The solution 550 ml total and first we will find the amount of alcohol. 3% = 0.03 550 ml x .03 = 16.5 ml alcohol
Then to find the amount of water used, we just have to subtract the amount of alcohol from the total volume
550 ml total - 16.5 ml alcohol = 533.5 ml water
Answer:
c) An element is made up of all the same type of atom
Explanation:
Atoms are the smallest unit of an element that consists of protons, electrons and neutrons in its structure. An element is the smallest part of a chemical substance that cannot be disintegrated i.e. it cannot be broken down further.
Atoms and elements are different in many ways but they are connected in the sense that an element contains only one type of atoms. For example, aluminum element is made up of only aluminum atoms. Different atoms form a molecule but same atoms form an element.
Answer:
2K + 2H2O → H2 + 2KOH
Explanation:
Find how many atoms you have on both sides then add 2 to both sides.
Reactant: Products:
K: 1+1=2 K: 1+1=2
H: 2+2=4 H: 3+1=4
O: 1+1=2 O: 1+1=2
Therefore it is balanced. Hope this helps
There are 0.566 moles of carbonate in sodium carbonate.
<h3>CALCULATE MOLES:</h3>
- The number of moles of carbonate (CO3) in sodium carbonate (Na2CO3) can be calculated by dividing the mass of carbonate in the compound by the molar mass of the compound.
- no. of moles of CO3 = mass of CO3 ÷ molar mass of Na2CO3
- Molar mass of Na2CO3 = 23(2) + 12 + 16(3)
- = 46 + 12 + 48 = 106g/mol
- mass of CO3 = 12 + 48 = 60g
- no. of moles of CO3 = 60/106
- no. of moles of CO3 = 0.566mol
- Therefore, there are 0.566 moles of carbonate in sodium carbonate.
Learn more about number of moles at: brainly.com/question/1542846
When a single compound breaks down into two or more compounds or elements in a chemical reaction then it is known as decomposition reaction.
The chemical symbol for sodium carbonate is 
.
The decomposition of sodium carbonate is:
The decomposition of sodium bicarbonate, will result in the formation of sodium oxide, 
and carbon dioxide, 
.
Hence, carbon dioxide, will produce with sodium oxide, on decomposition of .
