Balance the following equation Sa+ F2---> SF6

Equal volumes of two solutions, one containing a strong acid at pH 2 and the other containing a strong base at pH 12, are mixed.

mash [69]

Answer:

7

Explanation:

Assume we have 1 L of each solution.

Solution 1

$\text{[H$^{+}$]}= 10^\text{-pH} \text{ mol/L} = 10^{\text{-2}} \text{ mol/L}\\ \text{ moles of H}^{+} = \text{ 1 L solution} \times \dfrac{10^{-2}\text{ mol H}^{+}}{\text{1 L solution}} = 10^{-2}\text{ mol H}^{+}$

Solution 2

pH = 12

pOH = 14.00 - pOH = 14.00 - 12 = 2.0

$\text{[OH$^{-}$]}= 10^\text{-pOH} \text{ mol/L} = 10^{\text{-2}} \text{ mol/L}\\ \text{ moles of OH}^{-} = \text{ 1 L solution} \times \dfrac{10^{-2}\text{ mol OH}^{-}}{\text{1 L solution}} = 10^{-2}\text{ mol OH}^{-}$

3. pH after mixing

H⁺ + OH⁻ ⟶ H₂O

I/mol: 10⁻² 10⁻²

C/mol: -10⁻² -10⁻²

E/mol: 0 0

The H⁺ and OH⁻ have neutralized each other. The pH will be that of pure water.

pH = 7

8 0

3 years ago

Write a balanced nuclear equation for the beta decay of protactinium

PSYCHO15rus [73]

$_{91}^{231}Pa --\ \textgreater \ _{92}^{231}U + _{-1}^{0}e$

7 0

3 years ago

A chemistry student must write down in her lab notebook the concentration of a solution of potassium chloride. The concentration

DerKrebs [107]

Answer:

3.65 g / ml correct to 3 sig. fig.

Explanation:

The computation of the concentration required is shown below:

As we know that

[A] = mass of solute ÷ volume of solution

Before that first find the mass of solute

Given that

Initial weight = 5.55g

And,

Final weight = 92.7 g

So,

Mass of KCl is

= 92.7 - 5.55

= 87.15 g ~ 87.2 g

Now the KCi is fully dissolved, so the volume is 23.9 ml

So, concentration is

= 87.2 g ÷ 23.9 ml

= 3.65 g / ml correct to 3 sig. fig.

6 0

3 years ago

May someone assist, please...? I don't know how to do chemistry work...

EastWind [94]

Answer:

1) 90.0 mL

2) 11.25 M

3) 0.477 M

4) 144 mL

Explanation:

The main formula that will be used for all these calculations is:

C₁V₁ = C₂V₂

C stands for concentration and V stands for volume and the subscripts 1 and 2 indicate an initial concentration or volume and a final concentration or volume.

For each problem, it's best to start by figuring out what you have and what you need to find. Figure out if you're looking for an initial value or a final value.

1) We need to find the initial volume. So, take what values you have and plug them in and then solve for whatever variable:

5.00 M · V₁ = 500.0mL · 0.900 M - divide by 5.00

C₁ = 90.0 mL

2) This time we're finding the initial concentration:

20.0mL · C₁ = 150.0mL · 1.50 M - divide by 20.0mL

C₂ = 11.25 M

3) Now we're finding the final concentration:

12.00mL · 3.50 M = 88.0mL · C₂ - divide by 88.0mL

C₂ = 0.477 M

4) Finally, we're looking for the final volume:

9.0mL · 8.0 M = 0.50 M · V₂ - divide by 0.50 M

V₂ = 144mL

6 0

3 years ago

A second-order reaction has a rate law: rate = k[a]2, where k = 0.150 m−1s−1. If the initial concentration of a is 0.250 m, what

Cerrena [4.2K]

Rate law for the given 2nd order reaction is:

Rate = k[a]2

Given data:

rate constant k = 0.150 m-1s-1

initial concentration, [a] = 0.250 M

reaction time, t = 5.00 min = 5.00 min * 60 s/s = 300 s

To determine:

Concentration at time t = 300 s i.e. $[a]_{t}$

Calculations:

The second order rate equation is:

$1/[a]_{t} = kt +1/[a]$

substituting for k,t and [a] we get:

1/[a]t = 0.150 M-1s-1 * 300 s + 1/[0.250]M

1/[a]t = 49 M-1

[a]t = 1/49 M-1 = 0.0204 M

Hence the concentration of 'a' after t = 5min is 0.020 M

7 0

3 years ago