Answer:
60 grams of ice will require 30.26 calories to raise the temperature 1°C.
Explanation:
The amount of heat (Q) to raise the temperature of 60.0 g of ice by 1°C can be calculated from:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat released or absorbed by the system.
m is the mass of the ice (m = 60.0 g).
c is the specific heat capacity of ice (c = 2.108 J/g.°C).
ΔT is the temperature difference (ΔT = 1.0 °C).
∴ Q = m.c.ΔT = (60.0 g)(2.108 J/g.°C)(1.0 °C) = 126.48 J.
<em>It is known that 1.0 cal = 4.18 J.</em>
<em>∴ Q = (126.48 J)(1.0 cal / 4.18 J) = 30.26 cal.</em>
Answer:
Explanation:
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In this case, when performing units conversions involving two proportional factors we need to make sure we first convert to the base unit and then to the target one; thus, since 1 kg = 1000 g and 1 g = 1000 mg, we set up the following expression:
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speed and direction in which an object is moving. both speed and direction of motion. is a vector. two or more velocities add by velocity addition.
Answer:
it will float if the object is 1g/cm^3(water 's density ) because it is less dense
The balanced chemical reaction is:
N2 + 3H2 = 2NH3
We are given the amount of ammonia formed
from the reaction. This is where we start our calculations.
0.575 g NH3 (1 mol NH3 / 17.03 g NH3) (3 mol
H2 / 2 mol NH3) ( 2.02 g H2 / 1 mol H2) = 0.10 g H2
