Steal; Anion
This is the answer because phosphorous is very electronegative.
the answer is option D "people emphasized obtaining knowledge through scientific experiments" (on plato)
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:
= Initial volume = 100 mL
= Final volume = 1000 mL
= Initial concentration = 0.5 M
= Final concentration
We have the relation
The new concentration is .
