1)we need a balanced equation: CuSO₄ + Zn ---> ZnSO₄ + Cu
2) we need to convert the grams of CuSO₄ to moles using the molar mass.
molar mass CuSO₄= 63.5 + 32.0 + (4 x 16.0)= 160 g/mol
3) convert moles of CuSO₄ to moles of Cu
4) convert moles of Cu to grams using it's molar mass.
molar mass Cu= 63.5 g/mol
I did it step-by-step as the explanation but you can do all of this in one step.
Answer is: 8568.71 of baking soda.
Balanced chemical reaction: H₂SO₄ + 2NaHCO₃ → Na₂SO₄ + 2CO₂ + 2H₂O.
V(H₂SO₄) = 17 L; volume of the sulfuric acid.
c(H₂SO₄) = 3.0 M, molarity of sulfuric acid.
n(H₂SO₄) = V(H₂SO₄) · c(H₂SO₄).
n(H₂SO₄) = 17 L · 3 mol/L.
n(H₂SO₄) = 51 mol; amount of sulfuric acid.
From balanced chemical reaction: n(H₂SO₄) : n(NaHCO₃) = 1 :2.
n(NaHCO₃) = 2 · 51 mol.
n(NaHCO₃) = 102 mol, amount of baking soda.
m(NaHCO₃) = n(NaHCO₃) · M(NaHCO₃).
m(NaHCO₃) = 102 mol · 84.007 g/mol.
m(NaHCO₃) = 8568.714 g; mass of baking soda.
Full Question:
A flask containing 420 Ml of 0.450 M HBr was accidentally knocked to the floor.?
How many grams of K2CO3 would you need to put on the spill to neutralize the acid according to the following equation?
2HBr(aq)+K2CO3(aq) ---> 2KBr(aq) + CO1(g) + H2O(l)
Answer:
13.1 g K2CO3 required to neutralize spill
Explanation:
2HBr(aq) + K2CO3(aq) → 2KBr(aq) + CO2(g) + H2O(l)
Number of moles = Volume * Molar Concentration
moles HBr= 0.42L x .45 M= 0.189 moles HBr
From the stoichiometry of the reaction;
1 mole of K2CO3 reacts with 2 moles of HBr
1 mole = 2 mole
x mole = 0.189
x = 0.189 / 2 = 0.0945 moles
Mass = Number of moles * Molar mass
Mass = 0.0945 * 138.21 = 13.1 g
Answer:
4600s
Explanation:
For a first order reaction the rate of reaction just depends on the concentration of one specie [B] and it’s expressed as:
![-\frac{d[B]}{dt}=k[B] - - - -\frac{d[B]}{[B]}=k*dt](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BB%5D%7D%7Bdt%7D%3Dk%5BB%5D%20-%20-%20-%20%20-%5Cfrac%7Bd%5BB%5D%7D%7B%5BB%5D%7D%3Dk%2Adt)
If we have an ideal gas in an isothermal (T=constant) and isocoric (v=constant) process.
PV=nRT we can say that P = n so we can express the reaction order as a function of the Partial pressure of one component.
![-\frac{d[P(B)]}{P(B)}=k*dt](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BP%28B%29%5D%7D%7BP%28B%29%7D%3Dk%2Adt)
![-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=k*dt](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BP%28N_%7B2%7DO_%7B5%7D%29%5D%7D%7BP%28N_%7B2%7DO_%7B5%7D%29%7D%3Dk%2Adt)
Integrating we get:
![\int\limits^p \,-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=\int\limits^ t k*dt](https://tex.z-dn.net/?f=%5Cint%5Climits%5Ep%20%5C%2C-%5Cfrac%7Bd%5BP%28N_%7B2%7DO_%7B5%7D%29%5D%7D%7BP%28N_%7B2%7DO_%7B5%7D%29%7D%3D%5Cint%5Climits%5E%20t%20k%2Adt)
![-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])=k(t_{2}-t_{1})](https://tex.z-dn.net/?f=-%28ln%5BP%28N_%7B2%7DO_%7B5%7D%29%5D-ln%5BP%28N_%7B2%7DO_%7B5%7D%29_%7Bo%7D%29%5D%29%3Dk%28t_%7B2%7D-t_%7B1%7D%29)
Clearing for t2:
![\frac{-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])}{k}+t_{1}=t_{2}](https://tex.z-dn.net/?f=%5Cfrac%7B-%28ln%5BP%28N_%7B2%7DO_%7B5%7D%29%5D-ln%5BP%28N_%7B2%7DO_%7B5%7D%29_%7Bo%7D%29%5D%29%7D%7Bk%7D%2Bt_%7B1%7D%3Dt_%7B2%7D)
![ln[P(N_{2}O_{5})]=ln(650)=6.4769](https://tex.z-dn.net/?f=ln%5BP%28N_%7B2%7DO_%7B5%7D%29%5D%3Dln%28650%29%3D6.4769)
![ln[P(N_{2}O_{5})_{o}]=ln(760)=6.6333](https://tex.z-dn.net/?f=ln%5BP%28N_%7B2%7DO_%7B5%7D%29_%7Bo%7D%5D%3Dln%28760%29%3D6.6333)
A micelle refers to an aggregate of a surfactant molecule, which a dispersed in a liquied colloid. In an aqueous liquid, a micelle is arranged in such a way that its hydrophillic head region will be in contact with the surrounding solvents while the hydrophobic tail region will be embedded in its center.
The three dimensional shape of a micelle is A SPHERE.
