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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Answer:
They are in constant motion.
Explanation:
More energy\heat= more kinetic energy=more motion\movement
The answer is 64.907 amu.
The atomic mass of an element is the average of the atomic masses of its isotopes. The relative abundance of isotopes must be taken into consideration, therefore:
atomic mass of copper = atomic mass of isotope 1 * abundance 1 + atomic mass of isotope 2 * abundance 2
We know:
atomic mass of copper = 63.546 amu
The atomic mass of isotope 1 is: 62.939 amu
The abundance of isotope 1 is: 69.17% = 0.6917
The atomic mass of isotope 1 is: x
The abundance of isotope 2: 100% - 69.17% = 30.83% = 0.3083
Thus:
63.546 amu = 62.939 amu * 0.6917 + x * 0.3083
63.546 <span>amu = 43.535 amu + 0.3083x
</span>⇒ 63.546 amu - 43.535 amu = 0.3083x
⇒ 20.011 amu = 0.3083x
⇒ x = 20.011 amu ÷ 0.3083 = 64.907 amu
Answer:
Y = 62.5%
Explanation:
Hello there!
In this case, for the given chemical reaction whereby carbon dioxide is produced in excess oxygen, it is firstly necessary to calculate the theoretical yield of the former throughout the reacted 10 grams of carbon monoxide:

Finally, given the actual yield of the CO2-product, we can calculate the percent yield as shown below:

Best regards!