A. Flueorescece. <span>The light from these ultraviolet lamps reacts with the chemicals of a mineral and causes the mineral to glow.
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Answer:
27.99 dm³
Explanation:
Applying
PV = nRT................ Equation 1
Where P = Pressure, V = Volume, n = number of mole, R = molar gas constant, T = Temperature.
From the question, we were aksed to find V.
Therefore we make V the subject of the equation
V = nRT/P................ Equation 2
Given: n = 1.31 moles, T = 37°C = 310K, P = 904 mmHg = (904×0.001316) = 1.1897 atm
Constant: R = 0.082 atm.dm³/K.mol
Substitute these values into equation 2
V = (1.31×310×0.082)/(1.1897)
V = 27.99 dm³
Answer:
C₅H₁₀O₅
Explanation:
Let's consider a compound with the empirical formula CH₂O. In order to determine the molecular formula, we have to calculate "n", so that
n = molar mass of the molecular formula / molar mass of the empirical formula
The molar mass of the molecular formula is 150 g/mol.
The molar mass of the empirical formula is 12 + 2 × 1 + 16 = 30 g/mol
n = (150 g/mol) / (30 g/mol) = 5
Then, we multiply the empirical formula by 5.
CH₂O × 5 = C₅H₁₀O₅
The main information we have to use here is the density of gold. From literature, the density of gold at room temperature is 19.32 g/cm³. To determine the mass, let's calculate the volume first. A wire is in the shape of a cylinder. Thus, the volume would be
V = πd²h/4
V = π(0.175 cm)²(1×10⁵ cm)/4
V = 2,405.28 cm³
Density = mass/volume
19.32 g/cm³ = Mass/2,405.28 cm³
Mass = 46,470 g gold wire
