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love history [14]
3 years ago
7

¿Cómo se relaciona la química en la salud?

Chemistry
1 answer:
mars1129 [50]3 years ago
7 0

salud: Medicina: La Química nos proporciona vacunas , antibióticos y todo tipo de medicamentos que nos curan y protegen de las enfermedades.

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What mass of Cl2, in grams is contained in a 10.0L tank at 27oC and 3.50 atm pressure ? a. 1.42 grams b. 142 grams c. 1.01 kg d.
Evgen [1.6K]

Answer:

the correct answer is A

Explanation:

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2 years ago
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Answer:

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2 years ago
The ph of a 0.0100 m solution of the sodium salt of a weak acid is 11.00. what is the ka of the acid?
Vitek1552 [10]
The answer is Ka = 1.00x10^-10.
Solution:
When given the pH value of the solution equal to 11, we can compute for pOH as
     pOH = 14 - pH = 14 - 11.00 = 3.00
We solve for the concentration of OH- using the equation
     [OH-] = 10^-pOH = 10^-3 = x

Considering the sodium salt NaA in water, we have the equation
     NaA → Na+ + A- 
hence, [A-] = 0.0100 M

Since HA is a weak acid, then A- must be the conjugate base and we can set up an ICE table for the reaction
                             A- + H2O ⇌ HA + OH-
     Initial             0.0100            0       0
     Change        -x                    +x     +x
     Equilibrium    0.0100-x         x       x

We can now calculate the Kb for A-:
     Kb = [HA][OH-] / [A-] 
           = x<span>²</span> / 0.0100-x
Approximating that x is negligible compared to 0.0100 simplifies the equation to
     Kb = (10^-3)² / 0.0100 = 0.000100 = 1.00x10^-4

We can finally calculate the Ka for HA from the Kb, since we know that Kw = Ka*Kb = 1.0 x 10^-14:
     Ka = Kw / Kb 
           = 1.00x10^-14 / 1.00x10^-4
           = 1.00x10^-10
7 0
3 years ago
How much heat, in joules, is required to warm a metal disc from 19 °C to 33 °C? The
zalisa [80]

Answer:

Q = 96.6 j

Explanation:

Given data:

Heat required = ?

Initial temperature = 19°C

Final temperature = 33°C

Mass of disc = 3.0 g

Specific heat capacity = 2.3 J/g.°C

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 33°C - 19°C

ΔT = 14°C

Q = 3.0 g×2.3 J/g.°C × 14°C

Q = 96.6 j

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2 years ago
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