Answer:
The equilibrium temperature of the coffee is 72.4 °C
Explanation:
Step 1: Data given
Mass of cream = 15.0 grams
Temperature of the cream = 10.0°C
Mass of the coffee = 150.0 grams
Temperature of the coffee = 78.6 °C
C = respective specific heat of the substances( same as water) = 4.184 J/g°C
Step 2: Calculate the equilibrium temperature
m(cream)*C*(T2-T1) = -m(coffee)*c*(T2-T1)
15.0 g* 4.184 J/g°C *(T2 - 10.0°C) = -150.0g *4.184 J/g°C*(T2-78.6°C)
62.76T2 - 627.6 = -627.6T2 + 49329.36
690.36T2 = 49956.96
T2 = 72.4 °C
The equilibrium temperature of the coffee is 72.4 °C
Answer:
10.6 moles of CO₂ are produced in this combustion
Explanation:
The combustion reaction is:
2C₂H₆ (g) + 7O₂ (g) ⟶ 4CO₂ (g) + 6H₂O (g)
We assume the ethane as the limiting reactant because the excersise states that the O₂ is in excess.
We make a rule of three:
2 moles of ethane can produce 4 moles of CO₂
Therefore 5.30 moles of ethane will produce (5.3 . 4) /2 = 10.6 moles
Below picture contains the given models.
Answer: Option-B is the correct model.
Explanation: Solid is a state of matter in which the particles are closely packed, has definite volume and shape. Like liquids they don't flow, either they occupy the volume of container as that occupied by both gases and liquids. The inter-molecular forces between solid particles are very strong as compared to liquids ans gases. So, the model B has a particles very closely packed to each other.
Hi,
The answer is B, Urbanisation.
Hope this helps.
r3t40
Bestie since it’s Cl2, put 2 next to KCl. That means you need to 2 moles of K, so put 2 next to KI. That’s it