Answer:
volume
Explanation:
things expand when heat is applied taking up more space hence volume
Using your knowledge of symbols, determine which compound listed must contain both copper and oxygen.
1. malachite, Cu 2CO 3(OH) 2
2. calcite, CaCO 3
3. glucose, C 6H 12O 6
4. chalcopyrite, CuFeS 2
Answer:
The correct option is Malachite which has both copper nd oxygen.
Explanation:
As we the symbol of copper is Cu while the symbols of oxygen is O. As the chemical formula of malachite is Cu₂CO₃(OH)₂. So it has Cu as well as O.
Glucose has only oxygen but no copper
Chalcopyrite has copper but no oxygen
Calcite has oxygen but no copper.
So correct answer is only malachite.
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Answer:
2Fe₂O₃
Explanation:
If the number of atoms of each element are the same on reactant and product sides, it is a balanced equation. This is not a balanced equation because the number of atoms of each element are not the same on reactant and product sides.
To balance an equation, let's first list the atoms:
FeOThen, we count the number of atoms on both sides:
1 Fe 22 O 3We multiply the number of atoms for each element on either side by the number of molecules (aka coefficient, or "big number"):
4 Fe 26 O 3To balance the equation, we need the total number for Fe on the right side to be 4, and the total number for O on the right side to be 6. We can do that by adding a "big number", or coefficient in front, which when multiplied by the number of atoms for each element, makes a balanced equation.
2 is a good value: 2 × 2 = 4, and 2 × 3 = 6.
As such:
4Fe + 3O₂ → 2Fe₂O₃
Calculate the mass in grams of carbon dioxide produced from 11.2 g of octane (C8H18) in the reaction above.
18 + 25 O2 --> 16 CO2 + 18 H2O
89.6 g
17.3 g
34.5 g
46.2 g
Answer:
34.6g
Explanation:
Given parameters:
Mass of Octane = 11.2g
Reaction expression;
2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O
Mass of octane = 11.2g
Unknown:
Mass of carbon dioxide produced = ?
Solution:
From the balanced reaction equation;
2 mole of octane produced 16 moles of carbon dioxide
From the given specie, let us find the number of moles;
Number of moles = [tex]\frac{mass}{molar mass}[/tex]
Molar mass of C₈H₁₈ = 8(12) + 18(1) = 114g/mole
Number of moles of octane = [tex]\frac{11.2}{114}[/tex] = 0.098mole
2 mole of octane produced 16 moles of carbon dioxide
0.098 mole of octane will produce [tex]\frac{0.098 x 16}{2}[/tex] = 0.79mole of CO₂
Mass of CO₂ = number of moles x molar mass
Molar mass of CO₂ = 12 + 2(16) = 44g/mol
Mass of CO₂ = 0.79 x 44 = 34.6g