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4 years ago

What determines the strength of a base?

2 answers:

8 0

The higher the dissociation constant the stronger the acid or base. Since electrolytes are created as ions are freed into solution there is a relationship between the strength of an acid, a base, and the electrolyte it produces. Acids and bases are measured using the pH scale.

Anvisha [2.4K]4 years ago

5 0

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You raise a bucket of water from the bottom of a deep well. if your power output is 138 w , and the mass of the bucket and the w

inn [45]

Maybe 34.5

I'm not quite sure

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4 years ago

For copper, ρ = 8.93 g/cm3 and M = 63.5 g/mol. Assuming one free electron per copper atom, what is the drift velocity of electro

viktelen [127]

Answer:

$V_d$ = 1.75 × 10⁻⁴ m/s

Explanation:

Given:

Density of copper, ρ = 8.93 g/cm³

mass, M = 63.5 g/mol

Radius of wire = 0.625 mm

Current, I = 3A

Area of the wire, $A = \frac{\pi d^2}{4}$ = $A = \frac{\pi 0.625^2}{4}$

Now,

The current density, J is given as

$J=\frac{I}{A}=\frac{3}{ \frac{\pi 0.625^2}{4}}$ = 2444619.925 A/mm²

now, the electron density, $n = \frac{\rho}{M}N_A$

where,

$N_A$ =Avogadro's Number

$n = \frac{8.93}{63.5}(6.2\times 10^{23})=8.719\times 10^{28}\ electrons/m^3$

Now,

the drift velocity, $V_d$

$V_d=\frac{J}{ne}$

where,

e = charge on electron = 1.6 × 10⁻¹⁹ C

thus,

$V_d=\frac{2444619.925}{8.719\times 10^{28}\times (1.6\times 10^{-19})e}$ = 1.75 × 10⁻⁴ m/s

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3 years ago

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Peter left town A at 13:30 and travelled towards town B at an average speed of 40mph. At 13:45 Philip left town A for town B at

skad [1K]

Answer:

fbddb

Explanation:

dsbse

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3 years ago

Astronomers have found water on mars.

vodomira [7]

The thin atmosphere of Mars is thought to be due to the planet's lack of a magnetic field, which has allowed the Solar wind to blow away much of the gas the planet once had. Venus, despite still having a thick atmosphere of CO2, surprisingly has a similar problem

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2 years ago

The kinetic energy of a ball with a mass of 0.5 kg and a velocity of 10 m/s is j. (formula: )

andreyandreev [35.5K]

1/2 m V-squared = j. j = 25 joules.

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3 years ago

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