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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:
3.69 g
Explanation:
Given that:
The mass m = 325 g
The change in temperature ΔT = ( 1540 - 165)° C
= 1375 ° C
Heat capacity = 0.490 J/g°C
The amount of heat required:
q = mcΔT
q = 325 × 0.490 × 1375
q = 218968.75 J
q = 218.97 kJ
The equation for the reaction is expressed as:
Then,
1 mole of the ethyne is equal to 26 g of ethyne required for 1544 kJ heat.
Thus, for 218.97 kJ, the amount of ethyne gas required will be:
= 3.69 g