when copper sulfide is partially roasted in air (reaction with o2), copper sulfite is formed first. subsequently, upon heating,
1 answer:
The question is incomplete, complete question is:
When copper(I) sulfide is partially roasted in air (reaction with oxygen), copper(I) sulfite is formed first. subsequently, upon heating, the copper sulfite thermally decomposes to copper(I) oxide and sulfur dioxide. Write balanced chemical equations for these two reactions.
Answer:
The balanced chemical equations for these two reactions:
Explanation:
On partial roasting of copper sulfide in an air. The balanced chemical reaction is given as:
On further heating of copper(I) sulfite it get decomposes into copper oxide and sulfur dioxide. The balanced chemical reaction is given as:
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Answer:
20.(45)L or about 20.4545L
Explanation:
PV = nRT
Where:
P - pressure
V - volume
n - number of particle moles
R - a constant
T - temperature in K
We can assume the P and n (and definitely R) stay the same, so we infer that
2.15 x 10⁻³mL
Explanation:
Given parameter:
Volume of blood sample in uL = 2.15uL
Conversion uL → mL
micro- and milli- are both prefixes of sub-units.
liter is a unit of volume of a substance.
micro - is 10⁻⁶
milli- is of the order 10⁻³
The problem is converting from micro to milli:
if we multiply 10⁻⁶ by 10³ we would have our milli;
1000uL = 1mL
2.15uL : 2.15uL x = 2.15 x 10⁻³mL
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