False. An increase in temperature is an exothermic reaction. However, when a temperature decreases this is known as an endothermic reactionz
Answer:
The answer to your question is Ferns
Explanation:
Ferns are called nonflowering plants and produce spores instead of seeds.
The percentage of Chromium in Chromium Oxide is calculated as follow,
Step 1: Calculate Molar mass of Cr₂O₃,
Cr = 51.99 u
O = 16 u
So,
2(51.99) + 3(16) = 103.98 + 48 = 151.98 u
Step 2: Secondly divide molar mass of only chromium with total mass of Cr₂O₃ and multiply with 100.
i.e.
=
× 100
=
68.41 %
So, the %age composition of chromium in chromium oxide is
68.41 %.
Answer:
The formula of Al³⁺ and SO₄²⁻ is aluminum sulfate.
Explanation:
The formula for aluminum sulfate is Al₂(SO₄)₃. If we say in terms of ions. The ions are Al³⁺. It is a positive ion or the cation. Other ion is SO₄²⁻. It is sulfate ion. It is anion.
Aluminum sulphate is used in water purification and as a mordant in dyeing and printing textiles.
Hence, the formula of Al³⁺ and SO₄²⁻ is aluminum sulfate.
Answer:
See Explanation
Explanation:
The equation of the reaction;
KHSO4(aq) + KOH(aq) -------> K2SO4(aq) + H2O(l)
Number of moles of KHSO4 = 49.6 g/136.169 g/mol = 0.36 moles
Since the reaction is in a mole ratio of 1:1, 0.36 moles of K2SO4 is produced.
Number of moles of KOH = 25.3 g/56.1056 g/mol = 0.45 moles
Since the reaction is 1:1, 0.45 moles of K2SO4 is produced
Hence K2SO4 is the limiting reactant.
Mass of K2SO4 formed = 0.36 moles of K2SO4 * 174.26 g/mol = 62.7 g
So;
1 mole of KHSO4 reacts with 1 mole of KOH
0.36 moles of KHSO4 reacts with 0.36 * 1/1 = 0.36 moles of KOH
Amount of excess KOH = 0.45 moles - 0.36 moles = 0.09 moles
Mass of excess KOH = 0.09 moles * 56.1056 g/mol = 5 g of excess KOH
