78.4 L volume of container is required to hold 3.2 moles of gas.
Explanation:
- STP is defined as the standard temperature and pressure of a gas in room temperature conditions. At STP, one mole of the gas which has Avogadro's number of molecules in it will occupy a volume of 22.4 L.
- So, one mole of a substance or gas will occupy a volume of 22.4 L then the volume of the container needed for 3.2 moles of gas is calculated by multiplying 22.4 L, standard volume with the moles of the gas 3.2 moles.
- Hence, the answer would be 78.4 L.
Answer:
9.15 atm
Explanation:
Ideal gas equation of state PV=nRT
P in hPa, V in L, n in mol, R is a constant which is 83.1 hpa*L/mol*k, T in kelvin.
Plug in all the number, and we will get:
P*6.21=2.02*83.1*343
P =9271.6(in hpa)=9.15 atm
Answer:
C
Explanation:
'Ordered Arrangement' basically means that it is a solid at room temperature. Room temperature is approximately 15-20C so we are looking for melting and boiling points that are above room temperature so it hasn't/can't melt or boil at room temperature and would therefore be solid. Option C is the only one where both points are temperatures above room temperature therefore option C is the only one where the substance would be in an 'ordered arrangement' at room temperature.
Hope this helped!
<span>C. polar bonds and asymmetrical structure
If the molecule contains polar bonds but it has a symmetrical structure, the polar bonds will cancel each other out so the overall molecule will be non-polar.
On the other hand, if the molecule contains polar bonds but has an asymmetrical structure, then the polar bonds won't cancel each other out, so the overall molecule ends up being polar.
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The formula for solving current given with resistance and power source or voltage is shown below:
I = V/R
When two 5 ohms resistors are in series, we have:
I = 9 volts / (5+5 ohms)
I = 0.9 amperes
When it is being added with another 7.5 resistors, we have:
I = 9 volts / (5+5+7.5 ohms)
I = 0.529 ampere
The answer to the question is the letter "D. decrease; 0.51 amps".
