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3 years ago

Explain how to balance the chemical equation and classify its reaction type.

Chemistry

2 answers:

jonny [76]3 years ago

8 0

Answer:

#1. Balanced equation: 2C₅H₅ + Fe → Fe(C₅H₅)₂

#2. Type of reaction: Synthesis reaction

Explanation:

Balanced equations are equations that obey the law of conservation of mass.

When an equation is balanced the number of atoms of each element is equal on both side of the equation.

Equations are balanced by putting appropriate coefficients on the reactants and products.

In our case, we are going to put coefficients 2, 1 and 1.

Thus, the balanced equation will be;

2C₅H₅ + Fe → Fe(C₅H₅)₂

This type of a reaction is known as synthesis reaction, in which two or more reactants or compounds combine to form a single compound or product.

Brrunno [24]3 years ago

3 0

Answer: The balanced chemical equation is given below.

Explanation:

Law of conservation of mass states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.

This also means that total mass on the reactant side must be equal to the total mass on the product side.

The balanced chemical equation follows:The balanced chemical equation follows:

On reactant side:

Number of carbon atoms = 10

Number of hydrogen atoms = 10

Number of iron atoms = 1

On product side:

Number of carbon atoms = 10

Number of hydrogen atoms = 10

Number of iron atoms = 1

So, the balanced chemical equation is given above.

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How many moles of solute are contained in a 250.0-milliliter solution that has a concentration of 0.85 M?

adoni [48]

Answer : The correct option is, 0.21 moles

Solution : Given,

Molarity of the solution = 0.85 M = 0.85 mole/L

Volume of solution = 250 ml = 0.25 L $(1L=1000ml)$

Molarity : It is defined as the number of moles of solute present in one liter of the solution.

Formula used :

$Molarity=\frac{\text{moles of solute}}{\text{Volume of solution in liters}}$

Now put all the given values in this formula, we get the moles of solute of the solution.

$0.85mole/L=\frac{\text{moles of solute}}{0.25L}$

$\text{moles of solute}=0.2125=0.21moles$

Therefore, the moles of solute is, 0.21 moles

7 0

3 years ago

Read 2 more answers

For the following reaction, 4.21 grams of hydrogen gas are allowed to react with 10.6 grams of ethylene (C2H4) . hydrogen(g) + e

sergiy2304 [10]

Answer:a) 11.34 g of ethane $(C_2H_6)$ can be formed

b) $C_2H_4$ is the limiting reagent

c) 3.44 g of the excess reagent remains after the reaction is complete

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}$

1. $\text{Moles of} H_2=\frac{4.21}{2}=2.10moles$

2. $\text{Moles of} C_2H_4=\frac{10.6}{28}=0.378moles$

$H_2(g)+C_2H_4(g)\rightarrow C_2H_6(g)$

According to stoichiometry :

1 mole of $C_2H_4$ require 1 mole of $H_2$

Thus 0.378 moles of $C_2H_4$ will require= $\frac{1}{1}\times 0.378=0.378moles$ of $H_2$

Thus $C_2H_4$ is the limiting reagent as it limits the formation of product and $H_2$ is the excess reagent.

moles of $H_2$ left = (2.10-0.378) = 1.72 moles

mass of $H_2$ left= $moles\times {\text {Molar mass}}=1.72moles\times 2g/mol=3.44g$

According to stoichiometry :

As 1 mole of $C_2H_4$ give = 1 mole of $C_2H_6$

Thus 0.378 moles of $C_2H_4$ give = $\frac{1}{1}\times 0.378=0.378moles$ of $C_2H_6$

Mass of $C_2H_6=moles\times {\text {Molar mass}}=0.378moles\times 30g/mol=11.34g$

Thus 11.34 g of ethane is formed.

4 0

4 years ago

Write the name for each of the following chemical formulas. d. P3N5 (for doping semiconductors)

rewona [7]

Answer:

Triphorphorus Pentanitride

Explanation:

since there are three phosphorus use the prefix tri-, since there are five nitrogen use the prefix penta. tri=3, penta=5. then add a suffix -ide to the end of nitrogen. that will give you triphosphorus pentanitride.

3 0

2 years ago

Select the atoms whose ions have an electron configuration of nd5(n = 3, 4, 5 ...). For example, the 2+ ion for V ends the elect

Travka [436]

Ir = [Xe] 4f14 5d7 6s2
Rn = [Xe] 4f14 5d10 6s2 6p6 
Fe = [Ar] 3d6 4s2 
Pa = [Rn] 5f2 6d1 7s2 
Y = [Kr] 4d1 5s2 
Re = [Xe] 4f14 5d5 6s2 
So only Rhenium (Re) has an electronic configuration where the d orbital has 5 electrons in it.

4 0

3 years ago

Read 2 more answers

What is the limiting reactant in

MArishka [77]

Answer:

The limiting reactant is the K

Explanation:

This is the reaction:

2 K + F₂ → 2KF

2 moles of potassium react with 1 mol of fluorine

Let's determine the moles:

6.02x10²³ molecules are contained in 1 mol

10 molecules are contained in (10 . 1) / NA = 1.66x10⁻²³ moles

For 1 mol of flourine, we need 2 moles of K

If we have 1.66x10⁻²³ moles of Fl₂, we will need the double of K

For 2 moles of K, we need 1 mol of Fl₂

If we have 1.66x10⁻²³ moles of K, we will need the half of Fl₂

Then, the limiting reactant is the K

5 0

3 years ago