Answer:
30 L H2
Explanation:
- 10 L N2 x <u>3 L H2</u> = 30 L H2
. 1 L N2
Try to verify my answer, Stoichiometry is not easy for me.
Answer:
No
Explanation:
They do not have any sensing organs. They can't see, smell or move on their own. They rely on random motions from their host to move around.
The mass of sodium bicarbonate (NaHCO₃) used in the experiment is 1.997 g
<h3>Calculating mass </h3>
From the question we are to calculate the mass of NaHCO₃ (sodium bicarbonate) used in the experiment
From the given information
Mass of empty evaporating dish = 46.233g
Mass of evaporating dish + Sodium bicarbonate = 48.230g
∴ Mass of sodium bicarbonate (NaHCO₃) = [Mass of evaporating dish + Sodium bicarbonate] - [Mass of empty evaporating dish]
Mass of sodium bicarbonate (NaHCO₃) = 48.230g - 46.233g
Mass of sodium bicarbonate (NaHCO₃) = 1.997 g
Hence, the mass of sodium bicarbonate (NaHCO₃) used in the experiment is 1.997 g
Learn more on Calculating mass here: brainly.com/question/15268826
To determine the molar mass, you need to get the atomic mass of the molecule. To do this, check the periodic table for the atomic mass or average atomic weight of each element.
Mg = 24.305 x 1 = 24.305 amu
O = 15.9994 x 2 =31.9988 amu
H = 1.0079 x 2 = 2.0158 amu
Then, add all the components to get the atomic mass of the molecule.
24.305 amu + 31.9988 amu + 2.0158 amu = 58.3196 amu
The atomic mass is just equivalent to its molar mass.
So, the molar mass of Magnesium hydroxide (Mg(OH)2) is 58.3196 g/mol.
Answer: -
3.151 M
Explanation: -
Let the volume of the solution be 1000 mL.
At 25.0 °C, Density = 1.260 g/ mL
Mass of the solution = Density x volume
= 1.260 g / mL x 1000 mL
= 1260 g
At 25.0 °C, the molarity = 3.179 M
Number of moles present per 1000 mL = 3.179 mol
Strength of the solution in g / mol
= 1260 g / 3.179 mol = 396.35 g / mol (at 25.0 °C)
Now at 50.0 °C
The density is 1.249 g/ mL
Mass of the solution = density x volume = 1.249 g / mL x 1000 mL
= 1249 g.
Number of moles present in 1249 g = Mass of the solution / Strength in g /mol
= 
= 3.151 moles.
So 3.151 moles is present in 1000 mL at 50.0 °C
Molarity at 50.0 °C = 3.151 M
