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2 years ago

When [H+] = 4.0 × 10–9 M in water at 25°C, then

Chemistry

1 answer:

Salsk061 [2.6K]2 years ago

5 0

Answer:

a. Kw = 1.0 × 10⁻¹⁴

Explanation:

a. Let's consider the self-ionization of water.

2 H₂O(l) ⇄ H₃O⁺(aq) + OH⁻

The ion-product of water (Kw) at 25 °C is:

Kw = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴

c. Considering [H⁺] = [H₃O⁺] = 4.0 × 10⁻⁹ M, the concentration of OH⁻ is:

[OH⁻] = 1.0 × 10⁻¹⁴/[H₃O⁺] = 2.5 × 10⁻⁶ M

b. We can calculate the pOH using the following expression.

pOH = -log [OH⁻] = -log 2.5 × 10⁻⁶ = 5.6

d. We can calculate the pH using the following expression.

pH = -log [H⁺] = -log 4.0 × 10⁻⁹ = 8.4

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The equilibrium constant, Kc, for the following

umka21 [38]

Answer:

2CH2Cl2(g) Doublearrow CH4(g) + CCl4(g)

0.205 moles of CH2Cl2 is introduced. Let by the time an equilibrium is reached x moles each of CH4 and CCl4 are formed => remaining moles of CH2Cl2 are 0.205-x

i.e at equilibrium the concentration on CH2Cl2 is (0.205-2x) mol/L, CH4 is x mol/L, CCl4 is x mol/L

Now the equilibrium constant equation : K = [CH4][CCl4]/[CH2Cl2]^2 ([.] - stands for concentration of the term inside the bracket)

10.5 = x*x/(0.205-2x)^2

=> 10.5(4x^2-0.82x+0.042) = x^2

=>42x^2-8.61x+0.441=x^2

=>41x^2-8.61x+0.441 = 0

This is a Quadratic in x, solving for the roots, we get x = 0.0886 , x = 0.121

The second solution for x will lead 0.205-2x to become negative, so is an infeasible solution.

Therefore equilibrium concentrations of the products and reactants correspond to x=0.0886 and they are , [CH2Cl2] = 0.205-2*0.0886 =0.0278 mol/L , [CH4] = 0.0886 mol/L , [CCl4] = 0.0886 mol/L

4 0

2 years ago

If the vapor's volume were to be incorrectly recorded as 125ml, how will this error affect the calculated molar mass of the unkn

pishuonlain [190]

Since you didn't give the actual volume (or any of the experimental values) I can only tell you how to do it. Do the calculation using the real (determined) volume of the flask. Then, re-do the calculation with v = 125ml. Take the two values and calculate % error; m = measured vol; g = guessed vol.

<span>[mW (m) - mW (g)]/mW (m) x 100% </span>

<span>(they want % error so, if it is negative, just get rid of the sign) </span>

3 0

2 years ago

What reaction might we use to synthesize nickel sulfate,niso4? Write equations for acid base neutralization. Include states in y

il63 [147K]

Answer:

Here's what I get

Explanation:

1. Nickel sulfate

base + acid ⟶ salt + water

NiSO₄ is a salt of the base Ni(OH)₂ and the acid sulfuric acid.

Hydroxides of transition metals are insoluble; most sulfates are soluble.

$\underbrace{\hbox{Ni(OH)$_{2}$(s)}}_{\hbox{base}} + \underbrace{\hbox{H$_{2}$SO$_{4}$(aq)}}_{\hbox{acid}} \longrightarrow \, \underbrace{\hbox{NiSO$_{4}$(aq)}}_{\hbox{salt}} + \underbrace{\hbox{2H$_{2}$O(l)}}_{\hbox{water}}$

2. Carbonate + acid

Most carbonates are insoluble.

They react with acids to form carbonic acid (H₂CO₃), which decomposes into water and carbon dioxide.

$\rm NiCO_{3}(s) + H_{2}SO_{4}(aq) \longrightarrow \, NiSO_{4}(aq) + H_{2}O(l) + CO_{2}(g)$

5 0

3 years ago

pyrite is a mineral composed of 46.5 mass % iron and 53.5 mass % sulfur. determine the empirical formula for calcocite.

padilas [110]

The empirical formula for pyrite is FeS2.

HOW TO CALCULATE EMPIRICAL FORMULA:

The empirical formula represents the simplest whole number ratio of constituents element of a compound. The empirical formula of pyrite can be calculated as follows:

46.5 mass % Fe = 46.5g of Fe

53.5 mass % S = 53.5g of S

Next, we divide each element's mass value by its molar mass

Fe = 46.5g ÷ 56g/mol = 0.83mol

S = 53.5g ÷ 32g/mol = 1.67mol

Next, we divide each mole value by the smallest (0.83mol)

Fe = 0.83mol ÷ 0.83 = 1

S = 1.67mol ÷ 0.83 = 2.014

Approximately, the ratio of Fe to S is 1:2. Therefore, the empirical formula of pyrite is FeS2.

Learn more at: brainly.com/question/14044066?referrer=searchResults

3 0

1 year ago

To completely convert 9. 0 moles of hydrogen gas (h2) to ammonia gas, how many moles of nitrogen gas (n2) are required?

evablogger [386]

To completely convert 9. 0 moles of hydrogen gas (h2) to ammonia gas, 3.0 moles of nitrogen gas (n2) are required.

<h3>What are moles?</h3>

The mole is a SI unit of measurement that is used to calculate the quantity of any substance.

<h3 />

The given reaction is $\rm N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$

By the stoichiometry rule of ratio hydrogen: nitrogen

3 : 1

The reacted moles of nitrogen is equals to H/3 moles of reacted hydrogen

So, moles of nitrogen

$\rm Moles\; of\; nitrogen = \dfrac{9.0 }{3} =3.0\;mol$

Thus, 3.0 moles of nitrogen gas (n2) are required.

Learn more about moles

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3 0

2 years ago