If 4.34 moles of potassium react with excess water, how many moles of hydrogen gas would be produced at STP
Solved
Show answers
More tips
- H Health and Medicine Simple and Effective: How to Get Rid of Cracked Heels...
- O Other How to Choose the Best Answer to Your Question on The Grand Question ?...
- L Leisure and Entertainment History of International Women s Day: When Did the Celebration of March 8th Begin?...
- S Style and Beauty Intimate Haircut: The Reasons, Popularity, and Risks...
- A Art and Culture When Will Eurovision 2011 Take Place?...
- S Style and Beauty How to Choose the Perfect Hair Straightener?...
- F Family and Home Why Having Pets at Home is Good for Your Health...
- H Health and Medicine How to perform artificial respiration?...
- H Health and Medicine 10 Tips for Avoiding Vitamin Deficiency...
- F Food and Cooking How to Properly Cook Buckwheat?...
Answers on questions: Chemistry
- C Chemistry What is the symbiotic relationship of bighorn sheep and worms...
- P Physics If an investigator wanted to reproduce the results with a beam of protons, what would require to get a strong m = 1 reflection at ϕ = 50∘? If an investigator wanted to reproduce...
- H History FREE POINTS QUICK QUICK...
- H History The first state to vote to secede was A. South Carolina B. Maryland C. Missouri D. Georgia...
- E English There burning all the witches even if you aren t one they got their pitchforks and prooftheir receipts and reasonsthey burning all the witches even if you aren t oneso light me...
Ответ:
2.17 mol
Explanation:
Given data:
Number of moles of potassium = 4.34 mol
Moles of hydrogen gas produced at STP = ?
Solution:
Chemical equation:
2K + 2H₂O → 2KOH + H₂
Now we will compare the moles of hydrogen and potassium.
K : H₂
2 : 1
4.34 : 1/2×4.34 = 2.17 mol
So, from 4.34 moles of potassium 2.17 moles of hydrogen are produced at STP.
Ответ:
model a √ (x+b) + c = d
use a constant in place of each variable a, b, c, and d. you can use positive and negative constants in your equation.
a) a = -2, b = 3, c = -5, d = 7
=> -2√(x + 3) - 5 = 7
isolate the radical: - 2√(x+3) = 5 + 7
=> - 2√(x+3) = 12
=> √(x + 3) = -12/2
=> √x+3) = -6
square both sides
[√(x+3)]^2 =(-6)^2
=> x+3 = 36
=> = 36 - 3
=> x = 33
verify the validity of the solution, replacing 33 for x in the original equation:
=> -2√(33+ 3) - 5 = 7
=> -2 √(36) - 5 = 7
=> -2(6) - 5 = 7
=> -12 - 5 = 7
=> -17 = 7
which is not true. so, by solving the equation we obtained a solution which is no really a solution. that is what is called an extraneous solution, because it does not satisfy the equation, which means that you must discard that solution. in this case the conclusion is that the equation has not solution.
you could have prediceted it when you got √(x+3) = -6, given that the square root cannot lead to a negative number.
the reason to obtain the extraneous solution is that by squaring both sides you forced them to be positive.
that is why you must always verify the solutions of a radical equation and discard the extraneous ones.
b) a = 2, b = 3, c = -5, d = 7
=> 2√(x + 3) - 5 = 7
isolate the radical =>
2√(x + 3) = 12
=> √(x+3) = 12 / 2
=> √(x + 3) = 6
square both sides =>
x + 3 = 36
=> x = 36 - 3
=> x = 33
now replace the value found to verify the solution:
=> 2√(x + 3) - 5 = 7
=> 2 √(33 +3) - 5 = 7
=> 2√(36) - 5 = 7
=> 2(6) - 5 = 7
=> 12 - 5 = 7
=> 7 = 7
which proves that the solution is right.
part 2 and part three were answered in part one.
remember, some radical equations may have extraneous solutions because when you square the the sides you force to positive results which is not always valid.