Answer:
0.471 mol/L
Explanation:
First, we'll begin by by calculating the number of mole of KMnO4 in 26g of KMnO4.
This is illustrated below:
Molar Mass of KMnO4 = 39 + 55 + (16x4) = 39 + 55 + 64 = 158g/mol
Mass of KMnO4 from the question = 26g
Mole of KMnO4 =?
Number of mole = Mass/Molar Mass
Mole of KMnO4 = 26/158 = 0.165mole
Now we can obtain the concentration of KMnO4 in mol/L as follow:
Volume of the solution = 350mL = 350/1000 = 0.35L
Mole of KMnO4 = 0.165mole
Conc. In mol/L = mole of solute(KMnO4)/volume of solution
Conc. In mol/L = 0.165mol/0.35
conc. in mol/L = 0.471mol/L
Answer:
It would be an element my dear.
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Pure gold is an element. That is, it is not a combination of other materials. However, gold used in jewelry is usually a “mixture”, called an alloy mixed with other metals to harden it, to change the color as in copper makes the pink or rose gold that is currently enjoying great polarity.
Answer:
Electrons that occur together in an orbital are called an electron pair. An electron will always try to enter the orbital with the lowest energy. ... In other words, within one energy level, electrons will fill an s orbital before starting to fill p orbitals. The s subshell can hold 2 electrons.
Explanation: PLZ BRAIN LeST THANKS :)
Answer:
9.07 L
Explanation:
<em>Calculate the volume occupied at s.t.p by 6.89 g of NH₃ gas [H = 1.0, N = 14.0].</em>
Step 1: Given and required data
- Molar mass of NH₃ (M): 17.0 g/mol
Step 2: Calculate the moles (n) of NH₃
We will use the following epxression.
n = m / M
n = 6.89 g / (17.0 g/mol) = 0.405 mol
Step 3: Calculate the volume occupied by 0.405 moles of NH₃ at STP
At STP, 1 mole of NH₃ occupies 22.4 L (assuming ideal behavior).
0.405 mol × 22.4 L/1 mol = 9.07 L
