Answer:
94.325 g
Explanation:
We'll begin by converting 350 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
350 mL = 350 mL × 1 L /1000 mL
350 mL = 0.35 L
Next, we shall determine the number of mole of KC₂H₃O₂ in the solution. This can be obtained as follow:
Volume = 0.35 L
Molarity of KC₂H₃O₂ = 2.75 M
Mole of KC₂H₃O₂ =?
Molarity = mole /Volume
2.75 = Mole of KC₂H₃O₂ / 0.35
Cross multiply
Mole of KC₂H₃O₂ = 2.75 × 0.35
Mole of KC₂H₃O₂ = 0.9625 mole
Finally, we shall determine the mass of KC₂H₃O₂ needed to prepare the solution. This can be obtained as illustrated below:
Mole of KC₂H₃O₂ = 0.9625 mole
Molar mass of KC₂H₃O₂ = 39 + (12×2) +(3×1) + (16×2)
= 39 + 24 + 3 + 32
= 98 g/mol
Mass of KC₂H₃O₂ =?
Mass = mole × molar mass
Mass of KC₂H₃O₂ = 0.9625 × 98
Mass of KC₂H₃O₂ = 94.325 g
Thus, the mass of KC₂H₃O₂ needed to prepare the solution is 94.325 g
Answer:
C) mass.
Explanation:
The speed of a body is given by the relation between the displacement of a body in a given time. It can be considered the greatness that measures how fast a body moves.
Speed analysis is divided into two main topics: average speed and instantaneous speed. It is considered a vector quantity, that is, it has a module (numerical value), a direction (Ex .: vertical, horizontal) and a direction (Ex .: forward, upwards). However, for elementary problems, where there is displacement in only one direction, the so-called one-dimensional movement, it is advisable to treat it as a scalar quantity (with only numerical value).
The mass of an object is not an important factor in determining the speed of that object. However, time, direction and distance are important factors in determining speed.
Answer: KMnO4-
Explanation:
You're looking at one potassium plus a polyatomic ion.
So K plus MnO4, equals:
KMnO4-
It also has a molar mass of 158.04 g/mol, I don't know if you need that, but I thought it would be nice to include it.
Answer:
MgBr2
Explanation:
(Mg^2+) + (Br^1-)
For every Mg we need 2 Br to balance out the compound.
78 protons are found in platinum(pt).
