Chemical reaction: Ba(NO₃)₂ + H₂SO₄ → BaSO₄ + 2HNO₃.
V(H₂SO₄) = 250 mL ÷ 1000 mL/L = 0,25 L.
m(BaSO₄) = 0,55 g.
n(BaSO₄) = m(BaSO₄) ÷ M(BaSO₄).
n(BaSO₄) = 0,55 g ÷ 233,38 g/mol.
n(BaSO₄) = 0,00235 mol.
From chemical reaction: n(BaSO₄) : n(Ba(NO₃)₂) = 1 : 1.
n(Ba(NO₃)₂) = 0,00235 mol.
c(Ba(NO₃)₂) = n(Ba(NO₃)₂) ÷ V.
c(Ba(NO₃)₂) = 0,00235 mol ÷ 0,25 L.
c(Ba(NO₃)₂) = 0,0095 mol/L.
Explanation:
The given cell reaction is as follows.
Hence, reactions taking place at the cathode and anode are as follows.
At anode ; Oxidation-half reaction : 
...... (1)
At cathode; Reduction-half reaction : 
....... (2)
Hence, balance the half reactions by multiplying equation (1) by 2 and equation (2) by 3.
Therefore, net cell reaction is as follows.
Net reaction: 
Thus, we can conclude that the overall cell reaction is as follows.
1 mmol --------------------- 1000 <span>µmol
( mmol ) -------------------- </span> 38231 µmol
mmol = 38231*1 / 1000
mmol = 38231/ 1000
=> 38.231 mmol
Answer:
Explanation:
What we need to do here is to determine the ratios by using the Rydberg equation starting with the transition to n1 = 1, 2,3, etc and see which one fits the data. Remember the question states that they are series and the wavelengths will be for increasing energy levels.
1/λ = Rh x ( 1/n₁² - 1/n₂²)
Lyman series ( n₁=1 and n₂= 2,3 etc) for the first two lines, the ratios will be:
1/λ₁ /1/λ₂ =(1/1 -1/ 2²) / (1/1 -1/ 3²) ⇒ 0.84 ≠ 0.74 (the first ratio)
For Balmer series n₁ = 2 and n₂ = 3,4,5, etc
1/λ₁ /1/λ₂ =(1/4 -1/3²) / (1/4 -1/4²) ⇒ 0.741 = 0.741 (match!)
Lets use the third line to check our answer:
1/λ₁ /1/λ₂ =(1/4 -1/3²) / (1/4 -1/5²) = 0.66
Answer:
The same number of molecules, 6.0 × 10²³ molecules.
Explanation:
The amount of any given gas that can be stored in a container depends on the <u>temperature, pressure and volume </u>of the container. It does not depend on the nature (or identity) of the gas.
So if a 10-liter flask contains 6.0 × 10²³ molecules of hydrogen gas, it will contain the same amount of molecules of any other gas when temperature and pressure remain constant.
