Answer:
molarity of diluted solution = 1.25 M
Explanation:
Using,
C1V1 (Stock solution) = C2V2 (dilute solution)
given that
C1 = 2.50M
V1 = 250ML
C2 = ?
V2 = 500ML
2.50 M x 250 mL = C2 x 500 mL
C2 = (2.50 M x 250 mL) / 500 mL
C2 = 1.25 M
Hence, molarity of diluted solution = 1.25 M
Answer:
V₂ = 0.6 V.
Explanation:
- We can use the general law of ideal gas: <em>PV = nRT.</em>
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 is constant, and have different values of P, V and T:
<em>(P₁V₁T₂) = (P₂V₂T₁).</em>
<em></em>
V₁ = V, P₁ = P, T₁ = T.
V₂ = ??? V, P₂ = 1.25 P, T₂ = 0.75 T.
<em>∴ V₂ = (P₁V₁T₂)/(P₂T₁) =</em> (P)(V)(0.75 T)/(1.25 P)(T)<em> = 0.6 V.</em>
it depends how heavy and how far it travels
Answer:
1. 12.6 moles
2. 8.95 moles
3. 2A + 5B → 3C
4. 48 moles
Explanation:
1. 2Fe + 3Cl₂ → 2FeCl₃
We assume the chlorine in excess. Ratio is 2:2
2 moles of Fe, can produce 2 moles of chloride
12.6 moles of Fe will produce 12.6 moles of chloride.
2. 2Fe + 3Cl₂ → 2FeCl₃
For the same reaction, first of all we need to convert the mass to moles:
500 g . 1mol / 55.85 g = 8.95 mol
As ratio is 2:2, the moles we have are the same, that the produced
4. The reaction for the combustion is:
2C₂H₆ (g) + 7O₂ (g) → 4CO₂ (g) + 6H₂O (l)
We assume the oxygen in excess.
Ratio is 2:6, so 2 mol of ethane produce 6 moles of water
Therefore 16 moles of ethane may produce (16 .6) / 2 = 48 moles
When utilizing the gravimetric method, it is crucial to completely dissolve your sample in 10 mL of water. A quantitative technique called gravimetric analysis employs the selective precipitation of the component under study from an aqueous solution.
A group of techniques known as gravimetric analysis are employed in analytical chemistry to quantify an analyte based on its mass. Gravimetric analysis is a quantitative chemical analysis technique that transforms the desired ingredient into a substance (of known composition) that can be extracted from the sample and weighed. This is a crucial point to remember.
Gravimetric water content (g) is therefore defined as the mass of water per mass of dry soil. To calculate it, weigh a sample of wet soil, dry it to remove the water, and then weigh the dried soil (mdry). Dimensions of the sample Water is commonly forgotten despite having a density close to one.
To know more about gravimetry, please refer:
brainly.com/question/18992495
#SPJ4
