Transporting metals, ions, water-insoluble molecules, and hormones. .... When erythrocytes are removed from circulation,
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The law of conservation of energy states that energy can neither be created nor destroyed - only converted from one form of energy to another. This means that a system always has the same amount of energy, unless it's added from the outside. ... The only way to use energy is to transform energy from one form to another.
Answer:

Explanation:
Hello,
In this case, since the density is defined as the ratio between the mass and the volume as shown below:

We can compute the density of water for the given 43 g that occupy the volume of 43 mL:

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Explanation:
The given data is as follows.
Mass of antimony = 19.75 g
Molar mass of Sb = 121.76 g/mol
Therefore, calculate number of moles of Sb as follows.
Moles of Sb = 
= 
= 0.162 mol
Mass of oxygen given is 6.5 g and molar mass of oxygen is 16 g/mol. Hence, moles of oxygen will be calculated as follows.
Moles of oxygen = 
= 
= 0.406 mol
Hence, ratio of moles of Sb and O will be as follows
Sb : O
1 : 2.5
We multiply both the ratio by 2 in order to get a whole number. Therefore, the ratio will be 2 : 5.
Thus, we can conclude that the empirical formula of the given oxide is
.