Answer is: 8568.71 of baking soda.
Balanced chemical reaction: H₂SO₄ + 2NaHCO₃ → Na₂SO₄ + 2CO₂ + 2H₂O.
V(H₂SO₄) = 17 L; volume of the sulfuric acid.
c(H₂SO₄) = 3.0 M, molarity of sulfuric acid.
n(H₂SO₄) = V(H₂SO₄) · c(H₂SO₄).
n(H₂SO₄) = 17 L · 3 mol/L.
n(H₂SO₄) = 51 mol; amount of sulfuric acid.
From balanced chemical reaction: n(H₂SO₄) : n(NaHCO₃) = 1 :2.
n(NaHCO₃) = 2 · 51 mol.
n(NaHCO₃) = 102 mol, amount of baking soda.
m(NaHCO₃) = n(NaHCO₃) · M(NaHCO₃).
m(NaHCO₃) = 102 mol · 84.007 g/mol.
m(NaHCO₃) = 8568.714 g; mass of baking soda.
Answer:
A = 2-iodo-2,3-dimethylbutane
B = Ethanol
C = Iodoethane (also called ethyl-iodide)
Explanation:
2-Ethoxy-2,3-dimethylbutane reacts with conc. HI to cleave the oxy-functional group.
On one end, ethanol is formed and on the other hand, 2-iodo-2,3-dimethylbutane is formed.
But ethanol reacts further with conc HI to give iodoethane.
Therefore,
A = 2-iodo-2,3-dimethylbutane
B = Ethanol
C = Iodoethane (also called ethyl-iodide)
This is all shown in the attached image.
Hope this Helps!!!
Answer:
b. 7.5 x 10^-3
Explanation:
To solve this problem we need to keep in mind the <em>definition of molarity</em>:
- Molarity = moles of solute / liters of solution
With the above information in mind it is possible to calculate the moles of solute, given the volume (10 mL) and concentration (0.75 M) of the solution:
- First we<u> convert 10 mL to L</u> ⇒ 10 mL / 1000 = 0.01 L
Then we <u>calculate the moles of AgNO₃</u>:
- moles of solute = Molarity * Liters of solution
- 0.01 L * 0.75 M = 7.5x10⁻³ mol AgNO₃
<em>One mole of AgNO₃ contains one mole of Ag⁺</em>, thus the number of Ag⁺ moles is also 7.5x10⁻³.
Answer:
Hi there!
It is difficult to remove chlorine out of sodium chloride because in order to separate them you must put that energy into the salt.
Chlorine gas is toxic, and extremely irritating to the eyes and mucous membranes.
Answer:
C. NaOH(s) + HCl(aaq) rightarrow NaCl(s) + H2O(aq)
Explanation:
A neutralization reaction may be defined as the reaction where an acid when it reacts with some base, the products formed are a salt and water.
Simply, we can write,
acid + base ---> salt + water
Here in option (C),
NaOH is sodium hydroxide is an base
HCl is hydrochloric acid
NaCl is sodium chloride and is the common salt
H20 is water
So, option C). NaOH(s) + HCl(aaq) -----> NaCl(s) + H2O(aq) is a neutralization reaction.
