Answer:BaCl (aq)+ K2CO3 (aq)——> Ba2CO3 (aq) + KCl (aq)
Explanation:
The equation is already balanced and it is a double replacement reaction.
Answer:
If you are given a chemical equation and specific amounts for each reactant in grams, you have to follow these steps, in order, to determine how much product can possilby be made:
1. Convert each reactant into moles of the product.
2. Determine which reactant is the limiting reactant.
3. Convert the moles of product, from the limiting reactant, to grams.
Explanation:
150,000 grams. Multiply 100 by 50 by 30 to find the volume of the container, which is 150,000 cm^3. knowing that a milliliter is just one cubic centimeter, and that water is one gram per milliliter, you can conclude that it is 150,000 grams of water.
Answer:
4.48 grams of potassium hydroxide that the chemist must be weighing out.
Explanation:
The pH of the KOH solution = 13
pH + pOH = 14
pOH = 14 - pH = 14 - 13 = 1
![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
![1=-\log[OH^-]](https://tex.z-dn.net/?f=1%3D-%5Clog%5BOH%5E-%5D)
![[OH^-]=0.1 M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D0.1%20M)

1 mole of hydroxide ions are obtained from 1 mole of KOH. Then 0.1 mole of hydroxide ions will be obtained from :
of KOH
![[Molarity]=\frac{\text{Moles of solute}}{\text{Volume of solution(L)}}](https://tex.z-dn.net/?f=%5BMolarity%5D%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20solute%7D%7D%7B%5Ctext%7BVolume%20of%20solution%28L%29%7D%7D)
Volume of KOH solution = 800 mL = 0.800 L ( 1 mL = 0.001 L)

Moles of KOH = 0.1 M × 0.800 L = 0.08 mol
Mass of 0.08 moles of KOH :
0.08 mol × 56 g/mol = 4.48 g
4.48 grams of potassium hydroxide that the chemist must be weighing out.
It begins by trying to solve a problem and or making a hypothesis.