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Katena32 [7]
3 years ago
13

If 15 drops of ethanol from a medical dropper weight 0.60g, how many drops does it takes from a dropper to dispense 1.0ml of eth

anol? The density of ethanol is 0.80g/ml
Chemistry
1 answer:
Hoochie [10]3 years ago
8 0

Answer : The number of drops it takes from a dropper to dispense 1.0 ml of ethanol is, 20 drops

Solution : Given,

Density of ethanol = 0.80 g/ml

Mass of ethanol = 0.60 g

First we have to calculate the volume of ethanol.

Formula used : Density=\frac{Mass}{Volume}

0.80g/ml=\frac{0.60g}{Volume}

Volume=\frac{0.60g}{0.80g/ml}=0.75ml

The volume of ethanol is, 0.75 ml

Now we have to calculate the number of drops it takes from a dropper to dispense 1 ml of ethanol.

As, the number of drops in 0.75 ml of ethanol = 15

So, the number of drops in 1.0 ml of ethanol = \frac{15}{0.75}\times 1.0=20

Therefore, the number of drops it takes from a dropper to dispense 1.0 ml of ethanol is, 20 drops

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Ethene is converted to ethane by the reaction flows into a catalytic reactor at 25.0 atm and 250.°C with a flow rate of 1050. L/
Sphinxa [80]

Answer : The percent yield of the reaction is, 76.34 %

Explanation : Given,

Pressure of C_2H_4 and H_2 = 25.0 atm

Temperature of C_2H_4 and H_2 = 250^oC=273+250=523K

Volume of C_2H_4 = 1050 L per min

Volume of H_2 = 1550 L per min

R = gas constant = 0.0821 L.atm/mole.K

Molar mass of C_2H_6 = 30 g/mole

First we have to calculate the moles of C_2H_4 and H_2 by using ideal gas equation.

For C_2H_4 :

PV=nRT\\\\n=\frac{PV}{RT}

n=\frac{PV}{RT}=\frac{(25atm)\times (1050L)}{(0.0821L.atm/mole.K)\times (523K)}

n=611.34moles

For H_2 :

PV=nRT\\\\n=\frac{PV}{RT}

n=\frac{PV}{RT}=\frac{(25atm)\times (1550L)}{(0.0821L.atm/mole.K)\times (523K)}

n=902.46moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

C_2H_4+H_2\rightarrow C_2H_6

From the balanced reaction we conclude that

As, 1 mole of C_2H_4 react with 1 mole of H_2

So, 611.34 mole of C_2H_4 react with 611.34 mole of H_2

From this we conclude that, H_2 is an excess reagent because the given moles are greater than the required moles and C_2H_4 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of C_2H_6.

As, 1 mole of C_2H_4 react to give 1 mole of C_2H_6

As, 611.34 mole of C_2H_4 react to give 611.34 mole of C_2H_6

Now we have to calculate the mass of C_2H_6.

\text{Mass of }C_2H_6=\text{Moles of }C_2H_6\times \text{Molar mass of }C_2H_6

\text{Mass of }C_2H_6=(611.34mole)\times (30g/mole)=18340.2g

The theoretical yield of C_2H_6 = 18340.2 g

The actual yield of C_2H_6 = 14.0 kg = 14000 g      (1 kg = 1000 g)

Now we have to calculate the percent yield of C_2H_6

\%\text{ yield of }C_2H_6=\frac{\text{Actual yield of }C_2H_6}{\text{Theoretical yield of }C_2H_6}\times 100=\frac{14000g}{18340.2g}\times 100=76.34\%

Therefore, the percent yield of the reaction is, 76.34 %

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4 years ago
John Dalton believed which of the following about atoms?
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Answer : Option B) All atoms of a single substance are identical.


Explanation : The scientist John Dalton proposed the atomic theory which had the postulates as follows.


i) All matter/substances consists of indivisible particles known as atoms.

ii) Atoms of the same element/substance are similar in mass,shape and size, but differ from the atoms of other elements.

iii) Atoms obey the law of conservation of energy which says atoms cannot be created or destroyed.

iv) Atoms of different elements may combine with each other in a fixed, simple, whole number ratios to form any compound atoms.

v) Atoms of same element can combine in ratio with more than one to form two or more compounds.

vi) The atom is considered to be the smallest unit of matter that can take part in a chemical reaction

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