Answer:
You determine the number of protons by its atomic number, just like electrons.
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Answer:
The answer to your question is 242 ml
Explanation:
Data
HI 0.211 M Volume = x
KMnO₄ 0.354 M Volume = 24 ml
Balanced Chemical reaction
12HI + 2KMnO₄ + 2H₂SO₄ → 6I₂ + Mn₂SO₄ + K₂SO₄ + 8H₂O
Process
1.- Calculate the moles of KMnO₄ 0.354 M in 24 ml
Molarity = moles / volume (L)
moles = Molarity x volume (L)
moles = 0.354 x 0.024
moles = 0.0085
2.- From the balanced chemical reaction we know that HI and KMnO₄ react in the proportion 12 to 2. Then,
12 moles of HI --------------- 2 moles of KMnO₄
x --------------- 0.0085 moles of KMnO₄
x = (0.0085 x 12)/2
x = 0.051 moles of HI
3.- Calculate the milliliters of HI 0.211 M
Molarity = moles/volume
Volume = moles/molarity
Volume = 0.051/0.211
Volume = 0.242 L or Volume = 242 ml
Answer:
C₃H₈(g) + 6 H₂O(g) ⇒ + 10 H₂(g) + 3 CO₂(g)
Explanation:
Propane can be turned into hydrogen by the two-step reforming process.
In the first step, propane and water react to form carbon monoxide and hydrogen. The balanced chemical equation is:
C₃H₈(g) + 3 H₂O(g) ⇒ 3 CO(g) + 7 H₂(g)
In the second step, carbon monoxide and water react to form hydrogen and carbon dioxide. The balanced chemical equation is:
CO(g) + H₂O(g) ⇒ H₂(g) + CO₂(g)
In order to get the net chemical equation for the overall process, we have to multiply the second step by 3 and add it to the first step. Then, we cancel what is repeated.
C₃H₈(g) + 3 H₂O(g) + 3 CO(g) + 3 H₂O(g) ⇒ 3 CO(g) + 7 H₂(g) + 3 H₂(g) + 3 CO₂(g)
C₃H₈(g) + 6 H₂O(g) ⇒ + 10 H₂(g) + 3 CO₂(g)
Answer:
The molarity (M) of the following solutions are :
A. M = 0.88 M
B. M = 0.76 M
Explanation:
A. Molarity (M) of 19.2 g of Al(OH)3 dissolved in water to make 280 mL of solution.
Molar mass of Al(OH)3 = Mass of Al + 3(mass of O + mass of H)
= 27 + 3(16 + 1)
= 27 + 3(17) = 27 + 51
= 78 g/mole
= 78 g/mole
Given mass= 19.2 g/mole
Moles = 0.246
Volume = 280 mL = 0.280 L
Molarity = 0.879 M
Molarity = 0.88 M
B .The molarity (M) of a 2.6 L solution made with 235.9 g of KBr
Molar mass of KBr = 119 g/mole
Given mass = 235.9 g
Moles = 1.98
Volume = 2.6 L
Molarity = 0.762 M
Molarity = 0.76 M