Draw models to represent the formation of the<br> positive calcium ion and the negative bromide ion.

What is true when a reaction has reached equilibrium?

jolli1 [7]

Answer:

Explanation:

The rate of the forward reaction is less than the rate of the reverse reaction. ... The reaction rates of the forward and reverse reactions are equal

4 0

3 years ago

Green plants use light from the sun to drive photosynthesis. photosynthesis is a chemical reaction in which water (H2O) and carb

alexandr1967 [171]

Answer:

Mass of H₂O is 3.0g

Explanation:

The reaction equation is given as:

6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂

Parameters that are known:

Mass of CO₂ used = 7.3g

Unknown: mass of water consumed = ?

Solution

To solve this kind of problem, we simply apply some mole concept relationships.

First, we work from the known to the unknown. From the problem, we have 7.3g of CO₂ that was used. We can find the number of moles from this value using the expression below:

Number of moles of CO₂ = $\frac{mass}{molar mass}$

From this number of moles of CO₂, we can use the balanced equation to relate the number of moles of CO₂ to that of H₂O:

6 moles of CO₂ reacted with 6 moles of H₂O(1:1)

We can then use the mole relationship with mass to find the unknown.

Workings

>>>> Number of moles of CO₂ =?

Molar mass of CO₂ :

Atomic mass of C = 12g

Atomic mass of O = 16g

Molar mass of CO₂ = 12 + (2 x16) = 44gmol⁻¹

Number of moles of CO₂ = $\frac{7.3}{44}$ = 0.166moles

>>>>>> if 6 moles of CO₂ reacted with 6 moles of H₂O, then 0.166moles of CO₂ would produce 0.166moles of H₂O

>>>>>> Mass of water consumed = number of mole of H₂O x molar mass

Mass of H₂0 = 0.166 x ?

Molar mass of H₂O:

Atomic mass of H = 1g

Atomic mass of O = 16

Molar mass of H₂O = (2x1) + 16 = 18gmol⁻¹

Mass of H₂O = 0.166 x 18 = 3.0g

7 0

4 years ago

Using the equations

Anna [14]

Considering the Hess's Law, the enthalpy change for the reaction is 221.8 kJ/mol.

Hess's Law indicates that the enthalpy change in a chemical reaction will be the same whether it occurs in a single stage or in several stages. That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when it occurs in a single stage.

In this case you want to calculate the enthalpy change of:

C₂H₄ (g) + 6 F₂ (g) → 2 CF₄ (g) + 4 HF (g)

which occurs in three stages.

You know the following reactions, with their corresponding enthalpies:

Equation 1: H₂ (g) + F₂ (g) → 2 HF (g) ∆H° = -79.2 kJ/mol

Equation 2: C (s) + 2 F₂ (g) → CF₄ (g) ∆H° = 141.3 kJ/mol

Equation 3: 2 C(s) + 2 H₂ (g) → C₂H₄ (g) ∆H° = -97.6 kJ/mol

Because of the way formation reactions are defined, any chemical reaction can be written as a combination of formation reactions, some going forward and some going back.

<h3 /><h3>FIRST STEP</h3>

First, to obtain the enthalpy of the desired chemical reaction you need one mole of C₂H₄ (g) on reactant side and it is present in first equation. Since this equation has one mole of C₂H₄ (g) on the product side, it is necessary to locate it on the reactant side (invert it).

When an equation is inverted, the sign of ΔH° also changes.

<h3>SECOND STEP</h3>

Now, you need 2 moles of CF₄ (g) on the product side. The second equation has 1 mole of CF₄ (g) on the product side, so it is necessary to multiply it by 2 to obtain 2 moles of CF₄ (g).

Since enthalpy is an extensive property, that is, it depends on the amount of matter present, since the equation is multiply by 2, the variation of enthalpy also.

<h3>THIRD STEP</h3>

Finally, you need 4 moles of HF (g) on the product side. The first equation has 2 moles of HF (g) on the product side, so it is necessary to multiply it by 2 to obtain 4 moles of the compound.

Since the equation is multiply by 2, the variation of enthalpy also is multiplied by 2.

<h3>SUMMARY</h3>

In summary, you know that three equations with their corresponding enthalpies are:

Equation 1: 2 H₂ (g) + 2 F₂ (g) → 4 HF (g) ∆H° = -158.4 kJ/mol

Equation 2: 2 C (s) + 4 F₂ (g) → 2 CF₄ (g) ∆H° = 282.6 kJ/mol

Equation 3: C₂H₄ (g) → 2 C(s) + 2 H₂ (g) ∆H° = 97.6 kJ/mol

Adding or canceling the reactants and products as appropriate, and adding the enthalpies algebraically, you obtain:

C₂H₄ (g) + 6 F₂ (g) → 2 CF₄ (g) + 4 HF (g) ΔH°= 221.8 kJ/mol

Finally, the enthalpy change for the reaction is 221.8 kJ/mol.

Learn more about molar enthalpy:

7 0

2 years ago

NaClO3 > NaCl + O2 <br> Balance

tiny-mole [99]

Answer:

Balancing Strategies: To balance this reaction it is best to get the Oxygen atoms on the reactant side of the equation to an even number. Once this is done everything else falls into place. Put a "2" in front of the NaClO3. Change the coefficient in front of the O2.

6 0

3 years ago

What is the limiting reactant when 19.9 g CuO react with 2.02 g H2?

Harlamova29_29 [7]

Answer:

Explanation:

use the equation

moles = mass/mr

=19.9/79.5

=0.250moles of CuO

then do the same for

H = 2.02/1

=2.02

so CuO is the limiting reagent because there is less amount of it.

Hope this helps :)

4 0

3 years ago

Draw models to represent the formation of the positive calcium ion and the negative bromide ion.