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Arturiano [62]
2 years ago
5

Does the excess reactant get used up completely in a reaction??

Chemistry
1 answer:
Alex787 [66]2 years ago
4 0

Answer:

In a chemical reaction, reactants that are not used up when the reaction is finished are called excess reagents. The reagent that is completely used up or reacted is called the limiting reagent, because its quantity limits the amount of products formed.

Explanation:

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Question 11 (5 points)
LenKa [72]

Answer:

False

Explanation:

On the left side of the equation (Li + O2), there is 1 Li atom and 2 O atoms.

but on thw right side of the equation (Li2O,) there is 2 Li atoms and 1 O atom

8 0
3 years ago
The diameter of a biscuit is approximately 51 millimeters (mm). An atom of bismuth (Bi) is approximately 320. picometers (pm) in
astraxan [27]

Answer:

1.5e+8 atoms of Bismuth.

Explanation:

We need to calculate the <em>ratio</em> of the diameter of a biscuit respect to the diameter of the atom of bismuth (Bi):

\\ \frac{diameter\;biscuit}{diameter\;atom(Bi)}

For this, it is necessary to know the values in meters for any of these diameters:

\\ 1m = 10^{3}mm = 1e+3mm

\\ 1m = 10^{12}pm = 1e+12pm

Having all this information, we can proceed to calculate the diameters for the biscuit and the atom in meters.

<h3>Diameter of an atom of Bismuth(Bi) in meters</h3>

1 atom of Bismuth = 320pm in diameter.

\\ 320pm*\frac{1m}{10^{12}pm} = 3.20*10^{-10}m

<h3>Diameter of a biscuit in meters</h3>

\\ 51mm*\frac{1}{10^{3}mm} = 51*10^{-3}m = 5.1*10^{-2}m

<h3>Resulting Ratio</h3>

How many times is the diameter of an atom of Bismuth contained in the diameter of the biscuit? The answer is the ratio described above, that is, the ratio of the diameter of the biscuit respect to the diameter of the atom of Bismuth:

\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1*10^{-2}m}{3.20*10^{-10}m}

\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1}{3.20}\frac{10^{-2}}{10^{-10}}\frac{m}{m}

\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1}{3.20}\frac{10^{-2}}{10^{-10}}\frac{m}{m}

\\ Ratio_{\frac{biscuit}{atom}}= 1.5*10^{-2+10}

\\ Ratio_{\frac{biscuit}{atom}}= 1.5*10^{8}=1.5e+8

In other words, there are 1.5e+8 diameters of atoms of Bismuth in the diameter of the biscuit in question or simply, it is needed to put 1.5e+8 atoms of Bismuth to span the diameter of a biscuit in a line.

6 0
3 years ago
The radius of the smaller circle is 8 feet. The distance from the rim of the inner circle to the rim of the outer circle is 3 fe
postnew [5]

Answer:

178.98 sq. feet

Explanation:

The path and the garden has been shown in the figure below. The green area is the garden and the area in brown is the path.

It has been given that,

Radius of garden = 8 feet

So, the area of garden = 3.14 × 8 × 8 = 200.96 sq. feet

The total radius of the land including garden and path = 8 + 3 = 11 feet

So, the total are of land including garden and path = 3.14 × 11 × 11 = 379.94 sq. feet

So, the area of path = Total area of the land - area of garden

Area of path = 379.94 - 200.96 = 178.98 sq. feet

5 0
3 years ago
Determining what it is you want to know:
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Am so pro let it goo let it goooo
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2 years ago
you see a coin at the bottom of a fountain of water but when you reach for it it is in another place than it appears to be
Roman55 [17]
Objects in the fountain appear to be somewhere but isnt
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3 years ago
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