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nikklg [1K]
3 years ago
5

Measurements are commonly recorded as

Chemistry
1 answer:
kaheart [24]3 years ago
5 0

Kilometers, Meters and centimeters if metric

Feet, inches, yards and miles if customary ( u.s.)

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What is the pOH of a solution of HNO3 that has [OH-] = 9.50 10-9 M?
bekas [8.4K]

Answer:

The pOH of HNO₃ solution that ha OH⁻ concentration 9.50 ×10⁻⁹M is 8.

Explanation:

Given data:

[OH⁻] = 9.50 ×10⁻⁹M

pOH = ?

Solution:

pOH = -log[OH⁻]

Now we will put the value of OH⁻ concentration.

pOH = -log[9.50 ×10⁻⁹M]

pOH = 8

Thus the pOH of HNO₃ solution that ha OH⁻ concentration 9.50 ×10⁻⁹M is 8.

6 0
3 years ago
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How would the moon appear from earth if the moon did not rotate?
ivolga24 [154]
If the moon did not rotate we would see all hemispheres of the moon as it revolves around the Earth and not just the phases.
4 0
3 years ago
Draw a lewis structure for hnc and assign the non-zero formal charges to each atom. draw the lewis structures with the formal ch
Yuliya22 [10]
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3 0
3 years ago
Calculate δg∘rxn and e∘cell for a redox reaction with n = 3 that has an equilibrium constant of k = 4.4×10−2. you may want to re
lapo4ka [179]
a) First, to get ΔG°rxn we have to use this formula when:

ΔG° = - RT ㏑ K 

when ΔG° is Gibbs free energy 

and R is the constant = 8.314 J/mol K

and T is the temperature in Kelvin = 25 °C+ 273 =  298 K 

and when K = 4.4 x 10^-2

so, by substitution:

ΔG°= - 8.314 * 298 *㏑(4.4 x 10^-2)

      = -7739 J  = -7.7 KJ


b) then, to get E
° cell for a redox reaction we have to use this formula:

ΔE° Cell = (RT / nF) ㏑K

when R is a constant = 8.314 J/molK

and T is the temperature in Kelvin = 25°C + 273 = 298 K

and n = no.of moles of e- from the balanced redox reaction= 3

and F is Faraday constant = 96485 C/mol

and K = 4.4 x 10^-2

so, by substitution:

∴ ΔE° cell = (8.314 * 298 / 3* 96485) *㏑(4.4 x 10^-2)

              = - 2.7 x 10^-2 V
  
8 0
3 years ago
A certain gas is present in a 10.0 LL cylinder at 4.0 atmatm pressure. If the pressure is increased to 8.0 atmatm the volume of
ICE Princess25 [194]

Answer:

The gas obeys Boyle’s law and the value of k_i\&k_f both are equal to 40.0 atm L.

Explanation:

Initial volume of the gas = V_1=10.0 L

Initial pressure of the gas = P_1=4.0 atm

Final volume of the gas = V_2=5.0 L

Final pressure of the gas = P_2=8.0 atm

This law states that pressure is inversely proportional to the volume of the gas at constant temperature.  

PV=k

The equation given by this law is:

P_1V_1=P_2V_2

P_1\propto \frac{1}{V_1}

P_1V_1=k_i

k_i=4.0 atm\times 10.0 L = 40.0 atm L

P_2\propto \frac{1}{V_2}

P_V_2=k_f

k_f=8.0 atm\times 5.0 L = 40.0 atm L

k_i=k_f=40.0 atm L

The gas in the cylinder is obeying Boyle's law.

The gas obeys Boyle’s law and the value of k_i\&k_f both are equal to 40.0 atm L.

6 0
3 years ago
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