1 = ?
From combined gas equation,
(P1 * V1) / T1 = (P2 * V2) / T2
Solving for T1,
T1 = (P1 * V1 * T2) / (P2 * V2)
T1 = (99.10*10³ * 100*10^-3 * 305) / (133.7*10³ * 74.2*10^-3)
T1 = 3022550 / 9920.54
T1 = 304.67K
T1 = 304.7K
