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3 years ago

a molecule of ozone gas has has three atoms of oxygen . witch of these describes ozone most specifically?

2 answers:

8 0

The best answer choice you could go for is B pure substance, a pure substance is a compound that is made of one element, like a diamond.

Vitek1552 [10]3 years ago

7 0

Answer:

Element

Explanation:

Since all 3 atoms are oxygen, it is an element. It cannot be further broken down.

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Which equivalence factor should you use to convert from 3.48 x 1018 atoms of magnesium (Mg) to moles of magnesium? A (1 mol Mg/1

faltersainse [42]

1242.375Mg=1976Mg located

6 0

3 years ago

C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l), ΔH = –1.37 × 103 kJ For the combustion of ethyl alcohol as described in the above equati

babymother [125]

Answer:

The true statements are: I. The reaction is exothermic.

II. The enthalpy change would be different if gaseous water was produced.

Explanation:

The given chemical reaction: C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l), ΔH= -1.37×10³kJ

1. In an exothermic reaction, heat or energy is released from the system to the surrounding. Thus for an exothermic process the change in enthalpy is less than 0 or negative (ΔH < 0) .

Since the enthalpy change for a combustion reaction is negative. Therefore, the given reaction is exothermic.

2. The change in enthalpy (ΔH) of a reaction is equal to difference of the sum of standard enthalpy of formation (ΔHf°) of the products and the reactants.

ΔHr° = ∑ n.ΔHf°(products) − ∑ n.ΔHf°(reactants)

As the value of ΔHf° of water in gaseous state and liquid state is not the same.

Therefore, the enthalpy change of the reaction will be different, if gaseous water was present instead of liquid water.

3. An oxidation-reduction reaction or a redox reaction involves simultaneous reduction and oxidation processes.

The given chemical reaction, represents the combustion reaction of ethanol.

Since combustion reactions are redox reactions. Therefore, the given combustion reaction is an oxidation-reduction reaction.

4. According to the ideal gas equation: P.V =n.R.T

Volume (V) ∝ n (number of moles of gas)

Since the number of moles (n) of gaseous reactants is 3 and number of moles of gaseous (n) products is 2.

Therefore, the volume occupied by 3 moles of the reactant gaseous molecules will be more than 2 moles product gaseous molecules.

3 0

3 years ago

How many moles of ethanol are produced starting with 500.g glucose?

Monica [59]

<h3>Answer:</h3>

5.55 mol C₂H₅OH

<h3>General Formulas and Concepts:</h3>

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Chemistry

Atomic Structure

Reading a Periodic Tables
Moles

Stoichiometry

Using Dimensional Analysis
Analyzing Reactions RxN

<h3>Explanation:</h3>

Step 1: Define

[RxN - Balanced] C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂

[Given] 500. g C₆H₁₂O₆ (Glucose)

[Solve] moles C₂H₅OH (Ethanol)

Step 2: Identify Conversions

[RxN] 1 mol C₆H₁₂O₆ → 2 mol C₂H₅OH

[PT] Molar mass of C - 12.01 g/mol

[PT] Molar Mass of H - 1.01 g/mol

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of C₆H₁₂O₆ - 6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol

Step 3: Stoichiometry

[DA] Set up conversion: $\displaystyle 500 \ g \ C_6H_{12}O_6(\frac{1 \ mol \ C_6H_{12}O_6}{180.18 \ g \ C_6H_{12}O_6})(\frac{2 \ mol \ C_2H_5OH}{1 \ mol \ C_6H_{12}O_6})$
[DA} Multiply/Divide [Cancel out units]: $\displaystyle 5.55001 \ mol \ C_2H_5OH$

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

5.55001 mol C₂H₅OH ≈ 5.55 mol C₂H₅OH

8 0

3 years ago

You mix 200. mL of 0.400M HCl with 200. mL of 0.400M NaOH in a coffee cup calorimeter. The temperature of the solution goes from

GuDViN [60]

Answer : The enthalpy of neutralization is, 56.012 kJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.08 mole of HCl neutralizes by 0.08 mole of NaOH

Thus, the number of neutralized moles = 0.08 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $200mL+200L=400mL$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 400mL=400g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 400 g

$T_{final}$ = final temperature of water = $27.78^oC=273+25.10=300.78K$

$T_{initial}$ = initial temperature of metal = $25.10^oC=273+27.78=298.1K$

Now put all the given values in the above formula, we get:

$q=400g\times 4.18J/g^oC\times (300.78-298.1)K$

$q=4480.96J$

Thus, the heat released during the neutralization = -4480.96 J

Now we have to calculate the enthalpy of neutralization.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -4480.96 J

n = number of moles used in neutralization = 0.08 mole

$\Delta H=\frac{-4480.96J}{0.08mole}=-56012J/mole=-56.012kJ/mol$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 56.012 kJ/mole

8 0

2 years ago

Any help with these questions would help i'm lost

SSSSS [86.1K]

Answer:

1. 31.25 mL

2. 1.98 g/L

3. 0.45 g/mL

Explanation:

For each of the problems, you need to perform unit conversions. You need to use the information given to you to convert to a specific unit.

1. You need volume (mL). You have density (g/mL) and mass (g). Divide mass by density. You will cancel out mL and be left with g.

(50.0 g)/(1.60 g/mL) = 31.25 mL

2. You are given grams and liters. You need to find density with units g/L. This means that you have to divide grams by liters.

(0.891 g)/(0.450 L) = 1.98 g/L

3. You have to find density again but this time with units g/mL. Divide the given mass by the volume.

(10.0 g)/(22.0 mL) = 0.45 g/mL

5 0

3 years ago