The gas sample consists of CH₄ and He.
Number of CH₄ moles - 4 g/ 16 g/mol = 0.25 mol
molar volume is where 1 mol of any gas occupies a volume of 22.4 L
Therefore 0.25 mol of CH₄ gas occupies a volume of 22.4 L/mol x 0.25 mol
Volume of methane - 5.6 L
Number of He moles - 2.00 g/ 4 g/mol = 0.5 mol
Volume occupied by He - 22.4 L /mol x 0.5 mol = 11.2 L
The total volume of the sample = 5.6 L + 11.2 L = 16.8 L
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Answer: Option (b) is the correct answer.
Explanation:
Since energy of reactants is less than the energy of products. Therefore, it means energy is absorbed during the reaction.
As the energy required to break the bonds in the reactants is greater than the energy released when products are formed.
Therefore, it is an endothermic reaction.
Thus, we can conclude that the statement, it is endothermic because the energy required to break bonds in the reactants is greater than the energy released when the products are formed is correct.
Answer:
The mass of tin is 164 grams
Explanation:
Step 1: Data given
Specific heat heat of tin = 0.222 J/g°C
The initial temeprature of tin = 80.0 °C
Mass of water = 100.0 grams
The specific heat of water = 4.184 J/g°C
Initial temperature = 30.0 °C
The final temperature = 34.0 °C
Step 2: Calculate the mass of tin
Heat lost = heat gained
Qlost = -Qgained
Qtin = -Qwater
Q = m*c*ΔT
m(tin)*c(tin)*ΔT(tin) = -m(water)*c(water)*ΔT(water)
⇒with m(tin) = the mass of tin = TO BE DETERMINED
⇒with c(tin) = the specific heat of tin = 0.222J/g°C
⇒with ΔT(tin) = the change of temperature of tin = T2 - T1 = 34.0°C - 80.0°C = -46.0°C
⇒with m(water) = the mass of water = 100.0 grams
⇒with c(water) = the specific heat of water = 4.184 J/g°C
⇒with ΔT(water) = the change of temperature of water = T2 - T1 = 34.0° C - 30.0 °C = 4.0 °C
m(tin) * 0.222 J/g°C * -46.0 °C = -100.0g* 4.184 J/g°C * 4.0 °C
m(tin) = 163.9 grams ≈ 164 grams
The mass of tin is 164 grams
Answer:
B
[(0.75)^3(0.25)]÷[(0.50)^2(0.75)]
Explanation:
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