The choices are true about the characteristic of a strong base, except for it having a concentration of above 1.0 M. Therefore, the answer is letter A. The concentration of the base is not a very important as to how strong really the base is.
We have that the name (not chemical symbol) of the main group element in period 5 and group 3A is
Metalloid boron (B)
From the Question we are told that
The element belongs to
Period 5 and group 3A
Generally
Group 3A of the periodic table includes the metalloid boron (B), aluminum (Al), indium (In), and thallium (Tl) gallium (Ga),
Period 5 is possessed by the metalloid boron (B) of the Group 3A
For more information on this visit
brainly.com/question/13025901?referrer=searchResults
We make use of ratios to solve this question.
moles NaOH = c · V = 0.1923 mmol/mL · 26.66 mL = 5.126718 mmol
moles H2SO4 = 5.126718 mmol NaOH · 1 mmol H2SO4 / 2 mmol NaOH = 2.563359 mmol
Hence
[H2SO4]= n/V = 2.563359 mmol / 42.45 mL = 0.06039 M
The answer to this question is [H2SO4] = 0.06039 M
Answer:
there are 3.63*10⁻² moles of ammonium anions
Explanation:
since the ammonium carbonate is
(NH₄)₂(CO₂)
and the ammonium carbonate dissociate according to
(NH₄)₂(CO₂) →2*NH₄⁺ + CO₂²⁻ ( we will neglect other carbonate species for simplicity)
then each mole of ammonium carbonate generates 2 moles of ammonium anions , then
n AC = 2* n A
where n represents moles AC represents ammonium carbonate and A represents ammonium anions
also we know that
n AC = m AC / M AC
where
m = mass of ammonium carbonate = 6.595 g
M = molecular weight of ammonium carbonate = 96.09 g/mol
then
n AC = m AC / M AC = 6.595 g/ 96.09 g/mol = 7.25*10⁻² moles of ammonium carbonate
thus
n AC = 2* n A → n A = n AC /2 = 7.25*10⁻² /2 = 3.63*10⁻² moles of ammonium anions
