A.
Only the Fe is unbalanced.
B.
Only the O is unbalanced
C.
Fe and O are both unbalanced.
D.
C and O are both unbalanced
These are two questions and two answers.
Question 1: Law of Conservation of Mass
Answer: option B. The total mass remains the same during a chemical reaction.
Explanation:
The law of conservation of mass is a universal law. It states that mass is mass is neither created or destroyed, but is is conserved.
In chemical reactions, that means that, always, the total mass of the reactants equals the total mass of the products or, as the option B. states, during a chemical reaction the total mass remains the same.
Since, in chemical reactions, the atoms are not modified (the atoms just bond in different form or with different atoms), that implies that total number of each kind of atoms in the reactants equals the total number of the same kind of atoms in the products.
That is the basis for balancing the chemical equations and for the stoicheometric calculations.
Question 2 . Which element(s) are not balanced in this equation?
Answer: option A. Only the Fe is unbalanced.
Explanation:
1) Given equation: Fe₂O₃ + 3 CO → Fe + 3 CO₂
2) Count the number of atoms of each kind on each side of the equation
i) Fe
reactant side: 2
product side: 1
Therefore, Fe is not balanced
ii) O
reactant side: 3 + 3 = 6
Product side: 3 × 2 = 6
Therefore, it is balanced
iii) C
reactant side: 3
product side: 3
Therefore, C is balanced.
3) Conclusion: Only the Fe is unbalanced.
For #1, the answer is B. The total mass remains the same during a chemical reaction.
For #2, given Fe2O3 + 3CO -> Fe + 3CO2
Fe is unbalanced, and O is also unbalanced
Therefore, Option C: Fe and O are both unbalanced, is true.
Answer:
C15H24O
Explanation:
TO GET THE EMPIRICAL FORMULA, WE NEED TO KNOW THE MASSES AND CONSEQUENTLY THE NUMBER OF MOLES OF EACH OF THE INDIVIDUAL CONSTITUENT ELEMENTS.
FIRSTLY, WE CAN GET THE MASS OF THE CARBON FROM THAT OF THE CARBON IV OXIDE. WE NEED TO KNOW THE NUMBER OF MOLES OF CARBON IV OXIDE GIVEN OFF. THIS CAN BE CALCULATED BY DIVIVDING THE MASS BY THE MOLAR MASS OF CARBON IV OXIDE. THE MOLAR MASS OF CARBON IV OXIDE IS 44G/MOL
The combustion of 1.376 g of butylated hydroxytoluene (BHT) produced 4.122 g CO2 and 1.350 g H2O. Calculations yield an empirical formula of CH2O, indicating one carbon, two hydrogen, and one oxygen atom.
To determine the empirical formula of butylated hydroxytoluene (BHT), we can follow these steps:
1. **Find moles of CO2 and H2O produced:**
\[ \text{moles of } CO_2 = \frac{\text{mass of } CO_2}{\text{molar mass of } CO_2} \]
\[ \text{moles of } H_2O = \frac{\text{mass of } H_2O}{\text{molar mass of } H_2O} \]
2. **Find the mole ratio:**
Divide the moles of each element (C, H, and O) in CO2 and H2O by the smallest number of moles.
3. **Write the empirical formula:**
Use the mole ratios to write the empirical formula.
Let's perform the calculations:
\[ \text{Molar mass of } CO_2 = 12.01 \, \text{(C)} + 2 \times 16.00 \, \text{(O)} = 44.01 \, \text{g/mol} \]
\[ \text{Molar mass of } H_2O = 2 \times 1.01 \, \text{(H)} + 16.00 \, \text{(O)} = 18.02 \, \text{g/mol} \]
\[ \text{moles of } CO_2 = \frac{4.122 \, \text{g}}{44.01 \, \text{g/mol}} \approx 0.0938 \, \text{mol} \]
\[ \text{moles of } H_2O = \frac{1.350 \, \text{g}}{18.02 \, \text{g/mol}} \approx 0.0749 \, \text{mol} \]
Divide by the smallest number of moles (0.0749) to get a ratio close to 1:1:
\[ \text{C} : \text{H} : \text{O} \approx 1.25 : 1 : 1 \]
The ratio is approximately 1:1:1, so the empirical formula is CH2O.
Learn more about empirical formula here:
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exothermic.
kinetic.
convective.
Answer is Health Claim