Answer:
The new pressure is 44.4 kPa.
Explanation:
We have,
Initial volume, ![V_1=22\ ml](https://tex.z-dn.net/?f=V_1%3D22%5C%20ml)
Initial pressure, ![P_1= 101.0\ kPa](https://tex.z-dn.net/?f=P_1%3D%20101.0%5C%20kPa)
It is required to find the new pressure when the volume is increased to 50 ml. The relationship between pressure and volume is known as Boyle's law.
![PV=k\\\\P_1V_1=P_2V_2](https://tex.z-dn.net/?f=PV%3Dk%5C%5C%5C%5CP_1V_1%3DP_2V_2)
is final pressure
![P_2=\dfrac{P_1V_1}{V_2}\\\\P_2=\dfrac{22\times 101\times 10^3}{50}\\\\P_2=44440\ Pa\\\\P_2=44.4\ kPa](https://tex.z-dn.net/?f=P_2%3D%5Cdfrac%7BP_1V_1%7D%7BV_2%7D%5C%5C%5C%5CP_2%3D%5Cdfrac%7B22%5Ctimes%20101%5Ctimes%2010%5E3%7D%7B50%7D%5C%5C%5C%5CP_2%3D44440%5C%20Pa%5C%5C%5C%5CP_2%3D44.4%5C%20kPa)
So, new pressure is 44.4 kPa.
An occluded front forms when a warm air mass is caught between two cooler air masses.
Stoichiometry time! Remember to look at the equation for your molar ratios in other problems.
31.75 g Cu | 1 mol Cu | 2 mol Ag | 107.9 g Ag 6851.65
⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻ → ⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻ = 107.9 g Ag
∅ | 63.5 g Cu | 1 mol Cu | 1 mol Ag 63.5
There's also a shorter way to do this: Notice the molar ratio from Cu to Ag, which is 1:2. When you plug in 31.75 into your molar mass for Cu, it equals 1/2 mol. That also means that you have 1 mol Ag because of the ratio, qhich you can then plug into your molar mass, getting 107.9 as well.
A.
→ ![4Fe(s) + 3CO_2(g)](https://tex.z-dn.net/?f=4Fe%28s%29%20%2B%203CO_2%28g%29)
B.
→ ![Fe(s) + 3Al_2O_3(s)](https://tex.z-dn.net/?f=Fe%28s%29%20%2B%203Al_2O_3%28s%29)
C.
→ ![Al_2(SO_4)_3(aq) + 3H_2(g)](https://tex.z-dn.net/?f=Al_2%28SO_4%29_3%28aq%29%20%2B%203H_2%28g%29)
What is a balanced chemical equation?
An equation that has an equal number of atoms of each element on both sides of the equation is called a balanced chemical equation.
A.
→ ![4Fe(s) + 3CO_2(g)](https://tex.z-dn.net/?f=4Fe%28s%29%20%2B%203CO_2%28g%29)
B.
→ ![Fe(s) + 3Al_2O_3(s)](https://tex.z-dn.net/?f=Fe%28s%29%20%2B%203Al_2O_3%28s%29)
C.
→ ![Al_2(SO_4)_3(aq) + 3H_2(g)](https://tex.z-dn.net/?f=Al_2%28SO_4%29_3%28aq%29%20%2B%203H_2%28g%29)
Learn more about the balanced chemical equation here:
brainly.com/question/15052184
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