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KonstantinChe [14]
3 years ago
13

What is the change in the boiling point of a solution made by dissolving 14.7 g of c6h12o6 into of water? the density of water i

s and ?
Chemistry
1 answer:
vovikov84 [41]3 years ago
8 0

First determine the molality of the solution. m = (w solute)/[(mm solute)(kg solvent)] -- mm means molar mass. So,

m = (14.7 g)/[(180 g/mol)(0.150 kg) =0.544 mol/kg = 0.544 m


Change in boiling point (∆t) = m x Kbp = (0.544 m)(0.512 °C/m) = 0.28°C


The solution will boil at 100.28°C at 1 atm.

the density if water is 1.00 g/ml and Kb= 0.512 C/m

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Answer:

An oxidizing agent, or oxidant, gains electrons and is reduced in a chemical reaction. ... A reducing agent is oxidized, because it loses electrons in the redox reaction.

Explanation:

An oxidizing agent, or oxidant, gains electrons and is reduced in a chemical reaction. ... A reducing agent is oxidized, because it loses electrons in the redox reaction.

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The atomic mass of hydrogen is one. True False
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Answer:

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A student isolated 7.2 g of 1-bromobutane reacting equimolar amounts of 1-butanol (10 ml) and NaBr (11.1 g) in the presence of s
Alla [95]

<u>Answer:</u> The percent yield of the 1-bromobutane is 48.65 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For NaBr:</u>

Given mass of NaBr = 11.1 g

Molar mass of NaBr = 103 g/mol

Putting values in equation 1, we get:

\text{Moles of NaBr}=\frac{11.1g}{103g/mol}=0.108mol

The chemical equation for the reaction of 1-butanol and NaBr is:

\text{1-butanol + NaBr}\rightarrow \text{1-bromobutane}

By Stoichiometry of the reaction

1 mole of NaBr produces 1 mole of 1-bromobutane

So, 0.108 moles of NaBr will produce = \frac{1}{1}\times 0.108=0.108 moles of 1-bromobutane

  • Now, calculating the mass of 1-bromobutane from equation 1, we get:

Molar mass of 1-bromobutane = 137 g/mol

Moles of 1-bromobutane = 0.108 moles

Putting values in equation 1, we get:

0.108mol=\frac{\text{Mass of 1-bromobutane}}{137g/mol}\\\\\text{Mass of 1-bromobutane}=(0.108mol\times 137g/mol)=14.80g

  • To calculate the percentage yield of 1-bromobutane, we use the equation:

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

Experimental yield of 1-bromobutane = 7.2 g

Theoretical yield of 1-bromobutane = 14.80 g

Putting values in above equation, we get:

\%\text{ yield of 1-bromobutane}=\frac{7.2g}{14.80g}\times 100\\\\\% \text{yield of 1-bromobutane}=48.65\%

Hence, the percent yield of the 1-bromobutane is 48.65 %

5 0
3 years ago
What is the total pressure in units of kPa in a 7.85 L container that contains 4.45 moles of N2 and 2.45 moles of O2 at a temper
dem82 [27]

Answer:

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7 0
3 years ago
Group the following electron configurations in pairs that would represent similar chemical properties at their atoms;
marishachu [46]

<u>Answer:</u> Pairs are:  (a) and (d), (b) and (f), (c) and (e)

<u>Explanation:</u>

In a periodic table, elements are arranged in 18 vertical columns known as groups and 7 horizontal rows known as periods.

Elements arranged in a group show similar chemical properties because of the presence of same number of valence electrons.

Valence electrons are defined as the electrons which are present in the outermost shell of an atom. Outermost shell has the highest value of 'n' that is principal quantum number.

For the given options:

  • <u>For a:</u>

The given electronic configuration is:  1s^22s^22p^63s^2

The number of valence electrons in the given configuration are 2

  • <u>For b:</u>

The given electronic configuration is:  1s^22s^22p^63s^3

The number of valence electrons in the given configuration are [2 + 3] = 5

  • <u>For c:</u>

The given electronic configuration is:  1s^22s^22p^63s^23p^64s^23d^{10}4p^6

The number of valence electrons in the given configuration are [2 + 6] = 8

  • <u>For d:</u>

The given electronic configuration is:  1s^22s^2

The number of valence electrons in the given configuration are 2

  • <u>For e:</u>

The given electronic configuration is:  1s^22s^22p^6

The number of valence electrons in the given configuration are [2 + 6] = 8

  • <u>For f:</u>

The given electronic configuration is:  1s^22s^22p^63s^23p^3

The number of valence electrons in the given configuration are [2 + 3] = 5

Electronic configuration of (a) and (d) will form a pair, (b) and (f) will form a pair, (c) and (e) will form a pair and will have similar chemical properties.

Hence, the pairs are:  (a) and (d), (b) and (f), (c) and (e)

4 0
3 years ago
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