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4 years ago

One of the signs of a chemical reaction is taking place

Chemistry

2 answers:

9966 [12]4 years ago

6 0

Any sort of changes including color temperature etc.

GalinKa [24]4 years ago

4 0

One sign of a chemical reaction is that gas is being created.

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An unknown gaseous substance has a density of 1.06 g/L at 31 °C and 371 torr. If the substance has the following percent composi

Anarel [89]

Answer:

C) C4H6 - Right answer

Explanation:

Let's combine the Ideal Gases Law with density to get the molecular formula for the unknown gas.

Density = mass / volume

1.06 g /L means that 1.06 grams of compound occupy 1 liter of volume.

P . V = n . R . T

Pressure in Torr must be converted to atm

760 Torr are 1 atm

371 Torr are __ (371 .1)/760 = 0.488 atm

0.488 atm . 1L = 1.06g/MM . 0.082 . 304K

(0.488 atm . 1L) / 0.082 . 304K = 1.06g/ MM

Mass / Molar mass = Moles → That's why the 1.06 g / MM

0.0195 mol = 1.06g / MM

1.06g/0.0195 mol = MM → 54.3 g/m

Now, let's use the composition

100 g of compound have 88.8 g of C

54.3 g of compound have ___ (54.3 . 88.8) /100 = 48 g of C

100 g of compound have 11.2 g of H

54.3 g of compound have __ (54.3 . 11.2)/100 = 6 g of H

48 g of C are included un 4 atoms

6 g of H are included in 6 atoms

4 0

3 years ago

Heat is the movement from areas of ___ temperature to areas of ___ temperature

crimeas [40]

Answer:

heat is the movement from areas of <u>high </u>temperature to areas of <u>low </u>temperature

4 0

3 years ago

A mixture of iron and sulfur can be separated by A.magnetic attraction

Elenna [48]

A. Magnetic attraction

6 0

3 years ago

When adjusted for any changes in ΔHΔH and ΔSΔS with temperature, the standard free energy change ΔG∘TΔGT∘Delta G_{T}^{\circ} at

STALIN [3.7K]

The equilibrium constant is 0.0022.

Explanation:

The values given in the problem is

ΔG° = 1.22 ×10⁵ J/mol

T = 2400 K.

R = 8.314 J mol⁻¹ K⁻¹

The Gibbs free energy should be minimum for a spontaneous reaction and equilibrium state of any reaction is spontaneous reaction. So on simplification, the thermodynamic properties of the equilibrium constant can be obtained as related to Gibbs free energy change at constant temperature.

The relation between Gibbs free energy change with equilibrium constant is ΔG° = -RT ln K

So, here K is the equilibrium constant. Now, substitute all the given values in the corresponding parameters of the above equation.

We get,

$1.22 * 10^{5} = - 8.314* 2400 * ln K$