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Stels [109]
1 year ago
10

Aqueous hydrobromic acid reacts with solid sodium hydroxide to produce aqueous sodium bromide and liquid water . What is the the

oretical yield of water formed from the reaction of of hydrobromic acid and of sodium hydroxide
Chemistry
1 answer:
natima [27]1 year ago
8 0

Mass of water produced is equal to 7.2 grams.

The reactants are reacting according to the following equation:

HBr + NaOH = NaBr + H2O

1 mole of HBr requires 1 mole of NaOH to be totally transformed into NaBr and water.

Now, let us calculate the amounts of HBr and NaOH in mols:

m(HBr) = 67.2g

m(NaOH) = 16g

M(HBr) = 80.9g/mol

M(NaOH) = 40g/mol

Following the simple relation

n(moles) = m(grams)/M(g/mol)

n(HBr) = 67.2g/(80.9g/mol)

n(NaOH) = 16g/(40g/mol)

n(H2O) = n(NaOH) = 0.40 mol, which is:

m(H2O) = n(H2O) x M(H2O)

0.40mol∗18g/mol=7.2g

Finally, the mass of water produced is equal to 7.2 grams.

Learn more about Hydrobromic acid here:

brainly.com/question/21544156

#SPJ4

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The hydrogen atom has just one electron, but many spectral lines. However it contains many shells and the movement of that electron from one shell to another causes the release of energy and also an emission of photons.

A spectral line are dark or bright lines formed within a specific frequency range which differ from other frequencies.Because of the difference of energy for the various shells, it produces different wavelengths and this is the reason for the many spectral line for hydrogen.

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At 298 K, the osmotic pressure of a glucose solution (C6H12O6 (aq)) is 12.1 atm. Calculate the freezing point of the solution. T
Anarel [89]

<u>Answer:</u> The freezing point of solution is -0.974°C

<u>Explanation:</u>

  • To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=iMRT

where,

\pi = osmotic pressure of the solution = 12.1 atm

i = Van't hoff factor = 1 (for non-electrolytes)

M = molarity of solute = ?

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature of the solution = 298 K

Putting values in above equation, we get:

12.1atm=1\times M\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\M=\frac{12.1}{1\times 0.0821\times 298}=0.495M

This means that 0.495 moles of glucose is present in 1 L or 1000 mL of solution

  • To calculate the mass of solution, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of solution = 1.034 g/mL

Volume of solution = 1000 mL

Putting values in above equation, we get:

1.034g/mL=\frac{\text{Mass of solution}}{1000mL}\\\\\text{Mass of solution}=(1.034g/mL\times 1000mL)=1034g

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

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of glucose = 0.495 moles

Molar mass of glucose = 180.16 g/mol

Putting values in above equation, we get:

0.495mol=\frac{\text{Mass of glucose}}{180.16g/mol}\\\\\text{Mass of glucose}=(0.495mol\times 180.16g/mol)=89.18g

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

  • The equation used to calculate depression in freezing point follows:

\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Freezing point of pure solution = 0°C

i = Vant hoff factor = 1 (For non-electrolytes)

K_f = molal freezing point elevation constant = 1.86°C/m

m_{solute} = Given mass of solute (glucose) = 89.18 g

M_{solute} = Molar mass of solute (glucose) = 180.16  g/mol

W_{solvent} = Mass of solvent (water) = [1034 - 89.18] g = 944.82 g

Putting values in above equation, we get:

0-\text{Freezing point of solution}=1\times 1.86^oC/m\times \frac{89.18\times 1000}{180.16g/mol\times 944.82}\\\\\text{Freezing point of solution}=-0.974^oC

Hence, the freezing point of solution is -0.974°C

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Answer:

Explanation:

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It is formed by the sharing of electron pair between bonded atoms.  

The atom with larger electronegativity attract the electron pair more towards itself and becomes partial negative while the other atom becomes partial positive.

Non polar covalent bond:

It is the bond where both bonded atoms share the pair of electron equally.

For example:

Hydrogen gas (H₂) is non polar covalent compound because the electronegativity of both bonded atoms are same. No poles are created that's why this is non polar covalent compound.

Polar covalent bond:

It is the bond where both bonded atoms share the pair of electron unequally.

For example:

In water the electronegativity of oxygen is 3.44 and hydrogen is 2.2. That's why electron pair attracted more towards oxygen, thus oxygen becomes partial negative and hydrogen becomes partial positive and bond is polar.

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