Answer:
Assume that the sack was initially close to the sea level. Its weight will increase even though its mass stays the same.
Explanation:
The weight of an object typically refers to the size of the planet's gravitational attraction (a force) on this object. That's not the same as the mass of the object. The weight of an object at a position depends on the size of the gravitational field there; on the other hand, the mass of the object is supposed to be same regardless of the location- as long as the object stays intact.
Let
denote the strength of the gravitational field at a certain point. If the mass of an object is
, its weight at that point will be
.
Indeed,
on many places of the earth. However, this value is accurate only near the sea level. The equation for universal gravitation is a more general way for finding the strength of the gravitational field at an arbitrary height. Let
denote the constant of universal gravitation, and let
denote the mass of the earth. At a distance
from the center of the earth (where
.
The elevation of many places in Bhutan are significantly higher than that of many places in India. Therefore, a sack of potato in Bhutan will likely be further away from the center of the earth (larger
) compared to a sack of potato in India.
Note, that in the approximation, the value of
is (approximately, because the earth isn't perfectly spherical) inversely proportional to the distance from the center of the planet. The gravitational field strength
On the other hand, the weight of an object of fixed mass is proportional to the gravitational field strength. Therefore, the same bag of potatoes will have a smaller weight at most places in Bhutan compared to most places in India.
Answer:
F2 is the limiting reactant
27.6 grams of NaF is produced.
Explanation:
Balance the equation first.
2Na+ F2 ---> 2NaF
To find the limiting reactant, solve for how much NaF can be produced with Na and F2
12.5g F2 x (1 mole F2/ 38.00 grams F2)x (2 mole NaF/ 1 mole F2)
=0.658 moles NaF
16.2g Na x (1 mole Na/ 22.99 grams Na)x (2 mole NaF/ 2 mole Na)
=0.705 moles NaF
Since F2 produced the least NaF, F2 is the limiting reactant.
Now, to find how much NaF there is, use the moles solved above with F2 as the limiting reactant.
0.658 moles NaF x (41.99 grams NaF/ 1 mole NaF)= 27.6 moles NaF
27.6 moles of NaF would be theoretically produced.
Answer:

Explanation:
You look at the type of atom and their electronegativity difference.
If ΔEN <1.6, covalent; if ΔEN >1.6, ionic
Ar/Xe: Noble gases; no reaction
F/Cs: Non-metal + metal; ΔEN = |3.98 – 0.79| = 3.19; Ionic
N/Br: Two nonmetals; ΔEN = |3.04 - 2.98| = 0.