Answer:
13.5g of AgNO3 will be needed
Explanation:
Silver nitrate, AgNO3 contains 1 mole of silver, Ag, per mole of nitrate. To solve this problem we need to convert the mass of Ag to moles. Thee moles = Moles of AgNO3 we need. With the molar mass of AgNO3 we can find the needed mass:
<em>Moles Ag-Molar mass: 107.8682g/mol-</em>
8.6g * (1mol / 107.8682g) = 0.0797 moles Ag = Moles AgNO3
<em>Mass AgNO3 -Molar mass: 169.87g/mol-</em>
0.0797 moles Ag * (169.87g/mol) =
<h3>13.5g of AgNO3 will be needed</h3>
Answer:
709 g
Step-by-step explanation:
a) Balanced equation
Normally, we would need a balanced chemical equation.
However, we can get by with a partial equation, as log as carbon atoms are balanced.
We know we will need an equation with masses and molar masses, so let’s <em>gather all the information</em> in one place.
M_r: 30.07 236.74
C₂H₆ + … ⟶ C₂Cl₆ + …
m/g: 90.0
(i) Calculate the moles of C₂H₆
n = 90.0 g C₂H₆ × (1 mol C₂H₆ /30.07 g C₂H₆)
= 2.993 mol C₂H₆
(ii) Calculate the moles of C₂Cl₆
The molar ratio is (1 mol C₂Cl₆/1 mol C₂H₆)
n = 2.993 mol C₂H₆ × (1 mol C₂Cl₆/1 mol C₂H₆)
= 2.993 mol C₂Cl₆
(iii) Calculate the mass of C₂Cl₆
m = 2.993 mol C₂Cl₆ × (236.74 g C₂Cl₆/1 mol C₂Cl₆)
m = 709 g C₂Cl₆
The reaction produces 709 g C₂Cl₆.
Answer:
Needle and Scale Leaves. This kind of leaves are mostly found in 'evergreen' trees. ...
Lanceolate Leaf. ...
Oblong Leaf. ...
Linear Leaves. ...
Cordate Leaves.
Explanation:
Answer:
you can see the answer at the pic