Can someone help me with this?

Calculate the ph of the solution made by adding 0.50 mol of hobr and 0.30 mol of kobr to 1.00 l of water. the value of ka for ho

nikdorinn [45]

To fully understand the problem, we use the ICE table to identify the concentration of the species. We calculate as follows:

Ka = 2.0 x 10^-9 = [H+][OBr-] / [HOBr]

HOBr = 0.50 M
KOBr = 0.30 M = OBr-

 HOBr + H2O <-> H+ + OBr- 
I 0.50 - 0 0.30 
C -x x x
---------------------------------------------
E(0.50-x) x (0.30+x) 

Assuming that the value of x is small as compared to 0.30 and 0.50 

Ka = 2.0 x 10^-9 = x (0.30) / 0.50) 

x = 3.33 x 10^-9 = H+
pH = 8.48

3 0

3 years ago

Mn(OH)2(s) + MnO4(aq) → MnO42–(aq) (basic solution) When the equation is balanced with smallest whole number coefficients, what

PilotLPTM [1.2K]

Answer:

Hi, the given equation has some missing parts. Actual equation is- ' $Mn(OH)_{2}(s)+MnO_{4}^{-}(aq.)\rightarrow MnO_{4}^{2-}(aq.)$ '

balanced equation: $Mn(OH)_{2}(s)+4MnO_{4}^{-}(aq.)+6OH^{-}(aq.)\rightarrow 5MnO_{4}^{2-}(aq.)+4H_{2}O(l)$

Explanation:

$Mn(OH)_{2}(s)\rightarrow MnO_{4}^{2-}(aq.)$

Balance O and H in basic medium: $Mn(OH)_{2}(s)+6OH^{-}(aq.)\rightarrow MnO_{4}^{2-}(aq.)+4H_{2}O(l)$

Balance charge: $Mn(OH)_{2}(s)+6OH^{-}(aq.)-4e^{-}\rightarrow MnO_{4}^{2-}(aq.)+4H_{2}O(l)$ ........(1)

$MnO_{4}^{-}(aq.)\rightarrow MnO_{4}^{2-}(aq.)$

Balance charge: $MnO_{4}^{-}(aq.)+e^{-}\rightarrow MnO_{4}^{2-}(aq.)$ .....(2)

$[equation(2)\times 4]+[equation (1)]:$

$Mn(OH)_{2}(s)+4MnO_{4}^{-}(aq.)+6OH^{-}(aq.)\rightarrow 5MnO_{4}^{2-}(aq.)+4H_{2}O(l)$

$OH^{-}(aq.)$ is present on the left hand side of balanced equation and it's coefficient is 6

6 0

2 years ago

Según la Organización Mundial de la Salud, el nitrato de plata ( densidad = 4.35 g/cc ) es una sustancia con propiedades cáustic

OlgaM077 [116]

Answer:

737.52 mL de agua

Explanation:

En este caso solo debes usar la expresión de molaridad de una solución la cual es:

M = moles / V

Donde:

V: Volumen de solución.

Como queremos saber la cantidad de agua, queremos saber en otras palabras cual es la cantidad de solvente que se utilizó para preparar los 800 mL de disolución.

Una disolución se prepara con un soluto y solvente. El soluto lo tenemos, que es el nitrato de plata. Con la expresión de arriba, calculamos los moles de soluto, y luego su masa. Posteriormente, calculamos el volumen con la densidad, y finalmente podremos calcular el solvente de esta forma:

V ste = Vsol - Vsto

Primero calcularemos los moles de soluto:

moles = M * V

moles = 2 * 0.800 = 1.6 moles

Con estos moles, se calcula la masa usando el peso molecular reportado que es 169.87 g/mol:

m = moles * PM

m = 1.6 * 169.87 = 271.792 g

Ahora usando el valor de la densidad, calcularemos el volumen de soluto empleado:

d = m/V

V = m/d

V = 271.792 / 4.35

V = 62.48 mL

Finalmente, la cantidad de agua necesaria es:

V agua = 800 - 62.48

V agua = 737.52 mL

8 0

3 years ago

Which pair of elements, when combined together, do not form covalent compound.

Assoli18 [71]

K and Ar.
The two non metals

8 0

2 years ago

Read 2 more answers

. What is the molarity of a solution in which 6.50 mol of NaOH is dissolved in

netineya [11]

Answer:

For 6.50 mol NaOH - 1.44M.

For 6.50 g NaOH - C. 0.0361 M

Explanation:

For 6.50 mol NaOH.

Molarity = moles solute/L solution = 6.5 mol/4.5 L solution = 1.44 mol/L=1.44M

There is no such answer in given choices.

For 6.50 g NaOH

M(NaOH) = 23 + 16 + 1 = 40 g/mol

Molarity = 6.5g* 1 mol/40g *1/4.5L = 0.0361 mol/L = 0.0361 M

There is such answer. It is C.

4 0

3 years ago