Answer:
VH2SO4 = 145.3 mL
Explanation:
Mw BaO2 = 169.33 g/mol
⇒ mol BaO2 = 53.5g * ( mol BaO2 / 169.33 g BaO2) = 0.545 mol BaO2
⇒according to the reaction:
mol BaO2 = mol H2SO4 = 0.545 mol
⇒ V H2SO4 = 0.545 mol H2SO4 * ( L H2SO4 / 3.75 mol H2SO4 )
⇒V H2SO4 = 0.1453 L (145.3 mL)
Answer:
(i) specific heat
(ii) latent heat of vaporization
(iii) latent heat of fusion
Explanation:
i. Q = mcΔT; identify c.
Here, Q is heat, m is the mass, c is the specific heat and ΔT is the change in temperature.
The amount of heat required to raise the temperature of substance of mass 1 kg by 1 degree C is known as the specific heat.
ii. Q = mLvapor; identify Lvapor
Here, Q is the heat, m is the mass and L is the latent heat of vaporization.
The amount of heat required to convert the 1 kg liquid into 1 kg vapor at constant temperature.
iii. Q = mLfusion; identify Lfusion
Here, Q is the heat, m is the mass and L is the latent heat of fusion.
Here, Q is the heat, m is the mass and L is the latent heat of vaporization.
The amount of heat required to convert the 1 kg solid into 1 kg liquid at constant temperature.
Answer:
Explanation:
Just saw your request regarding answering this so here it is:
All of them belong of Group 1 in periodic table and thus are highly reactive! Pattern of reactivity for Group 1 (Alkali metals) increases as you move down the group as their radius keeps increasing and thus electrons can be easily lost. Thus, to ID the lumps, Sheena should look at their reactivity and she should get the following trend:
Most reactive: Potassium (K)
Intermediate: Sodium (Na)
Least reactive: Lithium (Li)
Hope it helps!
