The number of moles in the 4.30 g of SiH₄ is 1.338 moles.
The mole is a SI unit of measurement that is used to calculate the quantity of any substance.
To calculate the moles
The molar mass of SiH₄ is 32.12 g
The mass of SiH₄ is 43.0 g
Thus, the number of moles in the 4.30 g of SiH₄ is 1.338 moles.
Learn more about moles
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43.0 grams of SiH4 converts to 1.338 moles.
Yo sup??
Balancing a chemical equation is more like solving mathematical equation.
You just need do the trial and error thing.
Just plug in some random numbers so that the number of atoms on both the sides become the same....that's all.....rest will come by practise.
Hope this helps
Answer:
Explanation:
Balancing Chemical Equations
Let's take a look at this scale. We can see that it is unbalanced, with the right (red) side, weighing more than the left (blue) side. In order for the two sides to be balanced, we need to put a little more mass on the left side until they are the same mass.
Unbalanced Scale
Just like we want the scale to be balanced on both sides, a chemical equation should also be balanced on both sides. A chemical equation shows us the substances involved in a chemical reaction - the substances that react (reactants) and the substances that are produced (products). In general, a chemical equation looks like this:
Chemical Equation: General Form
According to the law of conservation of mass, when a chemical reaction occurs, the mass of the products should be equal to the mass of the reactants. Therefore, the amount of the atoms in each element does not change in the chemical reaction. As a result, the chemical equation that shows the chemical reaction needs to be balanced. A balanced chemical equation occurs when the number of the atoms involved in the reactants side is equal to the number of atoms in the products side.
Let's take a look at an equation representing a chemical reaction:
Balancing Chemical Equations: Example
In this chemical reaction, nitrogen (N2) reacts with hydrogen (H) to produce ammonia (NH3).The reactants are nitrogen and hydrogen, and the product is ammonia. If we look at this equation, we can see that the equation is not balanced.
Balancing Chemical Equations Example
The equation is not balanced because in the reactants side, there are 2 nitrogen (N) atoms and 2 hydrogen (H) atoms. In the products side, there are 1 nitrogen (N) atoms and 3 hydrogen (H) atoms. The number of the atoms is not balanced on both sides.
To balance the chemical equation above, we need to make use of coefficients. A coefficient is a number that we place in front of a chemical formula. In the chemical equation, to make the number of nitrogen (N) atoms equal on both sides, first, we place a coefficient of 2 in front of NH3.
Balancing Chemical Equations Example
Once we do that, the number of nitrogen (N) atoms on both sides is balanced. However, the number of hydrogen (H) atoms is not balanced on both sides. We need to make use of another coefficient in front of H2. This time, we put a coefficient of 3 in front of H2 to balance the chemical equation.
Balancing Chemical Equations Example
The equation above is now balanced. There are 2 nitrogen (N) atoms and 6 hydrogen (H) atoms on both the reactants and products side. Since there is no coefficient in front of N2, that means the coefficient is equal to 1.
Steps to Balance an Equation
Practice always makes perfect. In general, to balance an equation, here are the things we need to do:
Count the atoms of each element in the reactants and the products.
Use coefficients; place them in front of the compounds as needed.
The steps are simple, but it is a process of trial and error. Let's take a look at a few more example equations and techniques that can be used to balance each one.
Example 1
This is a reaction between methane (CH4) and oxygen (O2), producing carbon dioxide (CO2) and water (H2O).
Balancing Chemical Equations Example 1
The reaction shown is a combustion reaction: a compound reacts with oxygen and produces carbon dioxide and water. The technique is to balance the carbon (C) atoms first, then the hydrogen (H) atoms, and then the oxygen (O) atoms.
In this case, the carbon (C) atoms are already balanced. So now we look at the hydrogen (H) atoms. There are 4 hydrogen (H) atoms on the reactants side and 2 hydrogen (H) atoms on the products side. To balance them, we put a coefficient of 2 in front of H2O
Answer:
1) 6(1/2)
2) 8(4/5)
Explanation:
1) 7 - 1/2 = 7/1 - 1/2 = 14/2 - 1/2 = 13/2 = 6(1/2)
2) 11(2/5) - 2(3/5) = 57/5 - 13/5 = 44/5 = 8(4/5)