I just know the ph is between 7 and 8
Answer:
1328 K or 1055 °C
Explanation:
To find the new temperature, you need to use the Combined Gas Law:
P₁V₁ / T₁ = P₂V₂ / T₂
In this equation, "P₁", "V₁", and "T₁" represent the initial pressure, volume, and temperature. "P₂", "V₂", and "T₂" represent the final pressure, volume, and temperature. After converting the temperature (T₁) from Celsius to Kelvin, you can plug the values into the equation and simplify.
P₁ = 1.0 atm P₂ = 0.90 atm
V₁ = 200 L V₂ = 1000 L
T₁ = 22 °C + 273.15 = 295.15 K T₂ = ? K
P₁V₁ / T₁ = P₂V₂ / T₂
(1.0 atm)(200 L) / 295.15 K = (0.90 atm)(1000 L) / T₂
0.6776 = (0.90 atm)(1000 L) / T₂
(0.6776) x T₂ = (0.90 atm)(1000 L)
(0.6776) x T₂ = 900
T₂ = 1328 K
T₂ = 1055 °C
The water cycle does not ensure that we have water. We must HAVE water to cause water cycle.
Evaporation, then condensation, and finally precipitation.
This would happen in order.
Hope this helps!
Answer:
d = 0.93 g/cm³
Explanation:
Given data:
Mass of object = 28 g
Volume of object = 3cm×2cm×5cm
density of object = ?
Solution:
Volume of object = 3cm × 2cm ×5cm
Volume of object = 30 cm³
Density of object:
d = m/v
by putting values,
d = 28 g/ 30 cm³
d = 0.93 g/cm³
Answer:
0.2042 M is the original concentration of
(aq) in the titrating solution.
Explanation:
Mass of
= 0.217 g
Moles of ![As_2O_3=\frac{0.217 g}{198 g/mol}=0.001096 mol](https://tex.z-dn.net/?f=As_2O_3%3D%5Cfrac%7B0.217%20g%7D%7B198%20g%2Fmol%7D%3D0.001096%20mol)
1 mole of
have 2 mole of As and 1 mole of
have 1 mole of As.
So, from 1 mole of
we will have 2 moles of ![H_3AsO_3](https://tex.z-dn.net/?f=H_3AsO_3)
Then from 0.001096 mol of
:
of ![H_3AsO_3](https://tex.z-dn.net/?f=H_3AsO_3)
![2Ce^{4+}(aq)+H_3AsO_3(aq)+3H_2O(l)\rightarrow 2Ce^{3+}(aq)+H_3AsO_4(aq)+2H^+(aq)](https://tex.z-dn.net/?f=2Ce%5E%7B4%2B%7D%28aq%29%2BH_3AsO_3%28aq%29%2B3H_2O%28l%29%5Crightarrow%202Ce%5E%7B3%2B%7D%28aq%29%2BH_3AsO_4%28aq%29%2B2H%5E%2B%28aq%29)
According to reaction, 1 mole of
reacts with 2 mole of cerium (IV) ions,then 0.002192 mol of
of cerium (IV) ions.
Volume of the acidic cerium{IV) sulfate = 21.47 ml =0.02147 L
1 mL = 0.001 L
![concentration = \frac{Moles}{Volume(L)}](https://tex.z-dn.net/?f=concentration%20%3D%20%5Cfrac%7BMoles%7D%7BVolume%28L%29%7D)
![[Ce^{4+}]=\frac{0.004384 mol}{0.02147 L}=0.2042 M](https://tex.z-dn.net/?f=%5BCe%5E%7B4%2B%7D%5D%3D%5Cfrac%7B0.004384%20mol%7D%7B0.02147%20L%7D%3D0.2042%20M)
0.2042 M is the original concentration of
(aq) in the titrating solution.