How many ATOMS of boron are present in 3.61 grams of boron trifluoride ?
1 answer:
Hi there!
We can use the following conversions to solve:
Total mass --> amount of mols --> amount of atoms (Avogadro's number)
Begin by calculating the amount of boron trifluoride in 3.61 grams:
3.61 g * (1 mol BF₃ / 67.8 g) ≈ 0.0532 mol BF₃
Use avogadro's number to convert:
0.0532 mol * 6.02× 10²³atoms / 1 mol = 3.203 × 10²² atoms
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Answer:
Q = 143,921 J = 143.9 kJ.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to calculate the absorbed heat by considering this is a process involving sensible heat associated to the vaporization of water, which is isothermic and isobaric; and thus, the heat of vaporization of water, with a value of about 2259.36 J/g, is used as shown below:
Thus, we plug in the mass and the aforementioned heat of vaporization of water to obtain the following:
Regards!
Answer : The energy removed must be, 29.4 kJ
Explanation :
The process involved in this problem are :
The expression used will be:
where,
= heat released for the reaction = ?
m = mass of benzene = 94.4 g
= specific heat of solid benzene =
= specific heat of liquid benzene =
= enthalpy change for fusion =
Now put all the given values in the above expression, we get:
Negative sign indicates that the heat is removed from the system.
Therefore, the energy removed must be, 29.4 kJ