What would be the formula of the precipitate that forms when pb(no3)2 (aq) and k2so4 (aq) are mixed?

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

A mixture of water and graphite is heated to 600 K in a 1 L container. When the system comes to equilibrium it contains 0.17 mol

Kryger [21]

Answer:

0.44 moles

Explanation:

Given that :

A mixture of water and graphite is heated to 600 K in a 1 L container. When the system comes to equilibrium it contains 0.17 mol of H2, 0.17 mol of CO, 0.74 mol of H2O, and some graphite.

The equilibrium constant $K_c= \dfrac{[CO][H_2]}{[H_2O]}$

The equilibrium constant $K_c= \dfrac{(0.17 )(0.17)}{0.74}$

The equilibrium constant $K_c= 0.03905$

Some O2 is added to the system and a spark is applied so that the H2 reacts completely with the O2.

The equation for the reaction is :

$H_2 + \dfrac{1}{2}O_2 \to H_2O \\ \\ 0.17 \ \ \ \ \ \ \ \ \ \to0.17$

Total mole of water now = 0.74+0.17

Total mole of water now = 0.91 moles

Again:

$K_c= \dfrac{[CO][H_2]}{[H_2O]}$

$0.03905 = \dfrac{[0.17+x][x]}{[0.91 -x]}$

0.03905(0.91 -x) = (0.17 +x)(x)

0.0355355 - 0.03905x = 0.17x + x²

0.0355355 +0.13095 x -x²

x² - 0.13095 x - 0.0355355 = 0

By using quadratic formula

x = 0.265 or x = -0.134

Going by the value with the positive integer; x = 0.265 moles

Total moles of CO in the flask when the system returns to equilibrium is :

= 0.17 + x

= 0.17 + 0.265

= 0.435 moles

=0.44 moles (to two significant figures)

3 0

3 years ago

Iodine 131, cesium 137, and strontium 90 are all considered isotopes. What are isotopes?

Nitella [24]

Answer:

Isotopes are variants of an element, where the number of neutrons is different but its number of protons stay the same.

Explanation:

If the number of protons had changed then it would just be a different element.

Electron number changing simply adds charge to the atom, and doesn't change it or anything.

PLS GIVE BRAINLIEST

3 0

2 years ago

Read 2 more answers

Draw the structure of bromous acid, hbro2. optimize formal charges.

ArbitrLikvidat [17]

STRUCTURE OF BROMOUS ACID: H–O–Br=O

In this structure, all the elements have a formal charge of zero. The formal charge of each element is calculated below: 

H: 1 – 1/2(2) – 0 = 0
O: 6 – 1/2(4) – 4 = 0
Br: 7 – 1/2(6) – 4 = 0
O: 6 – 1/2(4) – 4 = 0

8 0

3 years ago

Which of the following describes the greenhouse effect?

ankoles [38]

Earth's greenhouse gases trap heat in the atmosphere and warm the planet.

6 0

2 years ago

Read 2 more answers