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
Atoms have electrons filled in energy shells.
1. H - hydrogen atom has one electron in the First energy shell. Therefore hydrogen has a partially filled first energy shell
2.Li - Li electron configuration is 2,1
The outermost energy shell is the second energy shell in which there is only one electron
Therefore the second energy shell is partially filled. This is the correct answer
3. K - electron configuration is 2,8,8,1
The outermost energy shell is the fourth energy shell which is partially filled. The second energy shell is completely filled
4.Na - electron configuration is 2,8,1
The outermost energy shell is the third energy shell which is partially filled
Second energy shell is completely filled
From the given options Li is the only element with a partially filled second energy shell
Answer is Li
Answer:
the molar mass of any element can be determined by finding the atomic mass of the element on the periodic table for example, if the atomic mass of sulfer is 32.066 amu, then it's molar mass is 32.066 g / mol
Answer:
Other side
Opposite function
On both sides of the equation
In the numerator and not the denominator
Explanation:
To isolate a single variable when rearranging equations, move all other variables to the other side of the equation by using the opposite function on them and remembering to perform that operation on both sides of the equation. Make sure the rearrangement has the target variable in the numerator, not the denominator.
