Answer:
The pOH of HNO₃ solution that ha OH⁻ concentration 9.50 ×10⁻⁹M is 8.
Explanation:
Given data:
[OH⁻] = 9.50 ×10⁻⁹M
pOH = ?
Solution:
pOH = -log[OH⁻]
Now we will put the value of OH⁻ concentration.
pOH = -log[9.50 ×10⁻⁹M]
pOH = 8
Thus the pOH of HNO₃ solution that ha OH⁻ concentration 9.50 ×10⁻⁹M is 8.
I don’t really know but I’ll get someone to help u
4km I believe is the answer
<span>We can use the heat
equation,
Q = mcΔT </span>
<span>Where Q is
the amount of energy transferred (J), m is the mass of the
substance (kg), c is the specific heat (J g</span>⁻¹ °C⁻<span>¹) and ΔT is the temperature
difference (°C).</span>
Density = mass / volume
The density of water = 0.997 g/mL
<span>Hence mass of 1.25 L (1250 mL) of water = 0.997 g/mL x 1250 mL</span>
<span> = 1246.25 g</span>
Specific heat capacity of water = 4.186 J<span>/ g °C.</span>
Let's assume that there is no heat loss to the surrounding and the final temperature is T.
By applying the equation,
5430 J = 1246.25 g x 4.186 J/ g °C x (T - 23) °C
(T - 23) °C = 5430 J / 1246.25 g x 4.186 J/ g °C
(T - 23) °C = 1.04 °C
T = 1.04 °C + 23 °C
T = 24.04 °C
Hence, the final temperature of the water is 24.04 °C.
