Answer:
It's <em>True</em> very true darling
Answer:
2KClO3 》》2KCl +3O2
C+ O2》》CO2
number of C moles
Required O2 moles (According to the mole ratio )
Relevant to the first equation, find the moles the KClO3, which is used to produce that amount of O2 moles
Now you can find the mass of KClO3
I mentioned the useful steps which can guide you to get the answer.
Explanation:
Number of coulombs of positive charge in 250cm^3 water is 1.3×10^7 C
The volume of 250 cm^3 corresponds to a mass of 250 g since the density of water is 1.0 g/cm^3
This mass corresponds to 250/18 = 14 moles since the molar mass of water is 18. There are ten proton (each with charge q = +e) in each molecule of 
So,
Q = 14NA q =14(6.02×10^23)(10)(1.60×10^−19C) = 1.3×10^7 C.
Mass is the quantity of matter in a physical body. It is also a measure of the body's inertia, the resistance to acceleration when a net force is applied. An object's mass also determines the strength of its gravitational attraction to other bodies.
Learn more about mass here:
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Well you should study how different chemicals work, and make flashcards to carry around or make notes during class to study for later
<span>NaCl
First calculate the molar mass of NaCl and AgNO3 by looking up the atomic weights of each element used in either compound
Sodium = 22.989769
Chlorine = 35.453
Silver = 107.8682
Nitrogen = 14.0067
Oxygen = 15.999
Now multiply the atomic weight of each element by the number of times that element is in each compound and sum the results
For NaCl
22.989769 + 35.453 = 58.44277
For AgNO3
107.8682 + 14.0067 + 3 * 15.999 = 169.8719
Now calculate how many moles of each substance by dividing the total mass by the molar mass
For NaCl
4.00 g / 58.44277 g/mol = 0.068443 mol
For AgNO3
10.00 g / 169.8719 g/mol = 0.058868
Looking at the balanced equation for the reaction, there is a 1 to 1 ratio in molecules for the reaction. Since there is a smaller number of moles of AgNO3 than there is of NaCl, that means that there will be some NaCl unreacted, so the excess reactant is NaCl</span>
