First step is to balance the reaction equation. Hence we get
P4 + 5 O2 => 2 P2O5
Second, we calculate the amounts we start with
P4: 112 g = 112 g/ 124 g/mol – 0.903 mol
O2: 112 g = 112 g / 32 g/mol = 3.5 mol
Lastly, we calculate the amount of P2O5 produced.
2.5 mol of O2 will react with 0.7 mol of P2O5 to produce 1.4
mol of P2O5.
This is 1.4 * (31*2 + 16*5) = 198.8 g
Answer: The rate of disappearance of
is 
Explanation:
The given chemical reaction is:

The rate of the reaction for disappearance of
is given as:
![\text{Rate of disappearance of }NO_2=-\frac{\Delta [NO_2]}{\Delta t}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20disappearance%20of%20%7DNO_2%3D-%5Cfrac%7B%5CDelta%20%5BNO_2%5D%7D%7B%5CDelta%20t%7D)
Or,

where,
= final concentration of
= 0.00650 M
= initial concentration of
= 0.0100 M
= final time = 100 minutes
= initial time = 0 minutes
Putting values in above equation, we get:

Hence, the rate of disappearance of
is 
Answer:
a. 0.73
Explanation:
Given data
- Change in the temperature (ΔT): 3.0°C
- Mass of the metal (m): 17.5 g
- Specific heat of the metal (c): ?
We can determine the specific heat of the metal using the following expression.
Q = c × m × ΔT
c = Q / m × ΔT
c = 38.5 J / 17.5 g × 3.0°C
c = 0.73 J/g.°C
Answer:
[OH⁻] = 3.54 × 10⁻¹⁰ M
Solution:
pOH is related to [OH⁻] as,
pOH = - log [OH⁻]
Putting value of pOH,
9.45 = -log [OH⁻]
Solving for [OH⁻],
[OH⁻] = 10⁻⁹·⁴⁵ ∴ 10 = Antilog
[OH⁻] = 3.54 × 10⁻¹⁰ M
Mass of sugar=Mass of solution - mass of water
=1.07X355 - 355
= 24.85g