!!PLEASE ANSWER!! 24 PTS!!

All of the following are conductors of electricity except

olga55 [171]

Rubber it's not a conductor

4 0

2 years ago

How does the A Hreaction relate to the A He of molecules involved in a reaction?

igor_vitrenko [27]

Answer:

B. ΔHreaction = ΔH°f reactants- ΔH°f products

Explanation:

Using Hess's law, it is possible to sum ΔH of several related reactions to find ΔH of a particular reaction.

Having in mind Hess's law, ΔH°f is defined as the change in enthalpy during the formation of 1 mole of substance from its constituent elements (That is, pure elements, mono or diatomics, that have a ΔH° = 0).

For example, in ΔH°f of H₂O, the equation is:

H₂(g) + 1/2O₂(g) → H₂O(g)

The constituent elements with ΔH°f = 0 are H₂(g) and O₂(g).

Now, using Hess's law, you can sum the ΔH°f of substance in a reaction as, for example:

NaOH + HCl → H₂O + NaCl. ΔHr

The ΔH°f for each substance in the reaction is:

NaOH: Na + 1/2H₂ + 1/2O₂ → NaOH (1)

HCl: 1/2H₂ + 1/2Cl₂ → HCl (2)

H₂O: H₂ + 1/2O₂ → H₂O (3)

NaCl: Na + 1/2Cl₂ → NaCl (4)

The algebraic sum of (3) + (4) is -(ΔH°f reactants):

H₂ + 1/2O₂ + Na + 1/2Cl₂ → NaCl + H₂O ΔH°f reactants

This reaction - {(1)+(2)} ΔH°f products

NaOH + HCl → H₂O + NaCl.

ΔHr = ΔH°f reactants- ΔH°f products

In the example, we obtain this relationship that can be expanded for all reactions. Thus, right answer is:

<h3>B. ΔHreaction = ΔH°f reactants- ΔH°f products</h3>

8 0

2 years ago

Please use the below balanced equation to answer this question. 2H2(g) + O2(g) -> 2H2O(l) How many grams of water wil

Blizzard [7]

At STP (standard temperature and pressure conditions), 1 mol of any gas occupies 22.4 L
This rule is applied to O₂
22.4 L volume occupied by 1 mol
Therefore 83.4 L occupied by - 1/ 22.4 x 83.4 = 3.72 mol
stoichiometry of O₂ to H₂O is 1:2
then the number of moles of water produced - 3.72 mol x 2= 7.44 mol
mass of water produced - 7.44 mol x 18.01 g/mol = 134.1 g
correct answer is D

4 0

2 years ago

What is the concentration of a solution with a volume of 660L that contains 33.4g of AlCO3?

lana [24]

Answer:

The concentration of the solution is 5.8168 × $10^{-4}$ mol. $dm^{-3}$

Explanation:

Here, we want to calculate the concentration of the solution.

The unit of this is mol/dm^3

So the first thing to do here is to calculate the number of moles of the solute present, which is the number of moles of AlCO3

The number of moles = mass/molar mass

molar mass of AlCO3 = 27 + 12 + 3(16) = 27 + 12 + 48 = 87g/mol

Number of moles = 33.4/87 = 0.384 moles

This 0.384 moles is present in 660 L

x moles will be present in 1 dm^3

Recall 1 dm^3 = 1L

x * 660 = 0.384 * 1

x = 0.384/660 = 0.00058168 = 5.8168 * 10^-4 mol/dm^3

7 0

2 years ago

Can you help me on #14 - #16?

gtnhenbr [62]

15 is b 16 is b and im working on 14

6 0

2 years ago