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3 years ago

Which of the following compounds has the highest boiling point.

1 answer:

4 0

Answer:
1-<u>Pentanol </u>has the highest boiling point.

Explanation:
Boiling points of given compounds are as follow,

1) 2-Pentanone = 101 °C

2) 1-Pentanol = 138 °C

3) Pentane = 36.1 °C

4) Pentanal = 103 °C

Pentanol has greatest boiling point as compare to others because it involes a strongest intermolecular interactions called as hydrogen bond interactions. Remaining compounds lack the direct attachement of hydrogen to most electrnegative atom like oxygen which is present in 1-Pentanol.

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What is the angle between the carbon-sulfur bond and the carbon-nitrogen bond in the thiocyanate ( SCN− ) ion?

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The Bay of Fundy in Nova Scotia, Canada is reported to have the largest tides in the world with high tide and low tide occurring

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The rate in m/s is 5.2 * 10^-4 m/s.

We know that the speed is given as the ratio of the distance covered to the time taken. In this case, we have been told that the rate at which the tide rises is 6.08 ft per hour. We would now need to convert the rate from 6.08 ft per hour to m/s.

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We know that;

1 foot/hour = 8.5 * 10^-5 m/s

6.08 ft per hour = 6.08 ft per hour * 8.5 * 10^-5 m/s/1 foot/hour

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2 years ago

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3 years ago

Solid sodium hydroxide reacts with gaseous carbon dioxide to form solid sodium carbonate salt and liquid water. Formulate the ba

Bond [772]

Answer:

a) 0.925 mol Na2CO3 can be theoretically produced

b) 0.075 moles of the excess starting material remains

Explanation:

balaced chemical:

2 NaOH(s) + CO2(g) ↔ Na2CO3(s) + H2O(aq)

1.85n 1.00n Xn

moles theor. Na2CO3:

⇒ nNaCO3 = 1.85nNaOH * ( nNa2CO3 / 2nNaOH)

⇒ nNa2CO3 = 0.925nNa2CO3

moles of the excess:

⇒moles CO2 react = 1.85nNaOH * nCO2 / 2nNaOH = 0.925n CO2

⇒moles CO2 excess = 1.00n - 0.925n = 0.075n excess CO2

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2 years ago

Hydrazine (n2h4) is used as rocket fuels. it reacts with oxygen to form nitrogen and water. write the balanced equation for this

FrozenT [24]

The answer is: volume of nitrogen is 70L.

Chemical reaction: N₂H₄ + O₂ → N₂ + 2H₂O.

m(N₂H₄) = 100 g; mass of hydrazine.

M(N₂H₄) = 32 g/mol; molar mass of hydrazine.

n(N₂H₄) = m(N₂H₄) ÷ M(N₂H₄).

n(N₂H₄) = 100 g ÷ 32 g/mol.

n(N₂H₄) = 3.125 mol; amount of hydrazine.

m(O₂) = 100 g; mass of oxygen.

M(O₂) = 32 g/mol; molar mass of oxygen.

n(O₂) = 100 g ÷ 32 g/mol.

n(O₂) = 3.125 mol; amount of oxygen.

From chemical reaction: n(O₂) : n(N₂) = 1 : 1; n(O₂) = n(N₂).

n(N₂) = 3.125 mol; amount og nitrogen gas.

V(N₂) = n(N₂) · Vm.

Vm = 22.4 L/mol; molar volume.

V(N₂) = 3.125 mol · 22.4 L/mol.

V(N₂) = 70 L.

There is not excess reagent, because hydrazine and oxygen are all used in chemical reaction.

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3 years ago