Answer:
El termopar B presenta un mayor grado de dispersión y también es más preciso. ... (c) La estimación para T = 175 ° C es probablemente la más cercana al valor real, porque el ... (cm3). Flujo de masa. Velocidad. (kg / min). Diferencia. Duplicar. (Di). Yo y yo. 2. 1 ... atm de gas. 2. 2. 2 f. 3. 2 f f. 30 14,7 lb 20 pulg. 4 14,7 lb 24 pulg 392 lb 7,00 10 lb pulg.
Do you mean h=7.0+10? If so your answer is 70.
Let MM(x) be the molar mass of x.
MM(Pb) : MM(PbO)
=207.21 : 223.20 = 451.4 g : x g
cross multiply and solve for x
x=223.2/207.21*451.4
= 486.23 g
Percentage yield = 365.0/486.23= 0.75067 = 75.07% (rounded to 4 sign. fig.)
There is 0.02538502095915 Moles in 5 grams of gold.
Density of the gas is 3.05 × 10⁻³ g / cm³.
<u>Explanation:</u>
Volume of the cylinder = π r² h
where r is the radius and h is the height of the height or the length of the glass tube.
Here r = 4 cm and h = 27.4 cm
Volume of the cylinder = 3.14 × 4 × 4 × 27.4 = 1376.6 cm³
We have to find the mass of the gas by subtracting the mass of the tube filled with the substance from the mass of the empty tube.
Mass of the substance = 258.5 - 254.3 = 4.2 g
We have to find the density using the formula as,

Plugin the values as,
= 3.05 × 10⁻³ g / cm³
So the Density of the gas is 3.05 × 10⁻³ g / cm³.