Answer:
The value of the silver in the coin is 35.3 $
Explanation:
First of all, let's calculate the volume of the coin.
2π . r² . thickness = volume
r = diameter/2
r = 41 mm/2 = 20.5 mm
2 . π . (20.5 mm)² . 2.5 mm = 6601 mm³
Now, this is the volume of the coin, so we must find out how many grams are on it.
6601 mm³ / 1000 = 6.60 cm³
Let's apply density.
D = Mass / volume
10.5 g/cm³ = mass /6.60 cm³
10.5 g/cm³ . 6.60 cm³ = mass
69.3 g = mass
Each gram has a cost of 0.51$
69.3 g . 0.51$ = 35.3 $
Answer:
The atomic mass of second isotope is 7.016
Explanation:
Given data:
Average Atomic mass of lithium = 6.941 amu
Atomic mass of first isotope = 6.015 amu
Relative abundance of first isotope = 7.49%
Abundance of second isotope = ?
Atomic mass of other isotope = ?
Solution:
Total abundance = 100%
100 - 7.49 = 92.51%
percentage abundance of second isotope = 92.51%
Now we will calculate the mass if second isotope.
Average atomic mass of lithium = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
6.941 = (6.015×7.49)+(x×92.51) /100
6.941 = 45.05235 + (x92.51) / 100
6.941×100 = 45.05235 + (x92.51)
694.1 - 45.05235 = (x92.51)
649.04765 = x
92.51
x = 485.583 /92.51
x = 7.016
The atomic mass of second isotope is 7.016
Barfoed's test is a concoction test utilized for identifying the nearness of monosaccharides. It depends on the diminishment of copper(II) acetic acid derivation to copper(I) oxide (Cu2O), which frames a block red hasten.
Barfoed's reagent comprises of a 0.33 molar arrangement of unbiased copper acetic acid derivation in 1% acidic corrosive arrangement. The reagent does not keep well and it is, thusly, fitting to make it up when it is really required. May store uncertainly as per a few MSDS's.
Metallic elements can bend because they have the property of being ductile. This means that a solid material stretches under tensile stress. If a material is ductile then it may be stretched into a wire. Further, the material can also be malleable.
No of moles of naoh = 2.40 ÷ (23+16+1) = 0.06mol
no of moles of na2co3 = 0.06 ÷ 2 = 0.03mol
mass of na2co3 = 0.03 × (23×2+12+16×3) = 0.03 × 106 = 3.18g