The coefficient for NaNO₃ = 6
<h3>Further explanation
</h3>
Equalization of chemical reaction equations can be done using variables. Steps in equalizing the reaction equation:
• 1. gives a coefficient on substances involved in the equation of reaction such as a, b, or c etc.
• 2. make an equation based on the similarity of the number of atoms where the number of atoms = coefficient × index between reactant and product
• 3. Select the coefficient of the substance with the most complex chemical formula equal to 1
Reaction
AI(NO₃)₃ +Na₂SO₄ →
Al₂(SO₄) +
NaNO₃
give coefficient
aAI(NO₃)₃ +bNa₂SO₄ →
Al₂(SO₄)₃ +c
NaNO₃
Al, left=a, right=2⇒a=2
N, left=3a, right=c⇒3a=c⇒3.2=c⇒c=6
Na, left=2b, right=c⇒2b=c⇒2b=6⇒b=3
The equation becomes :
2AI(NO₃)₃ +3Na₂SO₄ →
Al₂(SO₄)₃ +6NaNO₃
<u>Answer:</u> The average atomic mass of X is 28.09 amu
<u>Explanation:</u>
Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.
Formula used to calculate average atomic mass follows:
.....(1)
Mass of isotope 1 = 27.979 amu
Percentage abundance of isotope 1 = 92.21 %
Fractional abundance of isotope 1 = 0.9212
Mass of isotope 2 = 28.976 amu
Percentage abundance of isotope 2 = 4.70 %
Fractional abundance of isotope 2 = 0.0470
Mass of isotope 3 = 29.974 amu
Percentage abundance of isotope 3 = 3.09 %
Fractional abundance of isotope 3 = 0.0309
Putting values in equation 1, we get:
![\text{Average atomic mass of X}=[(27.979\times 0.9212)+(28.976\times 0.0470)+(29.974\times 0.0309)]](https://tex.z-dn.net/?f=%5Ctext%7BAverage%20atomic%20mass%20of%20X%7D%3D%5B%2827.979%5Ctimes%200.9212%29%2B%2828.976%5Ctimes%200.0470%29%2B%2829.974%5Ctimes%200.0309%29%5D)

Hence, the average atomic mass of X is 28.09 amu
Answer:
995.313KW
Explanation:
the explanation is in the picture
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Answer:
The coefficient is 1
Explanation:
CaO(s) + CO2(g) -> CaCO3(s)
In the balanced equation, the coefficient for CaO is 1
The coefficient represents the number of moles of a compound in the stoichiometry of the reaction
Answer:
4.67 kg
Explanation:
Given data
- Dimensions of the lead sheet: 75.0 cm by 55.0 cm by 0.10 cm
- Density of lead: 11.3 g/cm³
Step 1: Calculate the volume of the sheet
The volume of the sheet is equal to the product of its dimensions.

Step 2: Calculate the mass of the sheet
The density (ρ) is equal to the mass divided by the volume.
