Answer:
no
Explanation:
its made of a living organisim which in this case is oats
Here we have to calculate the amount of 
ion present in the sample.
In the sample solution 0.122g of ion is present.
The reaction happens on addition of excess BaCl₂ in a sample solution of potassium sulfate (K₂SO₄) and sodium sulfate [(Na)₂SO₄] can be written as-
K₂SO₄ = 2K⁺ +
(Na)₂SO₄=2Na⁺ +
Thus, BaCl₂+ = BaSO₄↓ + 2Cl⁻ .
(Na)₂SO₄ and K₂SO₄ is highly soluble in water and the precipitation or the filtrate is due to the BaSO₄ only. As a precipitation appears due to addition of excess BaCl₂ thus the total amount of ion is precipitated in this reaction.
The precipitate i.e. barium sulfate (BaSO₄)is formed in the reaction which have the mass 0.298g.
Now the molecular weight of BaSO₄ is 233.3 g/mol.
We know the molecular weight of sulfate ion () is 96.06 g/mol. Thus in 1 mole of BaSO₄ 96.06 g of ion is present.
Or. we may write in 233.3 g of BaSO₄ 96.06 g of ion is present. So in 1 g of BaSO₄ 
g of ion is present.
Or, in 0.298 g of the filtered mass (0.298×0.411)=0.122g of ion is present.
Explanation:
The solution of the lactic acd and sodium lactate is referred to as a buffer solution.
A buffer solution is an aqueous solution consisting of a mixture of a weak acid and its conjugate base, or vice versa. In this case, the weak acid is the lactic acid and the conjugate base is the sodium lactate.
Buffer solutions are generally known to resist change in pH values.
When a strong base (in this case, NaOH) is added to the buffer, the lactic acid will give up its H+ in order to transform the base (OH-) into water (H2O) and the conjugate base, so we have:
HA + OH- → A- + H2O.
Since the added OH- is consumed by this reaction, the pH will change only slightly.
The NaOH reacts with the weak acid present in the buffer sollution.
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This is bob. If you brainliest him. He will reward you with a certain script.
On the basis of the given data:
Volume (V) is 37.5 L
Temperature (T) is 307 K
Pressure (P) is 1.25 atm
In order to calculate the number of moles, the formula to be used is PV = nRT, Here R is 0.0821 Latm/mol/K
n = PV / RT
n = 1.25 atm × 37.5 L / 0.0821 Latm/mol/K × 307 K
n = 1.85 mol
