Answer:
The length of foil will be 8107.81 cm or 81.7081 m.
Explanation:
Given data:
Width of roll of foil = 302 mm
Height or thickness = 0.018 mm
Density of foil = 2.7 g/cm³
Mass of foil = 1.19 Kg
Length of foil = ?
Solution:
d = m/ v
v = length (l) × width (w) × height (h)
First of we will convert the Kg into gram and mm into cm.
one Kg = 1000 g
1.19 × 1000 = 1190 g
one cm = 10 mm
302 / 10 = 30.2 cm
0.018 / 10 = 0.0018 cm
Now we will put the values in formula:
d = m/ l× h× w
l = m / d × h× w
l = 1190 g / 2.7 g/cm³× 30.2 cm × 0.0018 cm
l = 1190 g/ 0.146772 g/cm
l = 8107.81 cm or 81.7081 m
As per the given chemical formula- Na2CO3.10H2O, one mole of the chemical compound contains 13 moles of oxygen atoms. Hence
Number of moles of oxygen atoms in one mole of Na2CO3.10H2O = 13
number of moles of oxygen atoms in 0.2 moles of Na2CO3.10H2O = 13 X 0.2 = 2.6
Now, one mole of a substance contains 6.022 X 10^23 particles of the substance. Thus
number of atoms of oxygen in one mole of oxygen atom = 6.022 X 10^23
number of moles of oxygen atoms in 2.6 moles of oxygen atoms = 2.6 X 6.022 X 10^23 = 15.657 X 10^23
= 1.566 X 10^24
Thus, there are 1.566 X 10^24 atoms of oxygen in 0.2 moles of Na2CO3.10H2O.
4.7
Answer:
The element that has been oxidized is the N
Explanation:
Zn²⁺(aq) + NH₄⁺(aq) → Zn(s) + NO₃⁻(aq)
See all the oxidation states:
Zn²⁺ → acts with +2
In ammonia, H acts with +1 and N with -3
Zn(s), acts with 0. In all the elements in ground state, the oxidation state is 0.
Zn changed from 2+ to 0. The oxidation number, has decreased.
This element has been reduced.
NO₃⁻ (aq) it's a ion, from nitric acid.
N acts with +5
O acts with -2
The global charge is -1
The N, has increased the oxidation state, so this element is the one oxidized.
