Answer:
Volume of water that must be added is 1.10 L
Explanation:
pH measures the acidity or the alkalinity of a substance
It is given by;
pH = -log[H+]
Using this we can find the concentration of H+ ions in the acid
pH = 2 = -log[H+]
Therefore;
[H+] = 10^-2
= 0.01 M
But, since 1 mole HNO₃ ionizes to give 1 mole of H+, then the concentration of HNO₃ is equal to the concentration of H+ ([HNO₃] = [H+])
Therefore;
Initial [HNO₃] = 0.01 M
Initial volume of HNO₃ = 11.1 mL or 0.0111 L
We can then use dilution equation to find the final volume after dilution.
The final pH is 4
Therefore, [H+] = 10^-4
= 0.0001 M
Thus, the final concentration of HNO₃ is 0.0001 M
Using dilution equation;
M1V1 =M2V2
Thus; V2 = M1V1÷ M2
= (0.01 M× 0.0111 L)÷ 0.0001 M
= 1.11 L
This means the final total volume will 1.11 L or 1110 ml
Therefore; The volume of water added = 1110 ml - 11.1 ml
= 1098.9 ml or
= 1.0989 L
= 1.10 L(2 d.p.)
Hence, The volume of water that must be added is 1.10 L
To answer the question, we assume that the given compound is an ideal gas that at STP, one mole of the substance will occupy 22.4 L. From the given volume, we determine the number of moles of substance.
7.8 L / (22.4 L /mole) = 0.3482 moles of cfa
Then, we multiply this number of moles by the molar weight of cfa which is equal to 88 g/mol.
Multiplying,
weight = (0.3482 moles of cfa) x (88 g/mol) = <em>30.64 grams</em>
Answer:
40% of the ammonia will take 4.97x10^-5 s to react.
Explanation:
The rate is equal to:
R = k*[NH3]*[HOCl] = 5.1x10^6 * [NH3] * 2x10^-3 = 10200 s^-1 * [NH3]
R = k´ * [NH3]
k´ = 10200 s^-1
Because k´ is the psuedo first-order rate constant, we have the following:
b/(b-x) = 100/(100-40) ; 40% ammonia reacts
b/(b-x) = 1.67
log(b/(b-x)) = log(1.67)
log(b/(b-x)) = 0.22
the time will equal to:
t = (2.303/k) * log(b/(b-x)) = (2.303/10200) * (0.22) = 4.97x10^-5 s
In nature reactions of ordinary molecular hydrogen are slow since it's a diatomic molecule whose atoms are held together by very strong covalent bonds.The reaction rate of hydrogen varies depending on temperature and the properties of the reactants, for instance under high temperatures above 500°C hydrogen reacts vigorously and with fluorine it reacts explosively even under low temperatures
Answer:6.719Litres of Cl2 gas.
Explanation:According to eqn of rxn
2Na +Cl2=2NaCl
P=689torr=689/760=0.91atm
T=39°C+273=312K
according to stoichiometry of the reaction,1Moles of Cl2 gives 2moles of NaCl
But 28g of NaCl was given,we have to convert this to moles by using the relation, n=mass/MW
MW of NaCl=23+35.5=58.5g/mol
n=28g(mass given of NaCl)/58.5
n=0.479moles of NaCl
Going back to the reaction,
if 1moles of Cl2 produces 2moles of NaCl
x moles of Cl2 will give 0.479moles of NaCl.
x=0.479*1/2
x=0.239moles of Cl2.
To find the volume, we use ideal ggas eqn,PV=nRT
V=nRT/P
V=0.239*0.082*312/0.91
V=6.719Litres
