Analyzing the formula for sulfuric acid reveals the molar proportions:
H: S: O
2: 1: 4
Next, we need to convert the provided mass of hydrogen into moles, calculated by:
Moles = mass / Mr
Moles = 7.27 / 1
Moles = 7.27
Thus, the number of moles for each element are:
S = 7.27 / 2 = 3.64 moles
O = 7.27 * 2 = 14.54 moles
Subsequently, the masses for sulfur and oxygen are:
S = 32 * 3.64 = 116.48 grams
O = 16 * 14.54 = 232.64 grams
To find the temperature at which the volume of the gas would be 0.550 L, given that it is 0.432 L at -20.0 °C, apply Charles’s Law.
The formula is v1/T1 = v2/T2
Known values:
V1 = 0.550 L
T1 = ?
T2 = -20°C + 273 = 253 K
V2 = 0.432 L
Rearranging for T1:
T1 = (V1 × T2) / V2
Calculating:
T1 = (0.55 L × 253) / 0.432 L = 322.11 K or 49.11°C
Answer:
Indeed, the chemist is capable of identifying the compound present in the sample.
Explanation:
In one mole of K₂O, potassium has a mass of 2 × 39.1 g = 78.2 g, while the total mass of K₂O is 94.2 g. The mass ratio of K compared to K₂O is calculated as 78.2 g / 94.2 g = 0.830.
For 1 mole of K₂O₂, potassium's mass remains the same at 78.2 g, but the total mass of K₂O₂ is 110.2 g. The mass ratio of K to K₂O₂ then equates to 78.2 g / 110.2 g = 0.710.
When the chemist measures the mass of K in relation to the overall sample, the mass ratio can be computed.
- If the mass ratio is 0.830, then it indicates a pure K₂O compound.
- If the mass ratio is 0.710, it indicates a pure K₂O₂ compound.
- If the mass ratio falls outside of 0.830 or 0.710, the sample is assessed to be a mixture.
Answer:
Explanation:
The relationship between the new temperature scale and the absolute temperature scale is defined as follows
Aw = 2 K
for K = 273.15 (the freezing point of water on the absolute scale)
Aw = 2 x 273.15 = 546.3 K
Each division of the new scale is equivalent to half that of each division on the absolute scale
each division of the new scale is minimal.
The value of R = 8.314 J per mole per K
Here, per K corresponds to 2Aw
Hence, the value of R in the new scale = 8.314/2 J per mole per Aw
= 4.157 J per mole per Aw
k = R / N
= 4.157 / 6.02 x 10²³
= .69 x 10⁻²³
= 6.9 x 10⁻²⁴ J per molecule per Aw .
Answer:
The enthalpy of the second intermediate equation is altered by halving its value and changing the sign.
Explanation:
Let's examine both the first and second intermediate reactions alongside the overall equation concerning the examined process;
First reaction;
Ca (s) + CO₂ (g) + ½O₂ (g) → CaCO₃ (s) ΔH₁ = -812.8 kJ
Second reaction;
2Ca (s) + O₂ (g) → 2CaO (s) ΔH₂ = -1269 kJ
Thus, the overall reaction becomes;
CaO (s) + CO₂ (g) → CaCO₃ (s) ΔH =?
According to Hess's law, which states that the total heat change in a reaction is equal to the sum of the heat changes for each step, we cannot simply sum the enthalpies for this overall reaction. Instead, we obtain the overall enthalpy by halving the second intermediate reaction's enthalpy and changing its sign before adding, as illustrated below;
Enthalpy of Intermediate reaction 1 + ½(-Enthalpy of Intermediate reaction 2) = Enthalpy of Overall reaction
