Answer is: <span>the coefficient of phosphoric acid is 12.
</span>Chemical reaction: P₄S₃ + NO₃⁻ + H⁺ → H₃PO₄ + SO₄⁻ + NO.
Reduction half reaction: NO₃⁻ + 4H⁺ + 3e⁻ → NO + 2H₂O /·38
Oxidation half reaction: P₄S₃ + 28H₂O → 4H₃PO₄ + 3SO₄²⁻ + 44H⁺ + 38e⁻ /·3.
38NO₃⁻ + 152H⁺ + 3P₄S₃ + 84H₂O → 38NO + 76H₂O + 12H₃PO₄ + 9SO₄²⁻ + 132H⁺.
Balnced chemical reaction:
3P₄S₃ + 38NO₃⁻ + 20H⁺ + 8H₂O → 12H₃PO₄ + 9SO₄²⁻ + 38NO.
The answer is: all true
<span>A. As the pressure of the gas increased, the volume of the gas decreased.
It is clear that if you compare the data on the left side. When the pressure increased the volume is decreased.
B. For all pairs of data of pressure and volume, P • V was appoximately the same.
The pressure is inversely related to the volume. You can take two data to prove it. Let use the first and second data
V * P= 1.03 * 50= 51.5
</span>V * P= <span>1.08 * 47.5= 51.3
C. For all pairs of data of pressure and volume, P • V mr001-1.jpg k for the same value k.
D. The regression equation was of the form V = kP–1 (which is the same as V = k/P).
The value of k can be expressed as k= P*V. If the equation is turned around, it could be expressed as V= k/P
The value of k is constant on different data, proved by the calculation on the second statement above. The value of k should be around 51.5
</span>
The implications are for this reader in this line is that the atom is belonging to some kind of system and it is good belonging to that system.
Hope this helps. :)
- Shelly O
You must react 5.3 g H_2 to produce 47 g H_2O.
<em>Step 1</em>. Calculate the <em>moles of H_2O</em>
Moles of H_2O = 47 g H_2O × (1 mol H_2O/18.02 g H_2O) =2.61 mol H_2O
<em>Step 2</em>. Calculate the <em>moles of H_2
</em>
Moles of H_2 = 2.61 mol H_2O × (2mol H_2/2 mol H_2O) = 2.61 mol H_2
<em>Step 3</em>. Calculate the <em>mass of H_2</em>
Mass of H_2 = 2.61 mol H_2 × (2.016 g H_2/1 mol H_2) = 5.3 g H_2