This is due to the solubility of CO₂ in blood is more than the solubility of Oxygen in blood
<u>Answer:</u> The final concentration of 
is 0.019767 M
<u>Explanation:</u>
Molarity is calculated by using the equation:
Initial moles of = 0.3 moles
Volume of solution = 3.1 L
For the given chemical equation:
<u>Initial:</u> 0.097
<u>At eqllm:</u> 0.097-x x x
The expression of 
for above equation follows:
![K_c=\frac{[POCl][Cl_2]}{[POCl_3]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BPOCl%5D%5BCl_2%5D%7D%7B%5BPOCl_3%5D%7D)
We are given:
Putting values in above expression, we get:
Neglecting the negative value of 'x' because concentration cannot be negative.
So, final concentration of = (0.097 - x) = (0.097 - 0.077233) = 0.019767 M
Hence, the final concentration of is 0.019767 M
Answer:
% composition of glucose = 85 %
Explanation:
Given Data
moles of glucose (C₆H₁₂O₆) = 1.3 mole
Total mass of mixture = 276 g
percent composition of glucose (C₆H₁₂O₆) = ?
Solution
Number of moles are given so we find out the mass of C₆H₁₂O₆
<em>mass = moles × molar mass</em>
Molar mass of glucose = 180.156 g/mol
mass = 1.3 mol × 180.156 g/mol
mass = 234.2 g
Now we find out the percent composition of glucose
% composition = (mass of compound ÷ mass of mixture) × 100
% composition = ( 234.2 g ÷ 276 g) × 100
% composition = 0.85 × 100
% composition = 85 %
Answer:
C: remaining at rest
Explanation:
Acceleration (a) is the change in velocity (Δv), remaining at rest is not changing the velocity.
Answer:
it is b because its releases heat in to all directions and not b because it staying inside and not releasing anything :)
Explanation:
