Y2O3 is the molecular formula. Yttrium (III) oxide is also called yttria. It is a white substance and is air-stable. The usual application for this compound is a starting material for inorganic compounds and in material science. IT is insoluble in water and has a high melting point.
Answer:
See below.
Explanation:
The mass of octane in the sample of gasoline is 0.02851 * 482.6 = 13.759 g of octane.
The balanced equation is:
2C8H18(l) + 25O2(g) ----> 16CO2(g) + 18H2O(g)
From the equation, using atomic masses:
228.29 g of octane forms 704 g of CO2 and 324.3 g of H2O
So the mass of CO2 formed from the combustion of 13.759 g of octane = (704 * 13.759) / 228.29
= 42.43 g of CO2.
Amount of water = 324.3 * 13.759) / 228.29
= 19.55 g of H2O.
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Answer:
8.55 × 10³ cal
Explanation:
Step 1: Given and required data
- Specific heat of water (c): 1 cal/g.°C
- Initial temperature: 22.7 °C
- Final temperature: 38.8 °C
Step 2: Calculate the temperature change (ΔT)
ΔT = Final temperature - Initial temperature = 38.8 °C - 22.7 °C = 16.1 °C
Step 3: Calculate the heat required (Q)
We will use the following expression.
Q = c × m × ΔT
Q = 1 cal/g.°C × 531 g × 16.1 °C = 8.55 × 10³ cal
