Iron, a metal, has which of the following features?

The pH of a 0.65M solution of hydrofluoric acid HF is measured to be 1.68. Calculate the acid dissociation constant Ka of hydrof

sergey [27]

Answer:

Kₐ = 6.7 x 10⁻⁴

Explanation:

First lets write the equilibrium expression, Ka , for the dissociation of hydrofluoric acid:

HF + H₂O ⇄ H₃O⁺ + F⁻

Kₐ = [ H₃O⁺ ] [ F⁻ ] /[ [ HF ]

Since we are given the pH we can calculate the [ H₃O⁺ ] ( pH = - log [ H₃O⁺ ] , and because the acid dissociates into a 1: 1 relation , we will also have [F⁻ ]. The [ HF ] is given in the question so we have all the information that is needed to compute Kₐ.

pH = -log [ H₃O⁺ ]

1.68 = - log [ H₃O⁺ ]

Taking antilog to both sides of this equation:

10^-1.68 = [ H₃O⁺ ] ⇒ 2.1 X 10⁻² M= [ H₃O⁺ ]

[ F⁻ ] = 2.1 X 10⁻² M

Solving for Kₐ :

Kₐ = ( 2.1 X 10⁻² ) x ( 2.1 X 10⁻² ) / 0.65 = 6.7 x 10⁻⁴

(Rounded to two significant figures, the powers of 10 have infinite precision )

4 0

2 years ago

Pressure gauge at the top of a vertical oil well registers 140 bars. The oil well is 6000 m deep and filled with natural gas dow

andreyandreev [35.5K]

Explanation:

(a) The given data is as follows.

Pressure on top ( $P_{o}$ ) = 140 bar = $1.4 \times 10^{7} Pa$ (as 1 bar = $10^{5}$ )

Temperature = $15^{o}C$ = (15 + 273) K = 288 K

Density of gas = $\frac{PM}{ZRT}$

$\frac{dP}{dZ} = \rho \times g$

$\frac{dP}{dZ} = \int \frac{PM}{ZRT}$

$\int_{P_{o}}^{P_{1}} \frac{dP}{dZ} = \frac{Mg}{ZRT} \int_{0}^{4700} dZ$

$ln (\frac{P_{1}}{P_{o}}) = \frac{18.9 \times 10^{-3} \times 9.81 \times 4700 m}{0.80 \times 8.314 J/mol K \times 288 K}$

= 0.4548

$P_{1} = P_{o} \times e^{0.4548}$

= $1.4 \times 10^{7} Pa \times 1.5797$

= $2.206 \times 10^{7} Pa$

Hence, pressure at the natural gas-oil interface is $2.206 \times 10^{7} Pa$ .

(b) At the bottom of the tank,

$P_{2} = P_{1} + \rho \times g \times h$

= 2.206 \times 10^{7} Pa + 700 \times 9.81 \times (6000 - 4700)[/tex]

= $309.8 \times 10^{5} Pa$

= 309.8 bar

Hence, at the bottom of the well at $15^{o}C$ pressure is 309.8 bar.

6 0

3 years ago

in heating a kettle of water on an electric stove, 3.34×10^3 J of thermal energy was provided by the element of the stove. yet,

insens350 [35]

Answer:

The percentage efficiency of the electrical element is approximately 82.186%

Explanation:

The given parameters are;

The thermal energy provided by the stove element, $H_{supplied}$ = 3.34 × 10³ J

The amount thermal energy gained by the kettle, $H_{absorbed}$ = 5.95 × 10² J

The percentage efficiency of the electrical element in heating the kettle of water, η%, is given as follows;

$\eta \% = \dfrac{H_{supplied} - H_{absorbed} }{H_{supplied}} \times 100$

Therefore, we get;

$\eta \% = \dfrac{3.34 \times 10^3 - 5.95 \times 10^2}{3.34 \times 10^3} \times 100 = \dfrac{549}{668} \times 100 \approx 82.186 \%$

The percentage efficiency of the electrical element, η% ≈ 82.186%.

4 0

2 years ago

The "air bags" that are currently installed in automobiles to prevent injuries in the event of a crash are equipped with sodium

stellarik [79]

Answer:

0.0177 L of nitrogen will be produced

Explanation:

The decomposition reaction of sodium azide will be:

$2NaN_{3}(s)--->2Na(s)+3N_{2}(g)$

As per the balanced equation two moles of sodium azide will give three moles of nitrogen gas

The molecular weight of sodium azide = 65 g/mol

The mass of sodium azide used = 100 g

The moles of sodium azide used = $\frac{mass}{molarmass}=\frac{100}{65}=1.54mol$

so 1.54 moles of sodium azide will give = $\frac{3X1.54}{2}=2.31$ mol

the volume will be calculated using ideal gas equation

PV=nRT

Where

P = Pressure = 1.00 atm

V = ?

n = moles = 2.31 mol

R = 0.0821 L atm / mol K

T = 25 °C = 298.15 K

Volume = $\frac{P}{nRT}=\frac{1}{2.31X0.0821X298.15}=0.0177L$

3 0

2 years ago

Convert 14.72 kg to ____ mg

navik [9.2K]

Answer:

14720000

Explanation:

1 kg = 1000000 mg

14.72 kg = 14.72 x 1000000

=14720000

Please Mark me brainliest

6 0

2 years ago