In the presence of nitric acid, UO 2+ undergoes a redox process. It is converted to UO 2 2+ and nitric oxide (NO) gas is produced according to the following unbalanced equation: H + ( a q ) + NO 3 − ( a q ) + UO 2+ ( a q ) → NO ( g ) + UO 2 2 + ( a q ) + H 2 O ( l ) If 2.55 × 10 2 mL NO( g ) is isolated at 29°C and 1.5 atm, what amount (moles) of UO 2+ was used in the reaction? (Hint: Balance the reaction by the oxidation states method.)
In the presence of nitric acid, UO 2+ undergoes a redox process. It is converted to UO 2 2+ and nitric oxide (NO) gas is produced according to the following unbalanced equation: H + ( a q ) + NO 3 − ( a q ) + UO 2+ ( a q ) → NO ( g ) + UO 2 2 + ( a q ) + H 2 O ( l ) If 2.55 × 10 2 mL NO( g ) is isolated at 29°C and 1.5 atm, what amount (moles) of UO 2+ was used in the reaction? (Hint: Balance the reaction by the oxidation states method.)
Solution Summary: The author explains the balance equation of a reaction, which is written according to law of conservation of mass.
In the presence of nitric acid, UO2+ undergoes a redox process. It is converted to UO22+ and nitric oxide (NO) gas is produced according to the following unbalanced equation:
H
+
(
a
q
)
+
NO
3
−
(
a
q
)
+
UO
2+
(
a
q
)
→
NO
(
g
)
+
UO
2
2 +
(
a
q
)
+
H
2
O
(
l
)
If 2.55 × 102 mL NO(g) is isolated at 29°C and 1.5 atm, what amount (moles) of UO2+ was used in the reaction? (Hint: Balance the reaction by the oxidation states method.)
Part A
2K(s)+Cl2(g)+2KCI(s)
Express your answer in grams to three significant figures.
Part B
2K(s)+Br2(1)→2KBr(s)
Express your answer in grams to three significant figures.
Part C
4Cr(s)+302(g)+2Cr2O3(s)
Express your answer in grams to three significant figures.
Part D
2Sr(s)+O2(g) 2SrO(s)
Express your answer in grams to three significant figures.
Thank you!
A solution contains 10-28 M TOTCO3 and is at pH 8.1. How much HCI (moles per liter of
solution) is required to titrate the solution to pH 7.0? (H2CO3: pKa1=6.35, pKa2=10.33)
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.