6CO2 + 6H2O → C6H12O6 + 6O2.
So,
These species react stoichiometrically. Thus, we can say, for example, that 1 mol A is stoichiometrically equivalent to 1 mol B in this reaction. I will say A = B.
A = B
A = C
A = D
B = C
B = D
C = D
As you can see, there are more than 4 distinct mole ratios for this reaction. Thus, the statement is false.
It will be extracted only 1/3 of NaCl less in 10 mL of water than in 30 mL of water.
If it is known that solubility of NaCl is 360 g/L, let's find out how many NaCl is in 30 mL of water:
360 g : 1 L = x g : 30 mL
Since 1 L = 1,000 mL, then:
360 g : 1,000 mL = <span>x g : 30 mL
Now, crossing the products:
x </span>· 1,000 mL = 360 g · 30 mL
x · 1,000 mL = 10,800 g mL
x = 10,800 g ÷ 1,000
x = 10.8 g
So, from 30 mL mixture, 10.8 g of NaCl could be extracted.
Let's calculate the same for 10 mL water instead of 30 mL.
360 g : 1 L = x g : 10 mL
Since 1 L = 1,000 mL, then:
360 g : 1,000 mL = <span>x g : 10 mL
Now, crossing the products:
x </span>· 1,000 mL = 360 g · 10 mL
x · 1,000 mL = 3,600 g mL
x = 3,600 g ÷ 1,000
<span>x = 3.6 g
</span>
<span>So, from 10 mL mixture, 3.6 g of NaCl could be extracted.
</span>
Now, let's compare:
If from 30 mL mixture, 10.8 g of NaCl could be extracted and <span>from 10 mL mixture, 3.6 g of NaCl could be extracted, the ratio is:
</span>3.6/10.8 = 1/3
Therefore, i<span>t will be extracted only 1/3 of NaCl less in 10 mL of water than in 30 mL of water.
</span>
The answer is D
Sodium loses an Electron and chlorine gains one
