The oxidation state of hydrogen gas is 0 and oxidation state of hydrogen cation is +1.
There’s an increase in oxidation number therefore it’s an oxidation reaction.
Oxidation reactions give out electrons. The masses and charges on both sides should be balanced
Half reaction is
H2 —> 2H+ +2e
<span>At standard temperature and pressure 22.4 l of an ideal gas would contain 1 mole. in order to find the change in moles we must look at the ideal gas law PV=nRT where P=Pressure V=volume n=Moles R= Gas constant T= Temperature. To simplify this equation we will be using the gas constant at .08206 L-atm/mol-K. We must first convert 100c to k which is 373.15. Then we can plug the values into our equation which gives us (2atm)(14.5 l)=(n)(.08206 L-atm/mol-K)(373.15). After some basic algebra we get the moles to equal roughly .95 which is .05 moles less than our original system.</span>
Answer:
No, the puddle was formed because of the sun, because if there was snow and it rained then it would have turned slippery or icy
Answer:
The partial pressure of NO2 = 0.152 atm
Explanation:
Step 1: Data given
Pressure NO2 = 0.500 atm
Total pressure at equilibrium = 0.674 atm
Step 2: The balanced equation
2NO2(g) → 2NO(g) + O2(g)
Step 3: The initial pressure
pNO2 = 0.500 atm
pNO = 0 atm
p O2 = 0 atm
Step 4: Calculate pressure at the equilibrium
For 2 moles NO2 we'll have 2 moles NO and 1 mol O2
pNO2 = 0.500 - 2x atm
pNO =2x atm
pO2 = xatm
The total pressure = p(total) = p(NO2) + p(NO) + p(O2)
p(total) = (0.500 - 2x) + 2x + x= 0.674 atm
0.500 + x = 0.674 atm
x = 0.174 atm
This means the partial pressure of NO2 = 0.500 - 2*0.174 = 0.152 atm
Answer is: volume of KBr is 357 mL.
c(KBr) = 0,716 M = 0,716 mol/L.
m(KBr) = 30,5 g.
n(KBr) = m(KBr) ÷ M(KBr).
n(KBr) = 30,5 g ÷ 119 g/mol.
n(KBr) = 0,256 mol.
V(KBr) = n(KBr) ÷ c(KBr).
V(KBr) = 0,256 mol ÷ 0,716 mol/L.
V(KBr) = 0,357 L · 1000 mL/L = 357 mL.
