Explanation:
here's the answer to your question about
9 grams of hydrogen gas (H2) will SC Johnson need to react in order to make 1 bottle of Windex.
Explanation:
Balance equation for the formation of ammonia from H2 gas.
N2 + 3H2 ⇒ 2 
Given
mass of ammonia in 1 bottle of windex = 51 gram
atomic mass of ammonia 17.01 gram/mole
number of moles = 
number of moles = 
= 3 moles of ammonia is formed.
in 1 bottle of windex there are 3 moles of ammonia 0r 51 grams of ammonia.
From the equation it can be found that:
3 moles of hydrogen reacted to form 2 moles of ammonia
so, x moles of hydrogen will react to form 3 moles of ammonia.
= 
x = 4.5 moles of hydrogen will be required.
to convert moles into gram formula used:
mass = atomic mass x number of moles (atomic mass of H2 is 2grams/mole)
= 2 x 4.5
= 9 grams of hydrogen.
Answer:
Na has the most similar configuration.
Explanation:
Na electron configuration: 1s²2s²2p⁶3s¹ or [Ne] 3s₁
Mg electron configuration: 1s²2s²2p⁶3s² or [Ne] 3s²
Be electron configuration: 1s²2s² or [He] 2s²
This is because Na and Mg are right next to each other in the same period (horizontal).
Answer:
pH = 3.3
Explanation:
Buffer solutions minimize changes in pH when quantities of acid or base are added into the mix. The typical buffer composition is a weak electrolyte (wk acid or weak base) plus the salt of the weak electrolyte. On addition of acid or base to the buffer solution, the solution chemistry functions to remove the acid or base by reacting with the components of the buffer to shift the equilibrium of the weak electrolyte left or right to remove the excess hydronium ions or hydroxide ions is a way that results in very little change in pH of the system. One should note that buffer solutions do not prevent changes in pH but minimize changes in pH. If enough acid or base is added the buffer chemistry can be destroyed.
In this problem, the weak electrolyte is HNO₂(aq) and the salt is KNO₂(aq). In equation, the buffer solution is 0.55M HNO₂ ⇄ H⁺ + 0.75M KNO₂⁻ . The potassium ion is a spectator ion and does not enter into determination of the pH of the solution. The object is to determine the hydronium ion concentration (H⁺) and apply to the expression pH = -log[H⁺].
Solution using the I.C.E. table:
HNO₂ ⇄ H⁺ + KNO₂⁻
C(i) 0.55M 0M 0.75M
ΔC -x +x +x
C(eq) 0.55M - x x 0.75M + x b/c [HNO₂] / Ka > 100, the x can be
dropped giving ...
≅0.55M x ≅0.75M
Ka = [H⁺][NO₂⁻]/[HNO₂] => [H⁺] = Ka · [HNO₂]/[NO₂⁻]
=> [H⁺] = 6.80x010⁻⁴(0.55) / (0.75) = 4.99 x 10⁻⁴M
pH = -log[H⁺] = -log(4.99 x 10⁻⁴) -(-3.3) = 3.3
Solution using the Henderson-Hasselbalch Equation:
pH = pKa + log[Base]/[Acid] = -log(Ka) + log[Base]/[Acid]
= -log(6.8 x 10⁻⁴) + log[(0.75M)/(0.55M)]
= -(-3.17) + 0.14 = 3.17 + 0.14 = 3.31 ≅ 3.3
Answer:
i can only help with 2 :( atimic mass is: 35.453 and number of electrons is:17
Explanation:
