Answer:
The mass of board is 1.52 g.
Explanation:
Given data:
Density of wood = 0.487 g/cm³
Length of board = 3.21 cm
Width of board = 12.2 cm
Height of board = 0.80 mm (0.80/10=0.08 cm)
Mass of board = ?
Solution:
We know that density of an object is equal to the mass of substance divided by its volume. So we can calculate the mass from density formula.
Formula:
D = m/v
d = density
m = mass
v = volume
Now we will calculate the volume first.
V = length × width × height
V = 3.21 cm × 12.2 cm × 0.08 cm
V = 3.13cm³
Now we will calculate the mass.
D = m/v
0.487 g/cm³ = m/ 3.13 cm³
m = 0.487 g/cm³ × 3.13 cm³
m = 1.52 g
The mass of board is 1.52 g.
<span>Air enters through the nose or mouth (which join to form the nasopharyngeal cavity). The air travels through the trachea which and the trachea splits into two bronchi. The air travels through the bronchi which split into smaller and smaller bronchioles. The tiny bronchioles and the air ends up in the miniscule alveoli, where the oxygen in the air diffuse into the bloodstream. Carbon dioxide diffuse from the bloodstream into the alveoli and the unwanted gas travels in reverse back to the nose and mouth, where it is breathed out.</span>
A scientific control is an experiment or observation designed to minimize the effects of variables other than the independent variable. This increases the reliability of the results, often through a comparison between control measurements and the other measurements.
Answer:
The volume of nitrogen oxide formed is 35.6L
Explanation:
The reaction of nitric acid with copper is:
Cu(s) + 4HNO₃ → Cu(NO₃)₂ + 2NO₂(g) + 2H₂O(l)
Moles of copper are:
Moles of nitric acid are:
As 1 mol of Cu reacts with 4 moles of HNO₃:
0.697 mol Cu × (4mol HNO₃ / 1mol Cu) = 2.79 moles of HNO₃ will react. That means Cu is limiting reactant.
Moles of NO₂ produced are:
0.697 mol Cu × (2mol NO₂ / 1mol Cu) = <em>1.394 moles of NO₂</em>
Using PV = nRT
<em>Where P is pressure (735torr / 760 = 0.967atm); n are moles (1.394mol); R is gas constant (0.082atmL/molK); T is temperature (28.2°C + 273.15 = 301.35K). </em>
Thus, volume is:
V = nRT / P
V = 1.394mol×0.082atmL/molK×301.35K / 0.967atm
V = 35.6L
<em>The volume of nitrogen oxide formed is 35.6L</em>
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