Answer:
rip there isnt a photo
Explanation:
i do know a lot about cells tho lol
Answer
Na OH reacts with H Cl and forms Na Cl and H₂O
NaOH + HCl → NaCl + H₂O
Here we can see that 1 mole of NaOH reacting with 1 mole of HCl and forming 1 mole of NaCl and 1 mole of H₂O
when NaOH and HCl are added together in equal amount then they will completely neutralize each other but NaOH is hygroscopic in nature which means it can absorb water from air so it will not be weighted accurately.
hence, for neutralization we will take extra NaOH.
Answer:
100ml of a stock 50% KNO3 solutions are needed to prepare 250ml of a 20% KNO3 solution.
Explanation:
In the given question it is mentioned that
S1=50%
V2=250ml
S2= 20%
We all know that
V1S1=V2S2
∴V1= V2×S2÷S1
∴V1= V2S2×1/S1
∴V1= 250×20÷50
∴V1= 100ml
Answer:
V₂ = 1.92 L
Explanation:
Given data:
Initial volume = 0.500 L
Initial pressure =2911 mmHg (2911/760 = 3.83 atm)
Initial temperature = 0 °C (0 +273 = 273 K)
Final temperature = 273 K
Final volume = ?
Final pressure = 1 atm
Solution:
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
by putting values,
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 3.83 atm × 0.500 L × 273 K / 273 K × 1 atm
V₂ = 522.795 atm .L. K / 273 K.atm
V₂ = 1.92 L
Answer:
Solubility of O₂(g) in 4L water = 3.42 x 10⁻² grams O₂(g)
Explanation:
Graham's Law => Solubility(S) ∝ Applied Pressure(P) => S =k·P
Given P = 0.209Atm & k = 1.28 x 10⁻³mol/L·Atm
=> S = k·P = (1.28 x 10⁻³ mole/L·Atm)0.209Atm = 2.68 x 10⁻³ mol O₂/L water.
∴Solubility of O₂(g) in 4L water at 0.209Atm = (2.68 x 10⁻³mole O₂(g)/L)(4L)(32 g O₂(g)/mol O₂(g)) = <u>3.45 x 10⁻² grams O₂(g) in 4L water. </u>
