The volume of a gas that is required yo react with 4.03 g mg at STP is 1856 ml
calculation/
- calculate the moles of Mg used
moles=mass/molar mass
moles of Mg is therefore=4.03 g/ 24.3 g/mol=0.1658 moles
- by use of mole ratio of Mg:O2 from the equation which is 2:1
the moles 02=0.1679 x1/20.0829 moles
- at STP 1 mole of a gas= 22.4 l
0.0895 moles=? L
- =0.0895 moles x22.4 l/ 1 mole=1.8570 L
into Ml = 1.8570 x1000=1856 ml approximately to 1860
Given :
A 10.99 g sample of NaBr contains 22.34% Na by mass.
To Find :
How many grams of sodium does a 9.77g sample of sodium bromine contain.
Solution :
By law of constant composition , in any given chemical compound, the elements always combine in the same proportion with each other.
Therefore , percentage of Na by mass in NaBr will be same for every amount .
Percentage of Na in 9.77 g NaBr is 22.34 % too .
Gram of Na = 
.
Hence , this is the required solution .
Answer:
287.30 g of FeCO₃
Solution:
The Balance Chemical Equation is as follow,
FeCl₂ + Na₂CO₃ → FeCO₃ + 2 NaCl
Step 1: Calculate Mass of FeCl₂ as,
Molarity = Moles ÷ Volume
Solving for Moles,
Moles = Molarity × Volume
Putting Values,
Moles = 2 mol.L⁻¹ × 1.24 L
Moles = 2.48 mol
Also,
Moles = Mass ÷ M.Mass
Solving for Mass,
Mass = Moles × M.Mass
Putting Values,
Mass = 2.48 mol × 126.75 g.mol⁻¹
Mass = 314.34 g of FeCl₂
Step 2: Calculate Mass of FeCO₃ formed as,
According to equation,
126.75 g (1 mole) FeCl₂ produces = 115.85 g (1 mole) FeCO₃
So,
314.34 g of FeCl₂ will produce = X g of FeCO₃
Solving for X,
X = (314.34 g × 115.85 g) ÷ 126.75 g
X = 287.30 g of FeCO₃
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<u>Answer:</u> The volume of acid should be less than 100 mL for a solution to have acidic pH
<u>Explanation:</u>
To calculate the volume of acid needed to neutralize, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is HCl
are the n-factor, molarity and volume of base which is NaOH
We are given:
Putting values in above equation, we get:
For a solution to be acidic in nature, the pH should be less than the volume of acid needed to neutralize.
Hence, the volume of acid should be less than 100 mL for a solution to have acidic pH
Answer is: pH of solution is 5,17.
Kb(NH₃) = 1,8·10⁻⁵.
c(NH₄Cl) = 0,084 M = 0,084 mol/L.
Chemical reaction: NH₄⁺ + H₂O → NH₃ + H₃O⁺.
Ka · Kb = 10⁻¹⁴.
Ka(NH₄⁺) = 10⁻¹⁴ ÷ 1,8·10⁻⁵.
Ka(NH₄⁺) = 5,55·10⁻¹⁰.
[H₃O⁺] = [NH₃] = x.
Ka(NH₄⁺) = [H₃O⁺] · [NH₃] ÷ [NH₄⁺].
5,55·10⁻¹⁰ = x² ÷ (0,084 M - x).
Solve quadratic equation: x = [H₃O⁺] = 6,8·10⁻⁶ M.
pH = -log[H₃O⁺].
pH = -log(6,8·10⁻⁶ M) = 5,17.
