3. The name of the functional group in the following compound

Assume that concentrated aqueous NH3 has a density of 0.252 g/mL (0.252 g of NH3 per mL of liquid). Calculate the volume of NH3

Kobotan [32]

The question is incomplete, here is the complete question:

Assume that concentrated aqueous NH₃ has a density of 0.252 g/mL (0.252 g of NH₃ per mL of liquid). Calculate the volume of NH₃ required to contain 0.442 mol?

Answer: The volume of ammonia required is 29.82 mL

Explanation:

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

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

Molar mass of ammonia = 17 g/mol

Moles of ammonia = 0.442 moles

Putting values in above equation, we get:

$0.442mol=\frac{\text{Mass of ammonia}}{17g/mol}\\\\\text{Mass of ammonia}=(0.442mol\times 17g/mol)=7.514g$

To calculate the volume of ammonia, we use the equation:

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

Density of ammonia = 0.252 g/mL

Mass of ammonia = 7.514 g

Putting values in above equation, we get:

$0.252g/mL=\frac{7.514g}{\text{Volume of ammonia}}\\\\\text{Volume of ammonia}=\frac{7.514g}{0.252g/mL}=29.82mL$

Hence, the volume of ammonia required is 29.82 mL

4 0

3 years ago

What is the ph (to nearest 0.01 ph unit) of a solution prepared by mixing 58.0 ml of 0.0179 m naoh and 60.0 ml of 0.0294 m ba(oh

Nina [5.8K]

Answer: -

12.59

Explanation: -

Strength of NaOH = 0.0179 M

Volume of NaOH = 58.0 mL = 58.0/1000 = 0.058 L

Number of moles = 0.0179 M x 0.058 L

= 1.04 x 10⁻³ mol

Mol of [OH⁻] given by NaOH = 1.04 x 10⁻³ mol

Strength of Ba(OH)₂ = 0.0294 M

Volume of Ba(OH)₂ = 60.0 mL = 60.0/1000 = 0.060 L

Number of moles = 0.0294 M x 0.060 L

= 1.76 x 10⁻³ mol

Mol of [OH⁻] given by Ba(OH)₂ =2 x 1.76 x 10⁻³ mol

Total [OH⁻] = 1.04 x 10⁻³ mol + 2 x 1.76 x 10⁻³ mol

= 4.56 x 10⁻³ mol

Total volume of the mixture = 58.0 + 60.0

= 118.0 mL

118.0 mL of the solution has 4.56 x 10⁻³ mol [OH⁻]

1000 mL of the solution has $\frac{4.56 x 10-3 mol x 1000mL }{118 mL}$

= 0.0386 mol

Using the relation

pOH = - log [OH-]

= - log 0.0386

= 1.41

Using the relation

pH + pOH = 14

pH = 14 - 1.41

= 12.59

5 0

3 years ago

Combustion analysis of 63.8 mg of a c, h and o containing compound produced 145.0 mg of co2 and 59.38 mg of h2o. what is the emp

Makovka662 [10]

: The empirical formula for the compound is C3H60 (see below) CO2 is the only product containing C, C produced = 145.0 mg CO2 x (1 g / 1000 mg) x (1 mole CO2 / 44.0 g CO2) x (1 mole C / 1 mole CO2) = 0.00330 moles C. H2O is the only product containing H, H produced = 59.38 mg H2O x (1 g / 1000 mg) x (1 mole H2O / 18.0 g H2O) x (2 moles H / 1 mole H2O) = 0.00660 moles H. Oxygen is in both and the unknown reacts with oxygen(in the air) 0.00330 moles C x (12.0 g C / 1 mole C) = 0.0396 g C = 39.6 mg C 0.00660 moles H x (1.01 g H / 1 mole H) = 0.00667 g H = 6.7 mg H Because the unknown weighed 63.8 mg and consists off justC, H, and O, then mass O = g unknown - g C - g H = 63.8 mg - 39.6 mg - 6.7 mg = 17.5 mg = 0.0175 g 0.0175 g O x (1 mole O / 16.0 g O) = 0.00109 moles O The mole ratio of C:H:O is: C = 0.00330 H = 0.00660 O = 0.00109 Divide by the smallest you get: C = 0.00330 / 0.00109 = 3.03 H = 0.00660 / 0.00109 = 6.06 O = 0.00109 / 0.00109 = 1.00

6 0

4 years ago

How many valence electrons do the alkaline earth metals have?

fenix001 [56]

Answer:

$\boxed{2}$