The visual representation is displayed in the following image.
For calculations, consider 100 grams of the compound:
ω(Cl) = 85.5% ÷ 100%.
ω(Cl) = 0.855; signifying the mass percentage of chlorine in the compound.
m(Cl) = 0.855 · 100 g.
m(Cl) = 85.5 g; this represents the mass of chlorine.
m(C) = 100 g - 85.5 g.
m(C) = 14.5 g; indicating the mass of carbon.
n(Cl) = m(Cl) ÷ M(Cl).
n(Cl) = 85.5 g ÷ 35.45 g/mol.
n(Cl) = 2.41 mol; this is the quantity of chlorine.
n(C) = 14.5 g ÷ 12 g/mol.
n(C) = 1.21 mol; this is the quantity of carbon.
n(Cl): n(C) = 2.41 mol: 1.21 mol = 2: 1.
The compound in question is identified as dichlorocarbene CCl₂.
Answer:
9.69g
Explanation:
To find the needed outcome, we first need to determine the number of moles of N2 present in 7.744L of the gas.
1 mole of gas takes up 22.4L at STP.
Thus, X moles of nitrogen gas (N2) will fill 7.744L, meaning
X moles of N2 = 7.744/22.4 = 0.346 moles
Next, we will convert 0.346 moles of N2 to grams to achieve the result sought. The calculation goes as follows:
Molar Mass of N2 = 2x14 = 28g/mol
Number of moles N2 = 0.346 moles
Find the mass of N2 =?
Mass = number of moles × molar mass
Mass of N2 = 0.346 × 28
Mass of N2 = 9.69g
Hence, 7.744L of N2 consists of 9.69g of N2
The result is 14.5 g L⁻¹.
Here, the problem indicates to reduce the units to one. The existing units are g/L. To achieve a singular unit format, we can move L to the numerator, which can be executed as per the exponent laws; specifically, 1 / aˣ = a⁻ˣ. Thus, we can express 1 / L as L⁻¹. Consequently, the simplified unit remains g L⁻¹. However, remember to leave a space between two different units. This ultimately depicts a unit of density.
The new pressure of the gas is calculated to be 40.7 kPa. Using the principle that P1 • V1 = P2 • V2, we can set 98.8 kPa (P1) multiplied by 21.7 mL (V1) equal to P2 (unknown pressure) multiplied by 52.7 mL (V2). To isolate P2, we rearrange the equation to P2 = (98.8 kPa • 21.7 mL) / 52.7 mL, resulting in P2 equal to 40.7 kPa.
The temperature difference after 3 hours is 5.16 K. Given that the moles of O₂ inhaled rate at 0.02 mole/min, which converts to 1.2 mole/hour, we know the average heat released during metabolism is 7.2 kJ/h·kg. Therefore, the amount of heat generated within 3 hours will be 7.2 kJ/h·kg multiplied by 3 hours, giving a result of 21.6 kJ/kg, or 21.6 x 10³ J/kg. Applying the formula Qp = Cp x ΔT, and taking the body's heat capacity to be 4.18 J/g·K, we find ΔT = 5.16 K.
