Answer:
I will ask when they made this graphics, how they measured the gas and how this affected the climate change
<u><em>BRAINLIEST</em></u>
The initial mass of sodium hydroxide is 3.3 g (answer C)
<u><em>calculation</em></u>
Step 1 : find the moles of iron (ii) hydroxide ( Fe(OH)₂
moles = mass÷ molar mass
from periodic table the molar mass of Fe(OH)₂ = 56 + [16 +1]2 = 90 g/mol
moles is therefore = 3.70 g÷ 90 g/mol = 0.041 moles
Step 2: use the mole ratio to calculate the moles of sodium hydroxide (NaOH)
from given equation NaOH : Fe(OH)₂ is 2 :1
therefore the moles of NaOH = 0.041 x 2 = 0.082 moles
Step 3: find mass of NaOH
mass = moles x molar mass
from the periodic table the molar mass of NaOH = 23 +16 +1 = 40 g/mol
mass = 0.082 moles x 40 g/mol = 3.3 g ( answer C)
The empirical formula for a compound is KClO3
Explanation
find the moles of each element
moles = % composition/molar mass
molar mass of of potassium =39g/mol ,chlorine = 35.5 g/mol, oxygen =16 g/mol
moles of potassium = 31.9 / 39 = 0.818 moles
moles of chlorine = 28.9/35.5 = 0.814 moles
moles of oxygen = 39.2/ 16 = 2.45 moles
find the mole ratio by dividing with the smallest mole = 0.814 moles
potassium = 0.818/0.814 =1
chlorine = 0.814/0.814 = 1
oxygen = 2.45 /0.814 =3
the empirical formula is therefore = KClO3
Answer:
Amount of excess Carbon (ii) oxide left over = 23.75 g
Explanation:
Equation of the reaction: Fe₂O₃ + 3CO ----> 2Fe + 3CO₂
Molar mass of Fe₂O₃ = 160 g/mol;
Molar mass of Carbon (ii) oxide = 28 g/mol
From the equation of reaction, 1 mole of Fe₂O₃ reacts with 3 moles of carbon (ii) oxide; i.e. 160 g of iron (iii) oxide reacts with 84 g (3 * 28 g) of carbon (ii) oxide
450 g of Fe₂O₃ will react with 450 * 84/180) g of carbon (ii) oxide = 236..25 g of carbon (ii) oxide
Therefore the excess reactant is carbon (ii) oxide.
Amount of excess Carbon (ii) oxide left over = 260 - 236.25
Amount of excess Carbon (ii) oxide left over = 23.75 g
