Answer is: an oxybromate compound is KBrO₄ (x = 4).
ω(Br) = 43.66% ÷ 100%.
ω(Br) = 0.4366; mass percentage of bromine.
If we take 100 grams of compound:
m(Br) = ω(Br) · 100 g.
m(Br) = 0.4366 · 100 g.
m(Br) = 43.66 g; mass of bromine.
n(Br) = m(Br) ÷ M(Br).
n(Br) = 43.66 g ÷ 79.9 g/mol,
n(Br) = 0.55 mol; amoun of bromine.
From chemical formula (KBrOₓ), amount of potassium is equal to amount of bromine: n(Br) = n(K).
m(K) = 0.55 mol · 39.1 g/mol.
m(K) = 21.365 g; mass of potassium in the compound.
m(O) = 100 g - 21.365 g - 43.66 g.
m(O) =34.97 g; mass of oxygen.
n(O) = 34.97 g ÷ 16 g/mol.
n(O) = 2.185 mol.
n(K) : n(Br) : n(O) = 0.55 mol : 0.55 mol : 2.185 mol /÷ 0.55 mol.
n(K) : n(Br) : n(O) = 1 : 1 : 4.
Explanation:
Steaming up or fogging happens when steam condenses on the mirror. Steam emerging from hot water can condense on a colder surface. That’s the reason you can see the result on a mirror instantaneously. Obviously, for a bathroom mirror to steam up, the steam that originates at the shower spray (or the bathtub) has to travel through the cooler air to reach the mirror. Since air tends to heat up easily, the mirror can steam up fast.
Answer:
1.1 × 10² g
Explanation:
First, we will convert 1.0 L to cubic centimeters.
1.0 L × (10³ mL/1 L) × (1 cm³/ 1 mL) = 1.0 × 10³ cm³
The density of water is 1.0 g/cm³. The mass corresponding to 1.0 × 10³ cm³ is:
1.0 × 10³ cm³ × (1.0 g/cm³) = 1.0 × 10³ g
1 mole of water (H₂O) has a mass of 18 g, consisting of 2 g of H and 16 g of O. The mass of Hydrogen in 1.0 × 10³ g of water is:
1.0 × 10³ g H₂O × (2 g H/18 g H₂O) = 1.1 × 10² g
(20/23) times 100 is 87%
now the answer asks for percentage ERROR
100-87 =13
so answer is C
