Answer:
209.3 Joules require to raise the temperature from 10 °C to 15 °C.
Explanation:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m × c × ΔT
Given data:
mass of water = 10 g
initial temperature T1= 10 °C
final temperature T2= 15 °C
temperature change =ΔT= T2-T1 = 15°C - 10°C = 5 °C
Energy or joules added to increase the temperature Q = ?
Solution:
We know that specific heat of water is 4.186 J/g .°C
Q = m × c × ΔT
Q = 10 g × 4.186 J/g .°C × 5 °C
Q = 209.3 J
Answer:
74 litre
Explanation:
using ideal gas eqation PV=nRT
here P(pressure)=81.8 kPa =81.8×10^3 Pa
moles=2.5
temperature=273.15+18=291.15K
Gas constant R=8.314m^3-Pa/K-mol
now, V=nRT/P = 8.314×2.5×291.5/81.8×10^3 ≈74litre
✌️;)
Answer:
a) HNO3 -> H+ + NO3- disassociation of Nitric Acid; to yield a Nitrate ion and a Proton, H+, or as a Hydronium ion H3O+
b) H2S04 -> Disassociation of Sulfuric Acid; simple way- 2H+ + SO4- -
c) H2S hydrogen sulphide in water is an acid; thus H+ HS- disassociation.
d) NaOH -> dissociation of Na+ + OH-; this is complete; sodium hydroxide is deliquescent, meaning it will draw water - EVEN from the air! Strong Base
e) Na2CO3 -> 2Na+ CO3- - Ionization of sodium carbonate - a salt
f) Na2S04 -> 2Na+ + SO4 - - ionization of sodium sulphate - a salt
g) NaCl -> Na+ + Cl- ionization of the salt, Sodium Chloride
Explanation:
Salts ionize at different rates; acids or bases dissociate; these are mostly strong acids and NaOH, a strong base.
Answer:
should be B
because the player has more mass than the puck
Explanation:
