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viva [34]
3 years ago
6

What is the [oh−] of a solution if its ph = 8.78? answer in units of m?

Chemistry
1 answer:
dexar [7]3 years ago
7 0
PH = -log([H+])
[H+] = 10^(-pH)

[H+] = 10^(-8.78) = 1.65*10^-9

[H+][OH-] = Kw
Kw = 1.0*10^-14 at 25 degrees celsius.

[OH-] = Kw/[H+] = (1.0*10^-14)/(1.65*10^-9) = 6.06*10^-6

The concentration of OH- ions is 6.1*10^-6 M.
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Heat & pressure. hope this helps
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What variable represents specific heat in the equation Q = mcAT?
Elan Coil [88]

Answer: A. The variable c

============================================

Explanation:

Q = heat transferred

m = mass

c = specific heat

\Delta T = delta T = change in temperature

4 0
3 years ago
Why is the air in a jet aircraft flying at high altitudes pressurized? *
saw5 [17]

Answer:

Aircraft cabins are therefore pressurized to maintained a similar pressure as that experienced at sea level to ensure normal breathing of passengers.

Explanation:

-Air becomes increasingly thinner with increasing altitudes.

-As such, oxygen becomes limited at higher altitudes and makes it difficult or almost impossible to breath  a condition called hypoxia.

-Aircraft cabins are therefore pressurized to maintained a similar pressure as that experienced at sea level to ensure normal breathing of passengers.

6 0
3 years ago
Realice las siguientes conversiones: a) 72°F a °C, b) 213.8°C a °F, c)180°C a K, d) 315K a °F, e) 1750°F a K, f) 0K a °F.
bonufazy [111]

Answer:

a) 72 °F= 22.22 °C

b)  213.8 °C=  416.84°F

c) 180 °C= 453.15 °K

d) 315 °K=  107.33 °F

e) 1750 °F= 1227.594 °K

f) 0 °K=  -459.67 °F

Explanation:

Para realizar el intercambio de unidades debes tener en cuenta las siguientes conversiones:

  • Fahrenheit a Celsius: C=\frac{F-32}{1.8}
  • Celsius a Fahrenheit: °F= °C*1.8 + 32
  • Celsius a Kelvin: °K= °C + 273.15
  • Kelvin a Fahrenheit: F= (K -273.15)*1.8 + 32
  • Fahrenheit a Kelvin:K=\frac{F-32}{1.8} + 273.15

Entonces se obtiene:

a) 72 °F= \frac{72-32}{1.8}=22.22 °C

b)  213.8 °C= 213.8*1.8 + 32= 416.84°F

c) 180 °C= 180°C + 273.15= 453.15 °K

d) 315 °K= (315 -273.15)*1.8 + 32= 107.33 °F

e) 1750 °F= \frac{1750-32}{1.8} + 273.15= 1227.594 °K

f) 0 °K= (0 -273.15)*1.8 + 32= -459.67 °F

7 0
2 years ago
a rectangular piece of aluminum foil measures 13.72 cm x 8.63 cm and has a mass of 3.1 g. Find how thick it is
Leona [35]

The two dimensions of aluminum foil are given 13.72 cm and 8.63 cm respectively with mass  3.1 g.

The density of aluminum is 2.7 g/cm^{3}. It is defined as mass per unit volume thus, volume of aluminum can be calculated as follows:

V=\frac{m}{d}

Putting the values.

V=\frac{3.1 g}{2.7 g/cm^{3}}=1.148 cm^{3}

The volume of cuboid is l\times b\times h, the length and breadth are given, height can be calculated as follows;

h=\frac{V}{l\times b}

Putting the values,

h=\frac{1.148 cm^{3}}{(13.72 cm)(8.63 cm)}=0.009695 cm

Or, approximately 9.7\times 10^{-3}cm

Therefore, thickness of aluminum foil is 9.7\times 10^{-3}cm

7 0
3 years ago
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