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3 years ago

How to balance chemical equations

1 answer:

8 0

To balance a chemical equation, you want the same amount of elements to equal the same on both sides.

Step1. Write equation out
[CH4 + Cl2 ---> CCl4 + HCl]
C:1; H:4; Cl:2 C:1; H:1; CL:5 /// Cl = 5 since 4Cl + 1Cl

The Carbon element is balanced, but Hydrogen isn't.
So to balance it we will add a coefficient behind HCl, so now

[CH4 + Cl2 ------> CCl4 + 4HCL]
C:1; H:4; Cl:2 C:1; H:4; CL:8 ///// Carbon and hydrogen are balanced, but now Chlorine is not. Now we balance that element by addind a coefficiant behind CL2////

[CH4 + 4Cl2 -----> CCl4 + 4HCl]
C:1; H:4; Cl:8 C:1; H:4; CL:8 ///// So now that we added a 4*Cl2, it equals to Cl:8. So now what most people want to see is if every element is at its lowest balance, so we see if we can any coefficient lower. Just like simplifying if possible.

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What is fire and how does it work

xenn [34]

Answer:

Fire is the result of applying enough heat to a fuel source, when you've got a whole lot of oxygen around. As the atoms in the fuel heat up, they begin to vibrate until they break free of the bonds holding them together and are released as volatile gases. These gases react with oxygen in the surrounding atmosphere.

Explanation:

3 0

3 years ago

Calculate the number of Li atoms in 7.8 mol of Li.

Gwar [14]

<h3>Answer:</h3>

4.7 × 10²⁴ atoms Li

<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

Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Stoichiometry

Using Dimensional Analysis

<h3>Explanation:</h3>

Step 1: Define

7.8 mol Li

Step 2: Identify Conversions

Avogadro's Number

Step 3: Convert

Set up: $\displaystyle 7.8 \ mol \ Li(\frac{6.022 \cdot 10^{23} \ atoms \ Li}{1 \ mol \ Li})$
Multiply/Divide: $\displaystyle 4.69716 \cdot 10^{24} \ atoms \ Li$

Step 4: Check

We are told to round to 2 sig figs. Follow sig fig rules and round.

4.69716 × 10²⁴ atoms Li ≈ 4.7 × 10²⁴ atoms Li

6 0

3 years ago

Diphosphorus pentoxide(P2O5) reacts with water to form phosphoric acid, a major industrial acid. In the laboratory, the oxide is

Akimi4 [234]

Answer: The mass of $4.65\times 10^{22}$ molecules of phosphorus pentoxide is 20.5 g

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to the molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given molecules}}{\text {avogadro's number}}=\frac{4.65\times 10^{22}}{6.023\times 10^{23}}=0.0772moles$

1 mole of $P_2O_5$ weigh = 283.9 g

Thus 0.0772 moles of $P_2O_5$ weigh = $\frac{283.9}{1}\times 0.0722=20.5g$

Thus the mass of $4.65\times 10^{22}$ molecules of phosphorus pentoxide is 20.5 g

3 0

3 years ago

Write the molecular equation and the net inonic equation for each of the following aqueous reactions. If no reaction occurs, wri

SIZIF [17.4K]

Answer: The equations are given below.

Explanation:

For the given options:

Option 1: $AgNO_3+NaI$

The molecular equation for this follows:

$AgNO_3(aq.)+NaI(aq.)\rightarrow AgI(s)+NaNO_3(aq.)$

Ionic form of the above equation follows:

$Ag^+(aq.)+NO_3^-(aq.)+Na^+(aq.)+I^-(aq.)\rightarrow AgI(s)+Na^+(aq.)+NO_3^-(aq.)$

As, sodium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation.

The net ionic equation for the above reaction follows:

$Ag^+(aq.)+I^-(aq.)\rightarrow AgI(s)$

Option 2: $Ba(NO_3)_2+K_2SO_4$

The molecular equation for this follows:

$Ba(NO_3)_2(aq.)+K_2SO_4(aq.)\rightarrow 2KNO_3(aq.)+BaSO_4(s)$

Ionic form of the above equation follows:

$Ba^{2+}(aq.)+2NO_3^-(aq.)+K^+(aq.)+SO_4^{2-}(aq.)\rightarrow 2K^+(aq.)+2NO_3^-(aq.)+BaSO_4(s)$

As, potassium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation.

The net ionic equation for the above reaction follows:

$Ba^{2+}(aq.)+SO_4^{2-}(aq.)\rightarrow BaSO_4(s)$

Option 3: $NH_4NO_3+K_2SO_4$

The molecular equation for this follows:

$2NH_4NO_3(aq.)+K_2SO_4(aq.)\rightarrow 2KNO_3(aq.)+H_2SO_4(aq.)+NH_3(g)$

Ionic form of the above equation follows:

$2NH_4^+(aq.)+2NO_3^-(aq.)+2K^+(aq.)+SO_4^{2-}(aq.)\rightarrow 2NH_3(g)+2H^+(aq.)+SO_4^{2-}(aq.)+2K^+(aq.)+2NO_3^-(aq.)$

As, nitrate, potassium and sulfate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation.

The net ionic equation for the above reaction follows:

$2NH_4^+(aq.)\rightarrow 2NH_3(g)+2H^+(aq.)$

Option 4: $LiCl+Al(NO_3)_3$

The molecular equation for this follows:

$3LiCl(aq.)+Al(NO_3)_3(aq.)\rightarrow 3LiNO_3(aq.)+AlCl_3(s)$

Ionic form of the above equation follows:

$3Li^+(aq.)+3Cl^-(aq.)+Al^{3+}(aq.)+NO_3^-(aq.)\rightarrow 3Li^+(aq.)+3NO_3^-(aq.)+AlCl_3(s)$

As, lithium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation.

The net ionic equation for the above reaction follows:

$Al^{3+}(aq.)+3Cl^-(aq.)\rightarrow AlCl_3(s)$

Hence, the molecular and ionic equations are given above.

5 0

3 years ago

A hot air balloonist puts 52000 L of air into their balloon at 500 Celsius and 975 atm. When they heat

Ksenya-84 [330]

Answer:

Explanation:

During the seventeenth and especially eighteenth centuries, driven both by a desire to understand nature and a quest to make balloons in which they could fly (Figure 1), a number of scientists established the relationships between the macroscopic physical properties of gases, that is, pressure, volume, temperature, and amount of gas. Although their measurements were not precise by today’s standards, they were able to determine the mathematical relationships between pairs of these variables (e.g., pressure and temperature, pressure and volume) that hold for an ideal gas—a hypothetical construct that real gases approximate under certain conditions. Eventually, these individual laws were combined into a single equation—the ideal gas law—that relates gas quantities for gases and is quite accurate for low pressures and moderate temperatures. We will consider the key developments in individual relationships (for pedagogical reasons not quite in historical order), then put them together in the ideal gas law

3 0

3 years ago