The RMS of O2 at 17 degrees is calculated as follows
RMs= ( 3RT/m)^1/2 where
R= ideal gas constant = 8.314
T= temperature= 17+273= 290 K
M= molar mass in KG = 32/1000= 0.032 Kg
Rms is therefore= sqrt (3x 8.314 x290/0.032 ) = sqrt( 226036.875
RMs=475.43
Answer:
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The compound potassium carbonate(K₂CO₃) is soluble in the water. The formulas for the ions that interact with the water is K⁺ and (CO₃)²⁻.
Dissociation of any compound is defined as the breaking of a compound into a simpler substance which is capable of recombining under different conditions. The compound potassium carbonate(K₂CO₃) is an ionic compound, and ionic compound is formed by a combination of cation and anion. When ionic compounds are dissolved in the water, it separates into the cations and anions.
Potassium carbonate(K₂CO₃) is a strong electrolyte. It means that it is completely dissociate in the water. Its dissociation in the water is shown as
K₂CO₃(aq)→2K⁺(aq)+(CO₃)²⁻(aq)
Therefore, the compound potassium carbonate forms K⁺ and (CO₃)²⁻ ions when interact with the water.
The question looks incomplete but the complete question is
The compound potassium carbonate is a strong electrolyte. Write the reaction when solid potassium carbonate is put into water
To know more about dissociation
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Answer:
1/3p0
Explanation:
The combined gas law:
P1V1/T1 = P2V2/T2, where P, V and T are Pressure, Volume, and Temperature. Temperature must always be in Kelvin. The subscriopts 1 and 2 are for initial (1) and final (2) conditions.
In this case, temperature is constant (adiabatically). V1 = 2.0L and V2 = 6.0L. I'll assume P1 = p0.
Rearrange the combined gas law to solve for final pressure, P2:
P1V1/T1 = P2V2/T2
P2 = P1*(V1/V2)*(T2/T1) [Note how I've arranged the volume and temoperature terms - as ratios. This helps us understand what the impact of raising or lowering one on the variables will do to the system].
No enter the data:
P2 = P1*(V1/V2)*(T2/T1): [Since T2 = T1, the (T2/T1) terms cancels to 1.]
P2 = p0*(2.0L/6.0L)*(1)
P2 = (1/3)p0
The final pressure is 1/3 the initial pressure.
Answer:
I don't think the question is complete
