Answer:
have the same number of atoms of each element in the reactants and in the products
Explanation:
<em>The basic principle in balancing a chemical equation would simply be to have the same number of atoms of each element in the reactants and in the products.</em>
<u>A balanced chemical equation is one that has the same number of atoms of each element on the reactant and the product's side of the equation.</u> For example, consider the equation below:
![H_2 + O_2 --> H_2O](https://tex.z-dn.net/?f=H_2%20%2B%20O_2%20--%3E%20H_2O)
On the reactant's side, there are 2 atoms of H and O while there are 2 atoms of H and 1 atom of O on the product's side. This is an imbalanced equation. In order for it to be balanced, the number of atoms of H and O on the reactant side must be equal to the number of H and O on the product side as below.
![2H_2 + O_2 --> 2H_2O](https://tex.z-dn.net/?f=2H_2%20%2B%20O_2%20--%3E%202H_2O)
Answer:
a) The theoretical yield is 408.45g of ![BaSO_{4}](https://tex.z-dn.net/?f=BaSO_%7B4%7D)
b) Percent yield = ![\frac{realyield}{408.45g}*100](https://tex.z-dn.net/?f=%5Cfrac%7Brealyield%7D%7B408.45g%7D%2A100)
Explanation:
1. First determine the numer of moles of
and
.
Molarity is expressed as:
M=![\frac{molessolute}{Lsolution}](https://tex.z-dn.net/?f=%5Cfrac%7Bmolessolute%7D%7BLsolution%7D)
- For the ![BaCl_{2}](https://tex.z-dn.net/?f=BaCl_%7B2%7D)
M=![\frac{1.75molesBaCl_{2}}{1Lsolution}](https://tex.z-dn.net/?f=%5Cfrac%7B1.75molesBaCl_%7B2%7D%7D%7B1Lsolution%7D)
Therefore there are 1.75 moles of ![BaCl_{2}](https://tex.z-dn.net/?f=BaCl_%7B2%7D)
- For the ![Na_{2}SO_{4}](https://tex.z-dn.net/?f=Na_%7B2%7DSO_%7B4%7D)
M=
}{1Lsolution}[/tex]
Therefore there are 2.0 moles of ![Na_{2}SO_{4}](https://tex.z-dn.net/?f=Na_%7B2%7DSO_%7B4%7D)
2. Write the balanced chemical equation for the synthesis of the barium white pigment,
:
![BaCl_{2}+Na_{2}SO_{4}=BaSO_{4}+2NaCl](https://tex.z-dn.net/?f=BaCl_%7B2%7D%2BNa_%7B2%7DSO_%7B4%7D%3DBaSO_%7B4%7D%2B2NaCl)
3. Determine the limiting reagent.
To determine the limiting reagent divide the number of moles by the stoichiometric coefficient of each compound:
- For the
:
![\frac{1.75}{1}=1.75](https://tex.z-dn.net/?f=%5Cfrac%7B1.75%7D%7B1%7D%3D1.75)
- For the
:
![\frac{2.0}{1}=2.0](https://tex.z-dn.net/?f=%5Cfrac%7B2.0%7D%7B1%7D%3D2.0)
As the
is the smalles quantity, this is the limiting reagent.
4. Calculate the mass in grams of the barium white pigment produced from the limiting reagent.
![1.75molesBaCl_{2}*\frac{1molBaSO_{4}}{1molBaCl_{2}}*\frac{233.4gBaSO_{4}}{1molBaSO_{4}}=408.45gBaSO_{4}](https://tex.z-dn.net/?f=1.75molesBaCl_%7B2%7D%2A%5Cfrac%7B1molBaSO_%7B4%7D%7D%7B1molBaCl_%7B2%7D%7D%2A%5Cfrac%7B233.4gBaSO_%7B4%7D%7D%7B1molBaSO_%7B4%7D%7D%3D408.45gBaSO_%7B4%7D)
5. The percent yield for your synthesis of the barium white pigment will be calculated using the following equation:
Percent yield = ![\frac{realyield}{theoreticalyield}*100](https://tex.z-dn.net/?f=%5Cfrac%7Brealyield%7D%7Btheoreticalyield%7D%2A100)
Percent yield = ![\frac{realyield}{408.45g}*100](https://tex.z-dn.net/?f=%5Cfrac%7Brealyield%7D%7B408.45g%7D%2A100)
The real yield is the quantity of barium white pigment you obtained in the laboratory.
Answer:
A drought poses a huge threat to all life. If a drought occurred the entire food chain would disintegrate within months. There would be no water for any animals or plants. Small mammals would not be able to eat plants, and reptiles would not be able to the small animals, and so on. A drought can destroy an ecosystem in a short amount of time.
Water is only being moved into the air through water vapor, so the air will become hotter than water. Land is also not moving in the atmosphere and absorbing heat like air is, so air will also be hotter than land, depending on what the land is made of.
Through precipitation, water in the atmosphere can return to the hydrosphere or percolate into the ground to become groundwater—part of the geosphere. ... Water in the biosphere can be released into the atmosphere through transpiration in plants, or respiration in animals.
Explanation:
We know that to relate solutions of with the factors of molarity and volume, we can use the equation:
![M_{1} V_{1} = M_{2} V_{2}](https://tex.z-dn.net/?f=%20M_%7B1%7D%20%20V_%7B1%7D%20%3D%20M_%7B2%7D%20%20V_%7B2%7D%20)
**
NOTE: The volume as indicated in this question is defined in L, not mL, so that conversion must be made. However it is 1000 mL = 1 L.
So now we can assign values to these variables. Let us say that the 18 M
![H_{2} SO_{4}](https://tex.z-dn.net/?f=%20H_%7B2%7D%20%20SO_%7B4%7D%20)
is the left side of the equation. Then we have:
![(18 M)(0.050 L)=(4.35M) V_{2}](https://tex.z-dn.net/?f=%2818%20M%29%280.050%20L%29%3D%284.35M%29%20V_%7B2%7D%20)
We can then solve for
![V_{2}](https://tex.z-dn.net/?f=%20V_%7B2%7D%20)
:
![V_{2}= \frac{(18M)(0.05L)}{4.35M}](https://tex.z-dn.net/?f=%20V_%7B2%7D%3D%20%5Cfrac%7B%2818M%29%280.05L%29%7D%7B4.35M%7D%20%20)
and
![V_{2} =0.21 L](https://tex.z-dn.net/?f=%20V_%7B2%7D%20%3D0.21%20L)
or
![210 mL](https://tex.z-dn.net/?f=210%20mL)
We now know that the total amount of volume of the 4.35 M solution will be
210 mL. This is assuming that the entirety of the 50 mL of 18 M is used and the rest (160 mL) of water is then added.