133.0 grams and 72.0 grams of Li2CO3 and Ca would be formed respectively.
From the equation of the reaction:
![2 Li + CaCO_3 ---> Li_2CO_3 + Ca](https://tex.z-dn.net/?f=2%20Li%20%2B%20CaCO_3%20---%3E%20Li_2CO_3%20%2B%20Ca)
The mole ratio of Li to Li2CO3 is 2:1 while that of Li to Ca is also 2:1.
Mole of 25.0 grams of Li = mass/molar mass
= 25/6.94
= 3.6 moles
Equivalent of mole of Li2CO3 = 3.6/2
= 1.8 moles
- Amount of Li2CO3 formed = mole x molar mass
= 1.8 x 73.89
= 133 grams
Equivalent mole of Ca = 3.6/2
= 1.8 moles
- Amount of Ca formed = 1.8 x 40
= 72 grams
More on stoichiometric calculations can be found here: brainly.com/question/2563006?referrer=searchResults
If your question means "How can you tell if a substance is a binary compound?"
<span>Then I'll answer this.
A binary compound is a chemical compound composed of only two elements.
</span>
Answer:
15 oxygens
Explanation:
Given the partially balanced reaction: ![\text{C}_{\text{5}}\text{H}_{\text{10}}+\text{O}_{\text{2}} \rightarrow \text{5C} \text{O}_{2}}+\text{5H}_\text{2}\text{O}](https://tex.z-dn.net/?f=%5Ctext%7BC%7D_%7B%5Ctext%7B5%7D%7D%5Ctext%7BH%7D_%7B%5Ctext%7B10%7D%7D%2B%5Ctext%7BO%7D_%7B%5Ctext%7B2%7D%7D%20%5Crightarrow%20%5Ctext%7B5C%7D%20%5Ctext%7BO%7D_%7B2%7D%7D%2B%5Ctext%7B5H%7D_%5Ctext%7B2%7D%5Ctext%7BO%7D)
The subscripts (small number to the right of each element symbol) are the number of atoms of the element within each compound/molecule, and the coefficients (numbers in front of each compound) represent the number of that molecule involved in one full reaction (if the equation were balanced).
The product side of the reaction is on the right of the arrow.
To determine the total number of Oxygens on the product side, we need to identify how many Oxygens are in each molecule (the subscript on the Oxygen), and then multiply times the number of that molecule that would be involved (Coefficient of the compound containing Oxygen). There are multiple compounds on the right side of the equation that contain Oxygen, so we'll need to add together the number of Oxygens each part contributes.
![\text{C}_{\text{5}}\text{H}_{\text{10}}+\text{O}_{\text{2}} \rightarrow \bold{5}\text{C} \bold{O_{2}}+\bold{5}\text{H}_\text{2}\bold{O}](https://tex.z-dn.net/?f=%5Ctext%7BC%7D_%7B%5Ctext%7B5%7D%7D%5Ctext%7BH%7D_%7B%5Ctext%7B10%7D%7D%2B%5Ctext%7BO%7D_%7B%5Ctext%7B2%7D%7D%20%5Crightarrow%20%5Cbold%7B5%7D%5Ctext%7BC%7D%20%5Cbold%7BO_%7B2%7D%7D%2B%5Cbold%7B5%7D%5Ctext%7BH%7D_%5Ctext%7B2%7D%5Cbold%7BO%7D)
![\text{\# reactant-side Oxygens}=\bold{5}\text{C} \bold{O_{2}}+\bold{5}\text{H}_\text{2}\bold{O}\\=5 \text{ CO}_{\text{2}}\text{ molecules} *\frac{\text{2 Oxygens}}{\text{1 CO}_{\text{2}}\text{ molecule}}+5 \text{ H}_{\text{2}}\text{O molecules} *\frac{1\text{ Oxygen}}{\text{1 H}_{\text{2}}\text{O molecule}}\\=10\text{ Oxygens}+5\text{ Oxygens}\\=15\text{ Oxygens}](https://tex.z-dn.net/?f=%5Ctext%7B%5C%23%20reactant-side%20Oxygens%7D%3D%5Cbold%7B5%7D%5Ctext%7BC%7D%20%5Cbold%7BO_%7B2%7D%7D%2B%5Cbold%7B5%7D%5Ctext%7BH%7D_%5Ctext%7B2%7D%5Cbold%7BO%7D%5C%5C%3D5%20%5Ctext%7B%20CO%7D_%7B%5Ctext%7B2%7D%7D%5Ctext%7B%20molecules%7D%20%2A%5Cfrac%7B%5Ctext%7B2%20Oxygens%7D%7D%7B%5Ctext%7B1%20CO%7D_%7B%5Ctext%7B2%7D%7D%5Ctext%7B%20molecule%7D%7D%2B5%20%5Ctext%7B%20H%7D_%7B%5Ctext%7B2%7D%7D%5Ctext%7BO%20molecules%7D%20%2A%5Cfrac%7B1%5Ctext%7B%20Oxygen%7D%7D%7B%5Ctext%7B1%20H%7D_%7B%5Ctext%7B2%7D%7D%5Ctext%7BO%20molecule%7D%7D%5C%5C%3D10%5Ctext%7B%20Oxygens%7D%2B5%5Ctext%7B%20Oxygens%7D%5C%5C%3D15%5Ctext%7B%20Oxygens%7D)
It is classified as a noble gas
Answer:
Sodium chloride solution:
First you need to calculate the mass of salt needed (done in the explanation), which is 58.44g. Then it have to be weighted in an analytical balance in a weighting boat and then transferred into a 2L volumetric flask that is going to be filled until the mark with distilled water.
Sulfuric acid dilution:
First you need to calculate the volume needed (done in the explanation), it is 16.6 mL. Using a graduated pipette one measures this volume and transfer it into a 2L volumetric flask that is already half filled with distilled water, and then one fills it until its mark.
Explanation:
Sodium chloride solution:
Each liter of a 0.500M solution has half mol, so 2L of said solution has 1 mol of salt. Sodium chloride molar mass is 58.44g/mol, so in 2L of solution there is 58.44g of salt. That`s the mass that`s going to be weighted and transferred to a 2L volumetric flask.
Sulfuric acid dilution:
This is the equation for dilution of solutions:
Where "c1" stands for the initial concentration (stock solution concentration), "v1" for the initial volume (volume of stock solution used), "c2" for the desired concentration and "v2" for the desired volume.
When we are diluting from a stock solution we want to know how much do we have to pipette from the stock solution into our volumetric flask. We do so by isolating the "v1" term from the dilution equation:
in this case that would be: