To the right
Because there is an unbalanced force in that direction
:)
Unit of M is also mole/L, where mole is the moles of solute and L is the volume of the solution. The latter is given: 158 mL or 0.158 L. So we need to find out the moles of NH4Br.
Moles of NH4Br = Mass of NH4Br/molar mass of NH4Br = 17.0g/(14+1*4+79.9)g/mol = 0.1736 mole.
So, the molarity of the solution = 0.1736mole/0.158L = 1.10 mole/L = 1.10 M
The formula used for determining gas pressure, volume and temperature interaction would be PV=nRT.
<span>• What is the temperature in Kelvins?
</span>You already right at this part. Kelvin temperature formula from celsius should be:
K= C+273.15=
<span>K= 27 +273.15 = 300.15
It is important to remember that the formula in this question is using Kelvin unit at temperature, not Celcius or Fahrenheit.
</span>
<span>• Assuming that everything else remains constant, what will happen to the pressure if the temperature decreases to -15 ºC?
</span>In this case, the temperature is decreased from 27C into -15C and you asked the change in the pressure.
Using PV=nRT formula, you can derive that the temperature will be directly related to pressure. If the temperature decreased, the pressure will be decreased too.
<span> If you increase the number of moles to 6 moles, increase temperature to 400K and reduce the volume to 25 L, what will the new pressure be?
</span>PV=nRT
P= nRT/V
P= 6 moles* <span>0.0821 L*atm/(mol*K) * 400K/25L= 7.8816 atm</span>
Answer:
0.171 M
Explanation:
Step 1: Given data
- Mass of H₃PO₄ (solute): 3.35 g
- Volume of solution (V): 200 mL
Step 2: Calculate the moles of solute
The molar mass of H₃PO₄ is 97.99 g/mol.
3.35 g × 1 mol/97.99 g = 0.0342 mol
Step 3: Convert "V" to liters
We will use the conversion factor 1 L = 1000 mL.
200 mL × 1 L/1000 mL = 0.200 L
Step 4: Calculate the molarity of the solution
We will use the definition of molarity.
M = moles of solute / liters of solution
M = 0.0342 mol/0.200 L = 0.171 M
Answer:
Explanation:
The air 9% mole% methane have an average molecular weight of:
9%×16,04g/mol + 91%×29g/mol = 27,8g/mol
And a flow of 700000g/h÷27,8g/mol = 25180 mol/h
In the reactor where methane solution and air are mixed:
In = Out
Air balance:
91% air×25180 mol/h + 100% air×X = 95%air×(X+25180)
Where X is the flow rate of air in mol/h = <em>20144 mol air/h</em>
<em></em>
The air in the product gas is
95%×(20144 + 25180) mol/h = 43058 mol air× 21%O₂ = 9042 mol O₂ ×32g/mol = <em>289 kg O₂</em>
43058 mol air×29g/mol <em>1249 kg air</em>
Percent of oxygen is: 
=<em>0,231 kg O₂/ kg air</em>
<em></em>
I hope it helps!
