Answer:
The original number of oxygen atoms was about 80 times larger than Avogadro's number. Since each sulfuric acid molecule contains 4 oxygen atoms, there are about 20 moles of sulfuric acid molecules.
<span>Answer: FALSE:
Explanation:
A reaction requires 22.4l of at STP. You have 32.0l of gas at 398k and 105.6 kpa.
1) STP stands for standard temperature and pressure.
2) Standard temperature is 0°C or 273.15 K
3) Standard pressure is 1 atm or 1013.25 kPa
4) use the ideal gas equation for both contidions
pV = n RT
=> n * R = pV /T
at STP n * R = 1031.25 kPa * 22.4 liter / 273.15 K = 84.5
at T = 398 K, p = 105.6 kPa, and V = 3.2.0 liter:
n * R = 105.6 kPa * 32.0 liter / 398 K = 8.49
Since R is a constant (the Universal Gases Constant), it is evident that the number of moles in the 32.0 liter of gas, at T = 398 K and P = 105.6 kPa is less than the number of moles of the 22.4 liter gas at STP.
There is not enough gas to carry out the reaction.
</span>
The answer is: image of the object.
Antoine Henri Becquerel (1852 – 1908) was a French physicist and the first person to discover evidence of radioactivity.
Becquerel wrapped fluorescing crystal (uranium salt potassium uranyl sulfate) in a cloth, along with the photographic plate and a copper Maltese cross.
Several days later, he discovered that a image of the cross appeared on the plate.
The uranium salt was emitting radiation.
Because of this discovery, Becquerel won a Nobel Prize for Physics in 1903, which he shared with Marie Curie and Pierre Curie.
Answer:
![\boxed {\boxed {\sf 6.4 \ atm}}](https://tex.z-dn.net/?f=%5Cboxed%20%7B%5Cboxed%20%7B%5Csf%206.4%20%5C%20atm%7D%7D)
Explanation:
We are asked to find the pressure of a gas in a can given a change in temperature. We will use Gay-Lussac's Law, which states the pressure of a gas is directly proportional to the temperature. The formula for this law is:
![\frac {P_1}{T_1}= \frac {P_2}{T_2}](https://tex.z-dn.net/?f=%5Cfrac%20%7BP_1%7D%7BT_1%7D%3D%20%5Cfrac%20%7BP_2%7D%7BT_2%7D)
Initially, the gas in the aerosol can has a pressure of 3.10 atmospheres at a temperature of 25 degrees Celsius.
![\frac { 3.10 \ atm}{25 \textdegree C}=\frac{P_2}{T_2}](https://tex.z-dn.net/?f=%5Cfrac%20%7B%203.10%20%5C%20atm%7D%7B25%20%5Ctextdegree%20C%7D%3D%5Cfrac%7BP_2%7D%7BT_2%7D)
The temperature is increased to 52 degrees Celsius, but the pressure is unknown.
![\frac { 3.10 \ atm}{25 \textdegree C}=\frac{P_2}{52 \textdegree C}](https://tex.z-dn.net/?f=%5Cfrac%20%7B%203.10%20%5C%20atm%7D%7B25%20%5Ctextdegree%20C%7D%3D%5Cfrac%7BP_2%7D%7B52%20%5Ctextdegree%20C%7D)
We are solving for the new pressure, so we must isolate the variable
. It is being divided by 52 degrees Celsius. The inverse operation of division is multiplication, so we multiply both sides of the equation by 52 °C.
![52 \textdegree C *\frac { 3.10 \ atm}{25 \textdegree C}=\frac{P_2}{52 \textdegree C} * 52 \textdegree C](https://tex.z-dn.net/?f=52%20%5Ctextdegree%20C%20%2A%5Cfrac%20%7B%203.10%20%5C%20atm%7D%7B25%20%5Ctextdegree%20C%7D%3D%5Cfrac%7BP_2%7D%7B52%20%5Ctextdegree%20C%7D%20%2A%2052%20%5Ctextdegree%20C)
![52 \textdegree C *\frac { 3.10 \ atm}{25 \textdegree C}=P_2](https://tex.z-dn.net/?f=52%20%5Ctextdegree%20C%20%2A%5Cfrac%20%7B%203.10%20%5C%20atm%7D%7B25%20%5Ctextdegree%20C%7D%3DP_2)
The units of degrees Celsius cancel.
![52 *\frac { 3.10 \ atm}{25}=P_2](https://tex.z-dn.net/?f=52%20%2A%5Cfrac%20%7B%203.10%20%5C%20atm%7D%7B25%7D%3DP_2)
![52 *0.124 \ atm = P_2](https://tex.z-dn.net/?f=52%20%2A0.124%20%5C%20atm%20%3D%20P_2)
![6.448 \ atm = P_2](https://tex.z-dn.net/?f=6.448%20%5C%20atm%20%3D%20P_2)
The original values of pressure and temperature have 2 and 3 significant figures. Our answer must be rounded to the least number of sig figs, which is 2. For the number we calculated, that is the tenths place. The 4 in the hundredth place tells us to leave the 4 in the tenths place.
![6.4 \ atm \approx P_2](https://tex.z-dn.net/?f=6.4%20%5C%20atm%20%5Capprox%20P_2)
The gas pressure in the can at 52 degrees Celsius is approximately <u>6.4 atmospheres.</u>