It is important to use low flame when evaporating water from a recovered filtrate because then the water and filtrate will not spatter and the filtrate can also be recovered after evaporating water.
If flame is not low then water as well as got spatter so it is important to use low flame so that the water and filtrate will not spatter.
Answer:
V = 0.63 L
Explanation:
To solve this problem, we need to use the Charle's law which is a law that involves temperature and volume, assuming we have a constant pressure. The problem do not state that the pressure is being altered, so we can safely assume that the pressure is constant (Maybe 1 atm).
Now, as the pressure is constant, the Charle's law is the following:
V₁ / T₁ = V₂ / T₂ (1) V is volume in Liter, and T is temperature in Kelvin.
Using this law with the given data, we solve for V₂:
V₂ = V₁T₂ / T₁
Before we use this expression, let's convert the temperatures to Kelvin:
T₁ = 19 + 273 = 292 K
T₂ = 250 + 273 = 523 K
Now, let's calculate the volume of the balloon:
V₂ = 0.35 * 523 / 292
<h2>
V₂ = 0.63 L</h2>
The balanced chemical reaction is expressed as follows:
<span>CuCl2 (aq) + 2AgNO3 (aq) → 2AgCl (s) + CuNO32 (aq)
To determine the </span><span>concentration of copper(II) chloride contaminant in the original groundwater sample, we use the final amount of silver chloride that was produced from the reaction and the relation of the substances from the chemical reaction. We calculate as follows:
mmol AgCl = 6.1 mg AgCl ( 1 mmol / 143.35 mg ) = 0.0426 mmol
mmol CuCl2 = </span>0.0426 mmol AgCl ( 1 mmol CuCl2 / 2 mmol AgCl ) = 0.0213 mmol CuCl2
concentration of CuCl2 in the original water sample = 0.0213 mmol CuCl2 / 200.0 mL = 1.0638 x 10^-4 mmol / mL or 1.0638 x 10^-4 mol/L
Yes, because if you have a flower and it reproduces then there's a HIGH chance that the next flower will be the same color because of traits! <span />
