This is an incomplete question, here is a complete question.
The Henry's law constant for oxygen dissolved in water is 4.34 × 10⁹ g/L.Pa at 25⁰C.If the partial pressure of oxygen in air is 0.2 atm, under atmospheric conditions, calculate the molar concentration of oxygen in air-saturated and oxygen saturated water.
Answer : The molar concentration of oxygen is, 
Explanation :
As we know that,
where,
= molar solubility of 
= ?
= partial pressure of = 0.2 atm = 1.97×10⁻⁶ Pa
= Henry's law constant = 4.34 × 10⁹ g/L.Pa
Now put all the given values in the above formula, we get:
Now we have to molar concentration of oxygen.
Molar concentration of oxygen = 
Therefore, the molar concentration of oxygen is,
11 with pemdas, you have to multiply all by 3
Answer:
two
Explanation:
The number of significant figures needed in the answer is 2.
This is because when finding the products of two numbers, the result is as accurate as the least number of significant figures of the numbers being multiplied.
Here the numbers being multiplied are;
1.31m
6.5m
1.31m has 3 significant figures
6.5m has 2 significant figures.
So, the product will have 2 significant figures
I: Current
V: Voltage
R: resistance
you’re welcome ;)
The answer is Three
!!!!!!
