There are 3 significant figures in this value, all values before and after the decimal point are significant. As there is a decimal point, the zeros trailing are also significant.
Answer:
Zn(s) → Zn⁺²(aq) + 2e⁻
Explanation:
Let us consider the complete redox reaction:
Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)
This is a redox reaction because, both oxidation and reduction is simultaneously taking place.
- Oxidation (loss of electrons or increase in the oxidation state of entity)
- Reduction (gain of electrons or decrease in the oxidation state of the entity)
- An element undergoes oxidation or reduction in order to achieve a stable configuration. It can be an octet configuration. An octet configuration is that of outer shell configuration of noble gas.
Here Zn(s) is undergoing oxidation from OS 0 to +2
And H in HCl (aq) is undergoing reduction from OS +1 to 0.
Therefore, for this reaction;
Oxidation Half equation is:
Zn(s) → Zn⁺²(aq) + 2e⁻
Reduction Half equation is:
2H⁺ + 2e⁻ → H₂(g)
Deuterium is a relatively uncommon form of hydrogen, but can be created from water.
- Heavy hydrogen commonly known as deuterium
- stable isotopes of hydrogen
- gets its name from the Greek word deuterons means second.
- has only one proton and one neutron
- nucleus of the hydrogen's deuterium atom is known as a deuteron containing one proton and one neutron.
- Deuterium forms chemical bonds that are stronger than regular hydrogen
- gas deuterium is colorless
- Deuterated water is used in Magnetic Resonance Spectroscopy.
- used in the determination of the isotopologue of various organic compounds.
- used in Infrared Spectroscopy.
To know more about Deuterium visit : brainly.com/question/27870183
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To Find :
The horizontal range of a projectile launched at 15 degrees to the horizontal with speed of 40 m/s.
Solution :
The horizontal range of a projectile is given by :
( Here, g is acceleration due to gravity = 10 m/s² )
Putting all value in above equation :
Therefore, the horizontal range of projectile is 80 m.
Answer:
6 x 10⁵ kg Hg
Explanation:
The mass of mercury in the entire lake is found by multiplying the concentration of the mercury by the volume of the lake.
The volume of the lake is calculated in cubic feet:
V = (SA)x(depth) = (100mi²)(5280ft/mi)² x (20ft) = 5.57568 x 10¹⁰ ft³
Cubic feet are then converted to mL (1cm³=1mL)
(5.57568 x 10¹⁰ ft³) x (12in/ft)³ x (2.54cm/in)³ = 1.578856752 x 10¹⁵ mL
The mass of mercury is then found:
m = CV = (0.4μg/mL)(1g/10⁶μg)(1kg/1000g) x (1.578856752 x 10¹⁵ mL) = 6 x 10⁵ kg Hg
