Answer:
a) 0.714g of bicarbonate of soda are required.
b) 0.221g of Al(OH)₃ are required
Explanation:
The reactions of HCl with bicarbonate of soda and aluminium hydroxide are:
HCl + NaHCO₃ → H₂O + NaCl + CO₂
3 HCl + Al(OH)₃ → 3H₂O + AlCl₃
The moles of HCl that we need neutralize are:
50mL = 0.050L * (0.17mol / L) = 0.0085 moles HCl
To solve these problem we need to find the moles of the antacid using the chemical reaction and its mass using its molar mass;
<em>a) </em><em>Moles NaHCO₃ = Moles HCl = 0.0085 moles </em>
The mass is -Molar mass NaHCO₃: -84g/mol-
0.0085 moles * (84g / mol) = 0.714g of bicarbonate of soda are required
b) 0.0085 moles HCl * (1mol Al(OH)₃ / 3mol HCl) = 2.83x10⁻³ moles Al(OH)₃
The mass is -Molar mass: 78g/mol-:
2.83x10⁻³ moles Al(OH)₃ * (78g/mol) =
<h3>0.221g of Al(OH)₃ are required</h3>
The graphics in the attachment is part of the question, which was incomplete.
Answer: Fr = 102N and angle of approximately 11°.
Explanation: From the attachment, it is observed that from the three forces acting on M, two are perpendicular. So to find them, we have to show their x- and y- axis components. From the graph:
Fx = 70+40-10 = 100
Fy = 40-20 = 20
Now, as the forces form a triangle, the totalforce is:
Fr = 
Fr = 
Fr = ≈ 102N
To determine the angle requested, we use:
arctg H = 
arctg H = 
H = tg 0.2 ≈ 11°.
The density of an ordinay rock is close to 3 g/cm^3 wihle the density of the paper clips is close to 8 g/cm^3 (the density of steel), then equal apparent volumes (same box) will contain different mass, being of course the mass of the box with paper clips much higher than that of the box with rocks.
Answer:
14.434 r.a.m.
Explanation:
- The atomic mass of an element is a weighted average of its isotopes in which the sum of the abundance of each isotope is equal to 1 or 100%.
∵ The atomic mass of N = ∑(atomic mass of each isotope)(its abundance)
∴ The atomic mass of N = (atomic mass of N-14)(abundance of N-14) + (atomic mass of N-16)(abundance of N-16)
atomic mass of N-14 = 14.0 r.a.m, abundance of N-14 = percent of N-14/100 = 78.3/100 = 0.783.
atomic mass of N-16 = 16.0 r.a.m, abundance of N-16 = percent of N-16/100 = 21.7/100 = 0.217.
∴ The atomic mass of N = (atomic mass of N-14)(abundance of N-14) + (atomic mass of N-16)(abundance of N-16) = (14.0 r.a.m)(0.783) + (16.0 r.a.m)(0.217) = 14.434 r.a.m.
