You must use 64.43 g H₂O.
<em>Balanced chemical equation</em>: H₂O + CO₂ → H₂CO₃
<em>Moles of CO₂</em> = 157.35 g CO₂ × (1 mol CO₂/44.01 g CO₂) = 3.5753 mol CO₂
<em>Moles of H₂O</em> = 3.5753 mol CO₂ × (1 mol H₂O/1 mol CO₂) = 3.5753 mol Fe
<em>Mass of H₂O</em> = 3.5753 mol H₂O × (18.02 g H₂O /1 mol H₂O) = 64.43 g H₂O
Carbon dioxide is a gaseous molecule made up of the elements, C and O. Each mole of carbon dioxide has one mole C and two mole oxygen atoms.
Molar mass of carbon dioxide (
)=
Percentage by mass of carbon = 
Percentage by mass of oxygen = 
Therefore C is 27.3 % and O is 72.7 % by mass in 1 mol CO
Answer:
The answer is 0.023 moles of phosphorus
Explanation:
The 15-15-15 fertilizer is a fertilizer of great versatility, made with nitrogen, phosphorus and potassium, which makes it one of the fertilizers most used for fertilizer in the sowing plant, thus covering the crop requirements from planting. .
This fertilizer consists of 14.25% phosphorus pentoxide (P2O5). Therefore, we have to remove 14.25% at 10 grams of 15-15-15 fertilizer to calculate the moles of phosphorus. As follows:
Grams of P2O5 = 10 g x 0.1425 = 1.425 g
We calculate the molecular weight of phosphorus. We use the periodic table:
Phosphorus molecular weight = 2 x 30.97 = 61.94 g/mol
Now we calculate the moles of phosphorus in the fertilizer:
Phosphorus moles = 1,425 g/61.94 g/mol = 0.023 moles
Answer:
Explanation:
Combustion reaction is given below,
C₂H₅OH(l) + 3O₂(g) ⇒ 2CO₂(g) + 3H₂O(g)
Provided that such a combustion has a normal enthalpy,
ΔH°rxn = -1270 kJ/mol
That would be 1 mol reacting to release of ethanol,
⇒ -1270 kJ of heat
Now,
0.383 Ethanol mol responds to release or unlock,
(c) Determine the final temperature of the air in the room after the combustion.
Given that :
specific heat c = 1.005 J/(g. °C)
m = 5.56 ×10⁴ g
Using the relation:
q = mcΔT
- 486.34 = 5.56 ×10⁴ × 1.005 × ΔT
ΔT= (486.34 × 1000 )/5.56×10⁴ × 1.005
ΔT= 836.88 °C
ΔT= T₂ - T₁
T₂ = ΔT + T₁
T₂ = 836.88 °C + 21.7°C
T₂ = 858.58 °C
Therefore, the final temperature of the air in the room after combustion is 858.58 °C
Labels for a hazardous chemical must contain:
• Name, Address and Telephone Number
• Product Identifier
• Signal Word
• Hazard Statement(s)
• Precautionary Statement(s)
• Pictogram(s)
So the answer is product identifier.
