A chemist has 5.0x10^25 molecules of ammonia (NH3)

How many grams of sodium acetate ( molar mass = 83.06 g/mol ) must be added to 1.00 Liter of a 0.200 M acetic acid solution to m

Pie

Answer: The mass of sodium acetate that must be added is 30.23 grams

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of acetic acid solution = 0.200 M

Volume of solution = 1 L

Putting values in above equation, we get:

$0.200M=\frac{\text{Moles of acetic acid}}{1L}\\\\\text{Moles of acetic acid}=(0.200mol/L\times 1L)=0.200mol$

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[\text{salt}]}{[\text{acid}]})$

$pH=pK_a+\log(\frac{[CH_3COONa]}{[CH_3COOH]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of acetic acid = 4.74

$[CH_3COONa]=?mol$

$[CH_3COOH]=0.200mol$

pH = 5.00

Putting values in above equation, we get:

$5=4.74+\log(\frac{[CH_3COONa]}{0.200})$

$[CH_3COONa]=0.364mol$

To calculate the mass of sodium acetate for given number of moles, we use the equation:

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

Molar mass of sodium acetate = 83.06 g/mol

Moles of sodium acetate = 0.364 moles

Putting values in above equation, we get:

$0.364mol=\frac{\text{Mass of sodium acetate}}{83.06g/mol}\\\\\text{Mass of sodium acetate}=(0.364mol\times 83.06g/mol)=30.23g$

Hence, the mass of sodium acetate that must be added is 30.23 grams

7 0

3 years ago

Calculate the wavelength for the transition from n = 4 to n = 2, and state the name given to the spectroscopic series to which t

finlep [7]

Answer:

The wavelength for the transition from n = 4 to n = 2 is 486nm and the name name given to the spectroscopic series belongs to The Balmer series.

Explanation

lets calculate -

Rydberg equation- $\frac{1}{\pi } =R(\frac{1}{n_1^2} -\frac{1}{n_2^2})$

where , $\pi$ is wavelength , R is Rydberg constant ( $1.097\times10^7$ ), $n_1$ and $n_2$ are the quantum numbers of the energy levels. (where $n_1=2 , n_2=4$ )

Now putting the given values in the equation,

$\frac{1}{\pi }=1.097\times10^7\times(\frac{1}{2^2} -\frac{1}{4^2} )$ $=2056875m^-^1$

Wavelength $\pi =\frac{1}{2056875}$

= $4.86\times10^-^7$ = 486nm

 Therefore , the wavelength is 486nm and it belongs to The Balmer series.

8 0

3 years ago

A 0.04358 g sample of gas occupies 10.0-ml at 292.0 k and 1.10 atm. upon further analysis, the compound is found to be 25.305% c

kirill [66]

Answer: A 0.04403 g sample of gas occupies 10.0-mL at 289.0 K and 1.10 atm. Upon further analysis, the compound is found to be 25.305% C and 74.695% Cl. What is the molecular formula of the compound? --------------------------------------... Seems like I did a problem very similar to this--this must be the "B" test. But the halogen was different. 25.305% C/12 = 2.108 74.695% Cl/35.5 = 2.104 So the empirical formula would be CH. However, there are many compounds which fit this bill, so we have to use the gas data. (And I made, in the previous problem, the simplifying assumption that 289C and 1.10 atm would offset each other, so I'll do that, too.) 0.044 grams/10 ml = x/22.4 liters 0.044g/0.010 liters = x/22.4 liters 22.4 liters/0.010 liters = 2240 (ratio) 2240 x .044 = 98.56 (actual atomic weight) CCl = 35.5+12 or 47.5, so two of those is 95 grams/mole. This is sufficiient to distinguish C2CL2, (dichloroacetylene) from C6CL6 (hexachlorobenzene) which would mass 3 times as much.

3 0

3 years ago

Read 2 more answers

HELP

Jobisdone [24]

Answer:

hello,

first one is 25.15

second is 301.55

Explanation:

honestly if you look up on the internet there is a converter to get you your answers

5 0