Answer :
The complete equation for the reaction of sulfuric acid and sugar is,
By the stoichiometry of the reaction,
1 mole of sucrose react with the 11 moles of sulfuric acid to give 12 moles of carbon and 11 moles of water.
In this reaction, sulfuric acid react with sucrose (sugar). It dehydrates the sugar molecules which means it eliminates the water.
Answer:
1.9 × 10² g NaN₃
1.5 g/L
Explanation:
Step 1: Write the balanced decomposition equation
2 NaN₃(s) ⇒ 2 Na(s) + 3 N₂(g)
Step 2: Calculate the moles of N₂ formed
N₂ occupies a 80.0 L bag at 1.3 atm and 27 °C (300 K). We will calculate the moles of N₂ using the ideal gas equation.
P × V = n × R × T
n = P × V / R × T
n = 1.3 atm × 80.0 L / (0.0821 atm.L/mol.K) × 300 K = 4.2 mol
We can also calculate the mass of nitrogen using the molar mass (M) 28.01 g/mol.
4.2 mol × 28.01 g/mol = 1.2 × 10² g
Step 3: Calculate the mass of NaN₃ needed to form 1.2 × 10² g of N₂
The mass ratio of NaN₃ to N₂ is 130.02:84.03.
1.2 × 10² g N₂ × 130.02 g NaN₃/84.03 g N₂ = 1.9 × 10² g NaN₃
Step 4: Calculate the density of N₂
We will use the following expression.
ρ = P × M / R × T
ρ = 1.3 atm × 28.01 g/mol / (0.0821 atm.L/mol.K) × 300 K = 1.5 g/L
Answer:
HNO₃ + NaOH ---> NaNO₃ + H₂O
Explanation:
This reaction appears to be a double-displacement reaction. In these reaction, the cation of one compound is swapped with the cation of another.
As such, the hydrogen cation (H⁺) from HNO₃ is swapped with the sodium cation (Na⁺) of NaOH.
Luckily, all of the cations have a +1 charge and the anions have a -1 charge. This means that no coefficients are necessary to balance the reaction.
The <u>complete balanced </u>equation is:
HNO₃ + NaOH ---> NaNO₃ + H₂O
The five levels of organization in a multicellular organism are cells, tissues, organs, organ systems and organisms. The level of complexity and functionality increases going from cells to organisms.
Answer:
30.3 g
Explanation:
At STP, 1 mol of any gas will occupy 22.4 L.
With the information above in mind, we <u>calculate how many moles are there in 32.0 L</u>:
- 32.0 L ÷ 22.4 L/mol = 1.43 mol
Then we <u>calculate how many moles would there be in 16.6 L</u>:
- 16.6 L ÷ 22.4 L/mol = 0.741 mol
The <u>difference in moles is</u>:
- 1.43 mol - 0.741 mol = 0.689 mol
Finally we <u>convert 0.689 moles of CO₂ into grams</u>, using its <em>molar mass</em>:
- 0.689 mol * 44 g/mol = 30.3 g
