Answer:
Part A: 5.899x10^-3 moles of Al
Part B: 1.573 g of AlBr3
Explanation:
Part A: We have to obtain the volume of the piece of aluminium; all sides of the square must be in cm. Then, use the density to obtain the mass.
0.059 is the volume of the Al udes for the reaction. Now, to oabtain the moles:
Part B: To obatin the mass of AlBr3, we need the balanced chemical equation:
2Al + 3Br2 → 2AlBr3
We assume bromine (Br2) is in excess, therefore, we calculate the aluminum bromide formed from the Al:
of Al
Answer :
The balanced chemical reaction will be,
Explanation :
Balanced chemical reaction : It is defined as the reaction in which the number of atoms of individual elements present on reactant side must be equal to the product side.
If the amount of atoms of each type on the left and right sides of a reaction differs then to balance the equation by adding coefficient in the front of the elements or molecule or compound in the chemical equation.
The coefficient tell us about that how many molecules or atoms present in the chemical equation.
When sulfuric acid react with potassium hydroxide then it react to give potassium sulfate and water as a product. This reaction is known as acid-base reaction.
The balanced chemical reaction will be,
We can first find the number of moles using the ideal gas law equation,
PV = nrT
where P - pressure - 1000 mmHg / 760 mmHg/atm = 1.32 atm
V - volume - 100 x 10⁻³ L
n - number of moles
r - universal gas constant - 0.08206 LatmK⁻¹mol⁻¹
T - temperature in Kelvin - 95 °C + 273 = 368 K
substituting these values
1.32 atm x 100 x 10⁻³ L = n x 0.08206 LatmK⁻¹mol⁻¹ x 368 K
n = 0.00437 mol
molar mass can be determined as follows
molar mass = mass present / number of moles
molar mass = 0.597 g / 0.00437 mol = 136.6 g/mol
molar mass of gas is 137 g/mol
I believe that the answer is Oxygen. Hope this helps. :)
Balance Chemical Equation is as follow,
<span> Cu + 2 AgNO</span>₃ → 2 Ag + Cu(NO₃)₂
According to Balance Equation,
2 Moles of Ag is produced by reacting = 1 Mole of Cu
So,
0.854 Moles of Ag will be produced by reacting = X Moles of Cu
Solving for X,
X = (0.854 mol × 1 mol) ÷ 2 mol
X = 0.427 Moles of Cu
Result:
0.854 Moles of Ag are produced by reacting 0.427 Moles of Cu.
