Sugar increases the viscosity of water
hope that helps
pH=6.98
Explanation:
This is a very interesting question because it tests your understanding of what it means to have a dynamic equilibrium going on in solution.
As you know, pure water undergoes self-ionization to form hydronium ions, H3O+, and hydroxide anions, OH−.
2H2O(l]⇌H3O+(aq]+OH−(aq]→ very important!
At room temperature, the value of water's ionization constant, KW, is equal to 10−14. This means that you have
KW=[H3O+]⋅[OH−]=10−14
Since the concentrations of hydronium and hydroxide ions are equal for pure water, you will have
[H3O+]=√10−14=10−7M
The pH of pure water will thus be
pH=−log([H3O+])
pH=−log(10−7)=7
Now, let's assume that you're working with a 1.0-L solution of pure water and you add some 10
Answer:
87.5 mi/hr
Explanation:
Because a = Δv / Δt (a = vf - vi/ Δt), we need to find the acceleration first to know the change in velocity so we can determine the final velocity.
vf = 60 mi/hr
vi = 0 mi/hr
Δt = 8 secs
a = vf - vi/ Δt
= 60 mi/hr - 0 mi/hr/ 8 secs
= 60 mi/hr / 8 secs
= 7.5 mi/hr^2
Now that we know the acceleration of the car is 7. 5 mi/hr^2, we can substitute it in the acceleration formula to find the final velocity when the initial velocity is 50 mi/hr after 5 secs.
vi = 50 mi/ hr
Δt = 5 secs
a = 7.5 mi/ hr^2
a = vf - vi/ Δt
7.5 = vf - 50 mi/hr / 5 secs
37.5 = vf - 50
87.5 mi/ hr = vf
According to the law of conservation of mass, the amount of BARIUM present of the reactants is the same as the amount present in the products (the precipitate).
(11.21 g BaSO4) / (233.4 g/mol BaSO4) = 0.0480 mol BaSO4 and original barium salt
(10.0 g) / (0.0480 mol) = 208.3 g/mol
So it must have been BaCl2, because the molar mass of Barium is 137 which leave 71 grams left. Since Barium is a +2 charge, it means the atom next to it must be twice. Chlorine mass is 35, which twice is 71