Answer:
120g
Explanation:
Step 1:
We'll begin by writing the balanced equation for the reaction.
Sn + 2HF —> SnF2 + H2
Step 2:
Determination of the number of mole HF needed to react with 3 moles of Sn.
From the balanced equation above,
1 mole of Sn and reacted with 2 moles of HF.
Therefore, 3 moles Sn will react with = 3 x 2 = 6 moles of HF.
Step 3:
Conversion of 6 moles of HF to grams.
Number of mole HF = 6 moles
Molar Mass of HF = 1 + 19 = 20g/mol
Mass of HF =..?
Mass = number of mole x molar Mass
Mass of HF = 6 x 20
Mass of HF = 120g
Therefore, 120g of HF is needed to react with 3 moles of Sn.
Answer: C) Either benzene or oxygen may limit the amount of product that can be formed
Explanation: Benzene and oxygen are the reactants of the equation. What type and the amount of reactants there are in a chemical reaction affects the outcome. Therefore, seeing as benzene and oxygen are the reactants, the answer is C).
Answer:
decreased by a factor of 10
Explanation:
pH is defined in such a way that;
pH= −log10(H)
Where H represents the concentration of Hydronium or Hydrogen ions
Given that pH is changed from 1 to 2,
By rearranging the above formula , we get 10−pH = H
- if pH=1,H=10−1=0.1M
- if pH=2,H=10−2=0.01M
Therefore, 0.1/0.01 = 10 and 0.1 > 0.01
Hence, the concentration of hydronium ions in the solution is decreased by a factor of 10
Answer:
A)
<u>4, 7, 4, 6</u>
B)
<u>12 moles</u>
Explanation:
__↑______↑
8.00 mol | 14.00 mol
________________
You can turn this into a system of variables which are solvable.
To do this, create variables for the coefficients of each compound in the reaction respectively.
Because to be balanced, the count of atoms in each element of the compound correspond to the coefficient of the variable in that compound so that the count of the left (reactant) side is set equal to the right (product) side.
a corresponds to the coefficient of the first compound, b corresponds to the coefficient of the second compound, c corresponds to the coefficient of the third compound, and d corresponds to the coefficient of the fourth compound.
(Reactant = Product)
Reactant: 1a [N] Product: 1c.
Reactant: 3a [H] Product: 2d.
Reactant: 2b [O] Product: 2c + 1d.
Thus the system is:
1a = 1c
3a = 2d
2b = 2c + 1d.
Then just use the substitution methods to solve.
