Answer:
The answer to your question is mass = 455.6 grams
Explanation:
Data
Dimensions = 68.6 yd x 12 in x 0.00035 in
density = 2.70 g/cm³
mass = ?
Process
1.- Convert dimensions to cm
1 yd ---------------- 91.44 cm
68.6 yd ------------ x
x = (68.6 x 91.44) / 1
x = 6227 cm
1 in ----------------- 2.54 cm
12 in --------------- x
x = (12 x 2.54) / 1
x = 30.48 cm
1 in -------------- 2.54
0.00035 in ------------- x
x = (0.00035 x 2.54) / 1
x = 0.000889 cm
2.- Find the volume of the foil
V = 6227 x 30.48 x 0.000889
V = 168.73 cm³
3.- Find the mass of the foil
mass = density x volume
mass = 2.70 x 168.73
mass = 455.6 g
use variable
1K₂MnF₆ + aSbF₅⇒ bKSbF₆ + cMnF₃ + dF₂
K, left=2,right=b⇒b=2
Mn, left=1, right=c⇒c=1
Sb, left=a, right=b⇒a=b=2
F, left=6.1+5a, right=6b+3c+2d
equation:
6+5(2) = 6(2)+3(1)+2d
16=15+2d
1=2d
d=0.5
So the reaction equation becomes:
1K₂MnF₆ + 2SbF₅⇒ 2KSbF₆ + 1MnF₃ + 0.5F₂ x2
2K₂MnF₆ + 4SbF₅⇒ 4KSbF₆ + 2MnF₃ + F₂
1. NaCl
formula weight - 58,4 g/mol.
number of moles - 0,14 mol
mass - 58,4 g/mol · 0,14 mol = 8,176 g.
2. KCl
formula weight - 74,55 g/mol.
number of moles - 0,005 mol
mass - 0,372 g.
3. C₆H₁₂O₆
formula weight - 180,1 g/mol.
number of moles - 0,0056 mol
mass - 180,1g/mol · 0,0056 mol = 1,01 g.
4. NaHCO₃
formula weight - 84 g/mol.
number of moles - 0,0262 mol
mass - 2,2 g.
5. CaCl₂·H₂O
formula weight - 147 g/mol.
number of moles - 0,001 mol
mass - 0,15 g.
6. MgSO₄
formula weight - 120,3 g/mol.
number of moles - 0,0008 mol
mass - 0,096 g.
7. NaH₂PO₄·H₂O
formula weight - 138 g/mol.
number of moles - 0,001 mol
mass - 0,14 g.
Answer:
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