Explanation:
The given data is as follows.
Current (I) = 3.50 amp, Mass deposited = 100.0 g
Molar mass of Cr = 52 g
It is known that 1 faraday of electricity will deposit 1 mole of chromium. As 1 faraday means 96500 C and 1 mole of Cr means 52 g.
Therefore, 100 g of Cr will be deposited by "z" grams of electricity.
z = 
= 185576.9 C
As we know that, Q = I × t
Hence, putting the given values into the above equation as follows.
Q = I × t
185576.9 C = 
t = 53021.9 sec
Thus, we can conclude that 100 g of Cr will be deposited in 53021.9 sec.
You have to use the equation PV=nRT.
P=pressure (in this case 1.89x10^3 kPa which equals 18.35677 atm)
1V=volume (in this case 685L)
n=moles (in this case the unknown)
R=gas constant (0.08206 (L atm)/(mol K))
T=temperature (in this case 621 K)
with the given information you can rewrite the ideal gas law equation as n=PV/RT.
n=(18.35677atm x 685L)/(0.08206atmL/molK x 621K)
n=246.8 moles
Fix ur transition, it sounds choppy
The balanced equation :
2NaHCO₃⇒CO₂ + Na₂CO₃+H₂O
<h3>Further explanation</h3>
Given
Reaction
NaHCO(s) --> _CO2+_NaCO(s)+_H2O
Required
The balanced equation
Solution
Maybe the equation should be like this :
NaHCO₃⇒CO₂ + Na₂CO₃+H₂O
Give a coefficient
NaHCO₃⇒aCO₂ + bNa₂CO₃+cH₂O
Make an equation
Na, left=1, right=2b⇒2b=1⇒b=1/2
H, left=1, right=2c⇒2c=1⇒c=1/2
C, left=1, right=a+b⇒a+b=1⇒a+1/2=1⇒a=1/2
The equation becomes :
NaHCO₃⇒1/2CO₂ +1/2Na₂CO₃+1/2H₂O x2
2NaHCO₃⇒CO₂ + Na₂CO₃+H₂O
Answer:
Option A. It has stayed the same.
Explanation:
To answer the question given above, we assumed:
Initial volume (V₁) = V
Initial temperature (T₁) = T
Initial pressure (P₁) = P
From the question given above, the following data were:
Final volume (V₂) = 2V
Final temperature (T₂) = 2T
Final pressure (P₂) =?
The final pressure of the gas can be obtained as follow:
P₁V₁/T₁ = P₂V₂/T₂
PV/T = P₂ × 2V / 2T
Cross multiply
P₂ × 2V × T = PV × 2T
Divide both side by 2V × T
P₂ = PV × 2T / 2V × T
P₂ = P
Thus, the final pressure is the same as the initial pressure.
Option A gives the correct answer to the question.
