Answer:
The larger piston
Explanation:
(i just took the same test)
Answer:- 3.68M.
Solution:- We have a 22.0% by mass solution of ethylene glycol and it's density is 1.04 gram per mL. It ask to calculate the molarity of the solution. We know that molarity is moles of solute per liter of solution. So, we need to figure out the moles of ethylene glycol and volume of solution.
Let's say we have 100 grams of the solution. Then mass of ethylene glycol would be 22.0 grams. Molar mass of ethylene glycol is 62.07 gram per mol.
Let's calculate it's moles first:

= 
From mass and density we calculate the volume of the solution and convert it to liters as:

= 0.0962 L

= 3.68M
So, the molarity of ethylene glycol solution is 3.68M.
Answer:
B
Explanation:
Carbon Dioxide is one of waste produced from Our cells during respiration. The circulatory system carries Carbon Dioxide from cells either in haemaglobin in RBC or plasm. The Carbon Dioxide being carried to blood cappilaries near alveoli before diffused into alveoli and then being removed away by respiratory system in the process of respiration while we exhale
Answer:
10.76 grams
Explanation:
Given that the amount of
is 25.0 grams.
Mass of 1 mole of
= 158 grams
The number of atoms in 1 mole of
is 4.
Mass of oxygen in 1 mole of
= 16\times 4 = 68 grams.
Here, 158 grams of
has 68 grams of oxygen
So, the amount of oxygen in 1 gram of
= 68/158 grams
Therefore, the amount of oxygen in 1 gram of
=
grams
=10.76 grams
Hence, 25 grams of
has 10.76 grams of oxygen.
Answer:
3.6 × 10²⁴ molecules
Explanation:
Step 1: Given data
Moles of methane (n): 6.0 moles
Step 2: Calculate the number of molecules of methane in 6.0 moles of methane
In order to convert moles to molecules, we need a conversion factor. In this case, we will use Avogadro's number: there are 6.02 × 10²³ molecules of methane in 1 mole of molecules of methane.
6.0 mol × 6.02 × 10²³ molecules/1 mol = 3.6 × 10²⁴ molecules