18. Focal length is a length which lies between

What equation in the blast furnace extraction of iron is not a redox reaction?

Misha Larkins [42]

The decomposition of limestone CaCO3 > CaO + CO2

8 0

3 years ago

A grade 8 students burnt white phosphorus in the air contained in a bell jar. what will be the volume of the renaming air in the

jok3333 [9.3K]

Try d cuz if im not mistaken if you burnt white phosphorus all the gas is going to take most the air

5 0

2 years ago

Read 2 more answers

A solution contains [Ba2+] = 5.0 × 10−5 M, [Zn2+] = 2.0 × 10−7 M, and [Ag+] = 3.0 × 10−5 M. Sodium oxalate (Na2C2O4) is slowly a

Jet001 [13]

Answer:

BaC₂O₄, then ZnC₂O₄, then Ag₂C₂O₄

Explanation:

1. Calculate the equilibrium concentrations of oxalate ion

Let [C₂O₄²⁻] = c

(a) Barium oxalate

BaC₂O₄ ⇌ Ba²⁺ + C₂O₄²⁻

E/mol·L⁻¹: 5.0 × 10⁻⁵ c

Ksp = [Ba²⁺][C₂O₄²⁻] = 5.0 × 10⁻⁵c = 1.5 × 10⁻⁸

c = (1.5 × 10⁻⁸)/(5.0 × 10⁻⁵) = 3.0 × 10⁻⁴ mol·L⁻¹

(b) Zinc oxalate

ZnC₂O₄ ⇌ Zn²⁺ + C₂O₄²⁻

E/mol·L⁻¹: 2.0 × 10⁻⁷ c

Ksp = [Zn²⁺][C₂O₄²⁻] = 2.0 × 10⁻⁷c = 1.35 × 10⁻⁹

c = (1.35 × 10⁻⁹)/(2.0 × 10⁻⁷) = 6.8 × 10⁻³ mol·L⁻¹

(c) Silver oxalate

Ag₂C₂O₄ ⇌ 2Ag⁺ + C₂O₄²⁻

E/mol·L⁻¹: 3.0 × 10⁻⁵ c

Ksp = [Ag⁺]²[C₂O₄²⁻] = (3.0× 10⁻⁵)²c = 9.0 × 10⁻¹⁰c = 1.1 × 10⁻¹¹

c = (1.1 × 10⁻¹¹)/(9.0 × 10⁻¹⁰) = 0.012 mol·L⁻¹

2. Decide the order of precipitation

BaC₂O₄ will precipitate when c > 3.0 × 10⁻⁴ mol·L⁻¹

ZnC₂O₄ will precipitate when c > 6.8 × 10⁻³ mol·L⁻¹

Ag₂C₂O₄ will precipitate when c > 0.028 mol·L⁻¹

This happens to be the order of increasing concentration of oxalate ion.

The order of precipitation is

BaC₂O₄, then ZnC₂O₄, then Ag₂C₂O₄

4 0

3 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

A cylindrical glass tube of length 27.4cm and radius 4 cm is filled with gas. The empty tube has a mass of 254.3 g. The tube fil

Temka [501]

Density of the gas is 3.05 × 10⁻³ g / cm³.

<u>Explanation:</u>

Volume of the cylinder = π r² h

where r is the radius and h is the height of the height or the length of the glass tube.

Here r = 4 cm and h = 27.4 cm

Volume of the cylinder = 3.14 × 4 × 4 × 27.4 = 1376.6 cm³

We have to find the mass of the gas by subtracting the mass of the tube filled with the substance from the mass of the empty tube.

Mass of the substance = 258.5 - 254.3 = 4.2 g

We have to find the density using the formula as,

$$ Density = \frac{mass}{volume}$

Plugin the values as,

$$ Density = \frac{4.2}{1376.6}$

= 3.05 × 10⁻³ g / cm³

So the Density of the gas is 3.05 × 10⁻³ g / cm³.

3 0

2 years ago