What is the ratio of moles of solute to liters of solution called?

Which formula is an empirical formula? (1) CH4 (2) C2 H6 (3) C3 H6 (4) C4 H10

rosijanka [135]

CH4 is an emprirical formula as it shows the simplest ratio of the numbers of different atoms present in the molecule. The empirical formula for CH4 is also the same as the molecular formula.
The other compunds can be simplified so they are not the empirical formula of compounds.
Hope this helps :).

3 0

3 years ago

The combustion of ethene in the presence of excess oxygen yields carbon dioxide and water: c2h4 (g) + 3o2 (g) → 2co2 (g) + 2h2o

meriva

Answer:

$\boxed{-267.5}$

Explanation:

You can calculate the entropy change of a reaction by using the standard molar entropies of reactants and products.

The formula is

$\Delta_{r} S^{\circ} = \sum_n {nS_{\text{products}}^{\circ} - \sum_{m} {mS_{\text{reactants}}^{\circ}}}$

The equation for the reaction is

C₂H₄(g) + 3O₂(g) ⟶ 2CO₂(g) + 2H₂O(ℓ)

ΔS°/J·K⁻¹mol⁻¹ 219.5 205.0 213.6 69.9

$\Delta_{r} S^{\circ} = (2\times213.6 + 2\times69.9) - (1\times219.5 + 3\times205.0)\\\\= 567.0 - 834.5 = \boxed{-267.5 \text{ J}\cdot\text{K}^{-1} \text{mol}^{-1}}$

3 0

3 years ago

Determine if the bond between each pair of atoms is pure covalent, polar covalent, or ionic. drag the appropriate items to their

dsp73

Types of Bonds can be predicted by calculating the difference in electronegativity.

If, Electronegativity difference is,

Less than 0.4 then it is Non Polar Pure Covalent

Between 0.4 and 1.7 then it is Polar Covalent

Greater than 1.7 then it is Ionic

For Br and Br,

E.N of Bromine = 2.96

________

E.N Difference 0.00 (Non Polar/Pure Covalent)

For N and O,

E.N of Oxygen = 3.44

E.N of Nitrogen = 3.04

________

E.N Difference 0.40 (Non Polar/Pure Covalent)

For P and H,

E.N of Hydrogen = 2.20

E.N of Phosphorous = 2.19

________

E.N Difference 0.01 (Non Polar/Pure Covalent)

For K and O,

E.N of Oxygen = 3.44

E.N of Potassium = 0.82

________

E.N Difference 2.62 (Ionic)

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

3 years ago

Match these items.

dolphi86 [110]

g - Measurement of Mass

m² - Measurement of Area

m³ - Measurement of Volume

km- Measurement of Length

5 0

3 years ago