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2 years ago

Explain with example the need of standard measurement system

Chemistry

2 answers:

steposvetlana [31]2 years ago

4 0

Answer:

A standard measurement system is important because it allows scientists to compare data and communicate with each other about their results. Metric System – is a measurement system based on the number 10.

cluponka [151]2 years ago

3 0

Answer:

A standard system of measurement is very important because it helps us from making mistakes at the moment of doing calculations and correctly working...

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Ethyl butyrate, CH3CH2CH2CO2CH2CH3, is an artificial fruit flavor commonly used in the food industry for such flavors as orange

SIZIF [17.4K]

Answer:

A. 10.0 grams of ethyl butyrate would be synthesized.

B. 57.5% was the percent yield.

C. 7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

Explanation:

$CH_3CH_2CH_2CO_2H(l)+CH_2CH_3OH(l)+H^+\rightarrow CH_3CH_2CH_2CO_2CH_2CH_3(l)+H_2O(l)$

Moles of butanoic acid = $\frac{7.60 g}{88 g/mol}=0.08636 mol$

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

$\frac{1}{1}\times 0.08636 mol=0.08636 mol$ of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 100%

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$100\%=\frac{\text{Experimental yield}}{10.0 g}\times 100$

Experimental yield = 10.0 g

10.0 grams of ethyl butyrate would be synthesized.

Theoretical yield of ethyl butyrate = 10.0 g

Experimental yield ethyl butyrate = 5.75 g

Percentage yield of the reaction = ?

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$=\frac{5.75 g}{10.0 g}\times 100=57.5\%$

57.5% was the percent yield.

Moles of butanoic acid = $\frac{7.60 g}{88 g/mol}=0.08636 mol$

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

$\frac{1}{1}\times 0.08636 mol=0.08636 mol$ of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 78.0%

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$78.0\%=\frac{\text{Experimental yield}}{10.0 g}\times 100$

Experimental yield = 7.80 g

7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

8 0

2 years ago

What does variable mean in a science project.

Paul [167]

Well, an independent variable ids the thing that stays the same and a dependent variable is the thing that changes.

6 0

3 years ago

Read 2 more answers

The equations given in the problem introduction can be added together to give the following reaction: overall: N2O4→2NO2 However

Murljashka [212]

Answer:

Reverse the $2 NO_2 \longrightarrow 2 NO + O_2$ reaction

Explanation:

Reactions:

$2 NO_2 \longrightarrow 2 NO + O_2$

$N_2O_4 \longrightarrow 2 NO + O_2$

Overall:

$N_2O_4 \longrightarrow 2 NO_2$

As can be seen, in the overall reaction we have $N_2O_4$ in the reactants like in the second reaction and $NO_2$ in the products. The $NO_2$ is in the first reaction but as a reactant so we need to reverse that reaction:

$2 NO + O_2 \longrightarrow 2 NO_2$

$N_2O_4 \longrightarrow 2 NO + O_2$

Combining:

$N_2O_4 + 2 NO + O_2\longrightarrow 2 NO + O_2 + 2 NO_2$

$N_2O_4 \longrightarrow 2 NO_2$

4 0

2 years ago

A balloon has a volume of 1.75L at a temperature of 298K. What will be the volume of the balloon if you take it out into the win

lozanna [386]

Answer:

1.52 L

Explanation:

P1V1T2=V2P2T1

V2=V1T2/T1

Fill in with given values then solve

We can check this by knowing that V and T at constant P have a proportional relationship. Hence, this is correct.

- Hope that helped! Please let me know if you need further explanation.

7 0

2 years ago

How can scientists detect blood doping?

aev [14]

Answer:

There are typically three ways that it is accomplished: use of erythropoietin (EPO) or synthetic oxygen carriers and blood transfusions. While transfusions of large volumes of blood or use of EPO can be detected, microdosing EPO or transfusing smaller volumes of packed red blood cells is much harder to detect.

6 0

2 years ago

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Explain with example the need of standard measurement system​

Explain with example the need of standard measurement system