Answer:
Pop for two minutes in the microwave, and enjoy the perfect balance of buttery and salty taste in every bite. Try this delicious salty snack for backyard barbeques, movie nights, birthday parties or an office snack. Choose ACT II Butter Popcorn for the best value in popcorn.
Explanation:
Answer:indirect causes of disease include genetics, lifestyle, and environmental factors
Explanation:
Answer:
final volume = 10.5 Liters N₂(g) at 21°C and 823Torr*
Explanation:
*Note=>No specified mass value of N₂(g) is defined in the problem. Therefore for a starting point, the gas sample is assumed to be 1.00 mole N₂(g) at STP conditions 22.4L
Determine volume of N₂(g) at 21°C(=294K) and 823 Torr (= 2.286 Atm).
Start with Volume of N₂(g) at 0°C and 1 Atm pressure => 22.4L and adjust to final volume of N₂(g) based upon 21°C(=294K) and 823 Torr (= 2.286 Atm).
V(final) = 22.4L(294K/273K)(360 Torr/823 Torr) = 22.4L(294/273)(360/823) = 10.55 Liters final volume.
Note: The volume of 1 mole (assumed) of any gas at STP (0°C/1 Atm) is 22.4 Liters. To convert to non-STP conditions, convert temperature and pressure factors (changes) that reflect what happens when the gas is expanded or decreased; but, these adjustments are taken independently for each variable of interest. The following notes explain.
For the increase in temperature from 0°C(=273K) to 21°C(=294K) one must apply a temperature ratio that will increase volume. That is, the change in volume due to the temperature change is 294K/273K. If a 273K/294K ratio were used the volume would have decreased. Not so for heating a sample of gas.
For the increase in pressure one should expect a decrease in volume. Therefore apply a pressure ration that will effectively decrease the volume of the gas. That is, to decrease a 22.4L sample at STP multiply the standard volume by a ratio of pressures that will decrease 22.4L to a smaller volume. That is, V(final by pressure effects) multiply by 360Torr/823Torr to decrease the STP VOLUME (22.4L) to the new non-standard volume. If 823Torr/360Torr were used, the final volume would not be smaller, but larger. Such is the physical effect of an increasing pressure change.
The x number in ^x indicates the number of electronics in an obit, e.g. 1s^2 means there are 2 electrons in the 1s orbit.
Total number of electrons in A = 2 + 2 + 6 +2 + 6 + 2 + 2
= 22
VALENCE electrons are those in the outer most shell.
In A, there are 4s^2 + 4p^2: 2 + 2 = 4
Answer should be C instead. So sorry for the mistake earlier!
Answer:
pH = 8.477
Explanation:
∴ ni 0.3 - -
nf - 0.3 0.3
∴ ni 0.50 - 0.3
nf 0.5 - X X 0.3 + X
∴ Ka = 2.0 E-9 = ([H3O+]*[OBr-]) / [HOBr]
⇒ Ka = 2.0 E-9 = ((X/1L)*(0.3 + X)/1 L) / ((0.5 - X)/1L)
⇒ 2.0 E-9 = ( 0.3X + X² ) / (0.5 - X)
⇒ X² + 0.3X - 1 E-9 = 0
⇒ X = 4.333 E-9 M
according Henderson-Hauselbach:
- pH = pk + Log [A-] / [HA]
∴ [OBr-] = 0.3 mol/ 1 L + 4.333 E-9 M = 0.300 M
∴ [HOBr] = 0.5 mol / 1 L - 4.33 E-9 M = 0.500 M
∴ pKa = - log Ka = - Log ( 2.0 E-9 ) = 8.6989
⇒ pH = 8.6989 + Log ( 0.300/0.500 )
⇒ pH = 8.477
