AB+XY->AY+XB
We know that the answer would be KCl because of the switching that takes place during a double displacement reaction. Just like Zn and MnO4 switched and combined, the remaining elements, K and Cl, will combine.
We know that the answer is simply KCl because both K and Cl have an ion of only +/-1, meaning when they cross, no suffixes are made, since their ions are only 1.
For example, if you combined Mg with Cl, you would get MgCl2, because Mg has an ion of +2.
I hope this helps!
The percentage of Chromium in Chromium Oxide is calculated as follow,
Step 1: Calculate Molar mass of Cr₂O₃,
Cr = 51.99 u
O = 16 u
So,
2(51.99) + 3(16) = 103.98 + 48 = 151.98 u
Step 2: Secondly divide molar mass of only chromium with total mass of Cr₂O₃ and multiply with 100.
i.e.
=

× 100
=
68.41 %
So, the %age composition of chromium in chromium oxide is
68.41 %.
Answer:
0.11 mol
Explanation:
<em>This is the chemical formula for acetic acid (the chemical that gives the sharp taste to vinegar): CH₃CO₂H. An analytical chemist has determined by measurements that there are 0.054 moles of oxygen in a sample of acetic acid. How many moles of hydrogen are in the sample?</em>
Step 1: Given data
- Formula of acetic acid: CH₃CO₂H
- Moles of oxygen in the sample of acetic acid: 0.054 moles
Step 2: Establish the appropriate molar ratio
According to the chemical formula of acetic acid, the molar ratio of H to O is 4:2.
Step 3: Calculate the moles of atoms of hydrogen
We will use the theoretical molar ratio for acetic acid.
0.054 mol O × (4 mol H/2 mol O) = 0.11 mol H
Answer:
Electronegativity is probably the biggest thing that plays into reactivity. Therefore, since fluorine has a higher electronegativity than chlorine, fluorine is more reactive.
Explanation:
I got it right
Answer:
<h3>The answer is 1.02 g/cm³</h3>
Explanation:
The density of a substance can be found by using the formula

From the question
mass of block = 25 g
volume = 24.5 cm³
The density of the block is

We have the final answer as
<h3>1.02 g/cm³</h3>
Hope this helps you