Which statement describes the process represented by this equation?
(1) A bond is formed as energy is absorbed.
(2) A bond is formed and energy is released.
(3) A bond is broken as energy is absorbed.
(4) A bond is broken and energy is released.
(2) Cu(s) + 2AgNO3(aq) ==>2Ag(s) + Cu(NO3)2(aq)
(3) 2H2(g) + O2(g) ==>2H2O(l)
(4) KOH(aq) + HCl(aq) ==>KCl(aq) + H2O(l)
Answer : The correct option is, (1)
Explanation :
The option (1) reaction is an example of decomposition reaction because in this reaction the larger compound, calcium carbonate decomposes into smaller molecule, calcium oxide and carbon dioxide.
Decomposition reaction : It is a type of reaction in which a larger compound decomposes into a smaller molecule or elements.
The option (2) reaction is an example of single displacement reaction because in this reaction the most reactive metal, copper displace the less reaction metal, silver.
The option (3) reaction is an example of combination reaction because in this reaction two reactants, hydrogen and oxygen react to give a single product as water.
The option (4) reaction is an example of double displacement reaction because in this reaction a positive cation and a negative anion of two reactants exchange their places to give two new products.
Hence, the correct option is, (1)
Answer:
when 1.00 g of magnesium reacts with 5.00 g of bromine, approximately 7.57 g of magnesium bromide is formed.
Explanation:
To find the mass of magnesium bromide formed when 1.00 g of magnesium reacts with 5.00 g of bromine, you need to first write a balanced chemical equation for the reaction between magnesium and bromine. The balanced equation for the formation of magnesium bromide (MgBr2) is as follows:
Mg + Br2 → MgBr2
Now, you can calculate the molar mass of each substance involved in the reaction:
Molar mass of Mg (magnesium) = 24.31 g/mol
Molar mass of Br2 (bromine) = 2 * 79.90 g/mol = 159.80 g/mol
Molar mass of MgBr2 (magnesium bromide) = 24.31 g/mol + 2 * 79.90 g/mol = 184.11 g/mol
Next, calculate the number of moles for each reactant:
Moles of Mg = Mass (1.00 g) / Molar mass (24.31 g/mol) = 0.0411 moles
Moles of Br2 = Mass (5.00 g) / Molar mass (159.80 g/mol) = 0.0313 moles (approximately, rounded to four decimal places)
Now, determine the limiting reactant. To do this, compare the mole ratio between Mg and Br2 in the balanced equation. The balanced equation shows that 1 mole of Mg reacts with 1 mole of Br2. Therefore, the limiting reactant is the one that is present in the smaller amount relative to the balanced equation's stoichiometry.
In this case, magnesium (0.0411 moles) is present in a smaller amount than bromine (0.0313 moles). So, magnesium is the limiting reactant.
Now that you know magnesium is the limiting reactant, you can calculate the mass of magnesium bromide formed using the stoichiometry of the balanced equation. According to the balanced equation, 1 mole of Mg produces 1 mole of MgBr2.
Moles of MgBr2 formed = Moles of Mg (limiting reactant) = 0.0411 moles
Now, calculate the mass of magnesium bromide formed:
Mass of MgBr2 = Moles of MgBr2 × Molar mass of MgBr2
Mass of MgBr2 = 0.0411 moles × 184.11 g/mol = 7.57 g
So, when 1.00 g of magnesium reacts with 5.00 g of bromine, approximately 7.57 g of magnesium bromide is formed.
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To solve this we use the equation,
M1V1 = M2V2
where M1 is the concentration of the stock solution, V1 is the volume of the stock solution, M2 is the concentration of the new solution and V2 is its volume.
2.5 M x V1 = 0.50 M x .100 L
V1 = 0.02 L or 20 mL
Answer:
20 milliliters
Explanation:
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