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3 years ago

Name a difference between a volvox and paramecium? plz help

Chemistry

1 answer:

Andrews [41]3 years ago

8 0

Volvox and paramecium live in the same living conditions. The visible difference between volvox and paramecium is the shape. Volvox is rounded, while paramecium is lying.

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You and your lab partner are asked to determine the density of an aluminum bar. The mass is known accurately to 4 significant fi

zmey [24]

Answer:

Average density for method A = 2.4 g/cm³

Average density for method B = 2.605 g/cm³

Explanation:

In order to calculate the average density for each method, we need to add the data for each method, and then divide the result by the number of measurements (in this case is 4 for both methods):

For Method A:

Σ = 2.2 + 2.3 + 2.7 + 2.4 = 9.6

Average = 9.6/4 = 2.4 g/cm³

For Method B:

Σ = 2.603 + 2.601 + 2.605 + 2.611 = 10.420

Average = 10.420/4 = 2.605 g/cm³

5 0

3 years ago

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

Combustion of hydrocarbons such as nonane () produces carbon dioxide, a "greenhouse gas." Greenhouse gases in the Earth's atmosp

maksim [4K]

Answer:

Part 1: C₉H₂₀ (l) + 14O₂ (g) ----> 9CO₂ (g) + 10H₂0 (g)

Part 2: Volume of CO₂ produced = 1223.21 L

Note: the complete second part of the question is given below:

2. Suppose 0.470 kg of nonane are burned in air at a pressure of exactly 1 atm and a temperature of 17.0 °C. Calculate the volume of carbon dioxide gas that is produced. Round your answer to 3 significant digits.

Explanation:

Part 1: Balanced chemical equation

C₉H₂₀ (l) + 14O₂ (g) ----> 9CO₂ (g) + 10H₂0 (g)

Part 2: volume of carbon dioxide produced

From the equation of the reaction;

At s.t.p., I mole of C₉H₂₀ reacts with 14 moles of O₂ to produce 9 moles of CO₂

molar mass of C₉H₂₀ = 128g/mol: molar mass of CO₂ = 44 g/mol, molar volume of gas at s.t.p. = 22.4 L

Therefore, 128 g of C₉H₂₀ produces 14 * 22.4 L of CO₂ i.e. 313.6 L of CO₂.

O.470 Kg of nonane = 470 g of nonane

470 g of C₉H₂₀ will produce 470 * (313.6/128) L of CO₂ = 1151.50 L of CO₂

Volume of CO₂ gas produced at 1 atm and 17 °C;

Using P₁V₁/T₁ = P₂V₂/T₂

V₂ = P₁V₁T₂/P₂T₁

where P₁ = 1 atm, V₁ = 1151.50 L, T₁ = 273 K, P₂ = 1 atm, T₂ = 17 + 273 = 290 K

Substituting the values; V₂ = (1 * 1151.5 * 290)/(1 * 273)

Therefore volume of CO₂ produced, V₂ = 1223.21 L of CO₂

3 0

4 years ago

Use the tabulated half-cell potentials to calculate the equilibrium constant (K) for the following balanced redox reaction at 25

o-na [289]

The equation Eºcell = 0.0592/n logK must be used to find n and also Eºcell
2 Al(s) + 3 Mg2+(aq) → 2 Al3+(aq) + 3 Mg(s) Al3+ +3e- --> Al Eº = -1.66 V Mg2+ +2e- -->Mg Eº = -2.37V
To balance the equation, 6 moles of electrons must be transferred (2 Al and 3 Mg). This will be the value of n in the equation.
To find Eºcell, you need the reduction potentials which should be given in a table, and given above. Eºcell = -1.66 - (-2.37) = 0.71 V log K = Eºcell x n/0.0592 = 0.71 x 6/0.0592 log K = 71.95 K = 10^71.95 K = 1.1x10^72

6 0

3 years ago

I need you help in this homework

Mrrafil [7]

Answer:

OPTION B ,COMPOUND CONTAINING AMMONIUM

4 0

3 years ago