Answer:
12.8 g of 
must be withdrawn from tank
Explanation:
Let's assume gas inside tank behaves ideally.
According to ideal gas equation- 
where P is pressure of , V is volume of , n is number of moles of , R is gas constant and T is temperature in kelvin scale.
We can also write, 
Here V, T and R are constants.
So, 
ratio will also be constant before and after removal of from tank
Hence, 
Here, 
and 
So, 
So, moles of must be withdrawn = (0.66 - 0.26) mol = 0.40 mol
Molar mass of = 32 g/mol
So, mass of must be withdrawn = 
Answer:
1.33 L.
Explanation:
- We can use the general law of ideal gas: PV = nRT.
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- If n and T are constant, and have different values of P and V:
<em>(P₁V₁) = (P₂V₂)</em>
<em></em>
Knowing that:
V₁ = 4.0 L, P₁ = 2.0 atm,
V₂ = ??? L, P₂ = 6.0 atm.
- Applying in the above equation
(P ₁V₁) = (P₂V₂)
<em>∴ V₂ = P ₁V₁/P₂</em> = (2.0 atm)(4.0 L)/(6.0 atm) =<em> 1.33 L.</em>
In photosynthesis, plants take in carbon dioxide and turn it into energy that comes out as oxygen.
Answer:
v = 37.9 ml
Explanation:
Given data:
Mass of compound = 1.56 kg
Density = 41.2 g/ml
Volume of compound = ?
Solution:
First of all we will convert the mass into g.
1.56 ×1000 = 1560 g
Formula:
D=m/v
D= density
m=mass
V=volume
v = m/d
v = 1560 g / 41.2 g/ml
v = 37.9 ml
(g solute/g solution)*100 = % mass/mass
30 g / 400 * 100
0,075 * 100
= 7,5% w/w
hope this helps!
