Given in the problem is the mass of the liquid (500 grams) and the volume of the liquid (1000 ml = 1000 cm^3).
We can use these two givens to calculate the density of the liquid using the following rule:
density = mass / volume
density = 500 / 1000 = 0.5 grams / cm^3
Comparing the calculated density with the choices we have, we can deduce that the liquid is most likely to be propane with density 0.494 g / cm^3
Answer:
6 C(s) + 3 O₂(g) + 2 Fe₂O₃(s) → 4 Fe(s) + 6 CO₂(g)
Explanation:
Iron can be formed in two steps.
Step 1: 2 C(s) + O₂(g) → 2 CO(g)
Step 2: Fe₂O₃(s) + 3 CO(g) → 2 Fe(s) + 3 CO₂(g)
In order to get the net chemical equation, we will multiply the first step by 3, the second step by 2, and then add them.
6 C(s) + 3 O₂(g) → 6 CO(g)
+
2 Fe₂O₃(s) + 6 CO(g) → 4 Fe(s) + 6 CO₂(g)
--------------------------------------------------------------------------------------------------
6 C(s) + 3 O₂(g) + 2 Fe₂O₃(s) + 6 CO(g) → 6 CO(g) + 4 Fe(s) + 6 CO₂(g)
6 C(s) + 3 O₂(g) + 2 Fe₂O₃(s) → 4 Fe(s) + 6 CO₂(g)
Answer:

Explanation:
Here in Calcium Chloride ionic bond is present in between calcium and chlorine atoms. As we know according to Octet rule calcium have two excess atoms and for matching nearest noble gas electronic configuration. It donate two electrons to gain more stability and form
, while chlorine is deficient from one electron to meet nearest noble gas electronic configuration therefore two chlorine atoms accept excess electron from calcium individually and form two
ions.

Hence aqueous solution of calcium chloride breaks the ionic bond pairing in one
and two
ions: 
Answer:
8.1 × 10² g
Explanation:
Step 1: Write the balanced equation
2 C₅₇H₁₁₀O₆ + 163 O₂ ⇒ 114 CO₂ + 110 H₂O
Step 2: Convert 1.6 lb of C₅₇H₁₁₀O₆ to g
We will use the conversion factor 1 lb = 453.592 g.
1.6 lb × 453.592 g/1 lb = 7.3 × 10² g
Step 3: Calculate the moles corresponding to 7.3 × 10² g of C₅₇H₁₁₀O₆
The molar mass of C₅₇H₁₁₀O₆ is 890.83 g/mol.
7.3 × 10² g × 1 mol/890.83 g = 0.82 mol
Step 4: Calculate the moles of water produced from 0.82 moles of C₅₇H₁₁₀O₆
The molar ratio of C₅₇H₁₁₀O₆ to H₂O is 2:110. The moles of H₂O produced are 110/2 × 0.82 mol = 45 mol
Step 5: Calculate the mass corresponding to 45 moles of H₂O
The molar mass of H₂O is 18.02 g/mol.
45 mol × 18.02 g/mol = 8.1 × 10² g