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Archy [21]
2 years ago
6

If you produce 14.5 g of O2 in the lab from a reaction that you showed by calculation could produce a maximum of 22.0 g of O2, w

hat is your percent yield?
Chemistry
2 answers:
Komok [63]2 years ago
5 0

Percent yield is simply the ratio of the actual amount produced over the theoretical amount that can be produced. In this case that would be:

percent yield = actual amount / theoretical amount

percent yield = (14.5 g / 22.0 g) * 100%

<span>percent yield = 65.91%</span>

almond37 [142]2 years ago
5 0

Answer: 65.9 %

Explanation:

The formula for calculating percentage yield is:

Percentage yield = \frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Given : Experimental yield = 14.5 g

Theoretical yield = 22.0 g

Putting in the values we get:

Percentage yield = \frac{14.5}{22.0}\times 100=65.9\%

Therefore, the percentage yield is 65.9 %.

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When an athlete gets hurt they apply sometimes an instant cold pack to the injury. The instant cold pack works by mixing two che
NeX [460]

Answer:

The evidence showing that there is a chemical reaction taking place is the instantaneous temperature drop once the cold pack is shaken.

Explanation:

When an athlete applies a cold pack to the injury, they shake it before, mixing the water and <em>ammonium-nitrate fertilizer</em> inside the cold pack. This mixing is an endothermic reaction, which means it absorbs heat. In turn, the temperature falls to 35 F for around 10 minutes.

3 0
2 years ago
Amy performed an experiment in lab. She improperly mixed the chemicals, and an explosion of light, sound, and heat occurred. Whe
agasfer [191]

The answer is D transformed

4 0
2 years ago
One of the most important chemical reactions is the Haber process, in which N2 and H2 are converted to ammonia which is used in
Lera25 [3.4K]

Answer:

c) 22

Explanation:

Let's consider the following balanced equation.

N₂(g) + 3 H₂(g) ----> 2 NH₃(l)

According to the balanced equation, 34.0 g of NH₃ are produced by 1 mol of N₂. For 170 g of NH₃:

170gNH_{3}.\frac{1molN_{2}}{34.0gNH_{3}} =5.00molN_{2}

According to the balanced equation, 34.0 g of NH₃ are produced by 3 moles of H₂. For 170 g of NH₃:

170gNH_{3}.\frac{3molH_{2}}{34.0gNH_{3}} =15.0molH_{2}

The total gaseous moles before the reaction were 5.00 mol + 15.0 mol = 20.0 mol.

We can calculate the pressure (P) using the ideal gas equation.

P.V = n.R.T

where

V is the volume (50.0 L)

n is the number of moles (20.0 mol)

R is the ideal gas constant (0.08206atm.L/mol.K)

T is the absolute temperature (400.0 + 273.15 = 673.2K)

P=\frac{n.R.T}{V} =\frac{20.0mol\times (0.08206atm.L/mol.K)\times 673.2K ) }{50.0L} =22.0atm

7 0
2 years ago
The concentration of Si in an Fe-Si alloy is 0.25 wt%. What is the concentration in kilograms of Si per cubic meter of alloy?
kodGreya [7K]

Answer : The concentration of Si in kilograms is, 19.55kg/m^3

Explanation :

As we are given that, the concentration of Si in an Fe-Si alloy is 0.25 wt% that means:

Weight of Si = 0.25 g = 0.00025 kg

Weight of Fe = 100 - 0.25 = 99.75 g = 0.09975 kg

Density of Si = 2.32g/cm^3=2.32\times 10^6g/m^3

Density of Fe = 7.87g/cm^3=7.87\times 10^6g/m^3

Now we have to calculate the concentration in kilograms of Si per cubic meter of alloy.

Concentration of Si in kilograms =  \frac{\text{Weight of Si in 100 g of alloy}}{\text{Volume of 100 g of alloy}}

Concentration of Si in kilograms =  \frac{\text{Weight of Si in 100 g of alloy}}{\frac{\text{Wight of Fe}}{\text{Density of Fe}}+\frac{\text{Wight of Si}}{\text{Density of Si}}}

Now put all the given values in this expression, we get:

Concentration of Si in kilograms = \frac{0.00025kg}{\frac{99.75g}{7.87\times 10^6g/m^3}+\frac{0.25g}{2.23\times 10^6g/m^3}}

Concentration of Si in kilograms = 19.55kg/m^3

Thus, the concentration of Si in kilograms is, 19.55kg/m^3

5 0
2 years ago
Suppose a soap manufacturer starts with a triglyceride that has the fatty acid chains arachidic acid, palmitic acid and palmitic
DIA [1.3K]

Answer:

Sodium arachidate; Sodium palmitate and Sodium palmitate

Explanation:

Triglycerides are esters of fatty acids with glycerol. In triglycerides, three fatty acid molecules are linked by ester bonds to each of the three carbon atoms in a glycerol molecule. The fatty acids may be same or different fatty acid molecules. Hydrolysis of triglycerides yields the three fatty acid molecules and glycerol.

Saponification is the process by which a base is used to catalyst the hydrolysis of the ester bonds in glycerides. The products of this base-catalyzed hydrolysis of triglycerides are the metallic salts of the three fatty acids and glycerol. The salts of the fatty acids are known as soaps.

For a triglyceride that has the fatty acid chains arachidic acid, palmitic acid and palmitic acid attached to the three backbone carbons glycerol, the saponification of the triglyceride with NaOH will yield the sodium salts or soaps of the three fatty acids as well as glycerol.

Arachidic acid will react with NaOH to yield sodium arachidate.

The two palmitic acid molecules will each react with NaOH to yield sodium palmitate.

8 0
1 year ago
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