Answer:
Is the fixed number of water molecules present in one formula unit of a substance
Explanation:
When copper sulphate is heated in a china dish, the blue colomed crystal changes to a white powder due to the removal of molecules of water of crystallization
Answer:
Q = 270 Joules (2 sig. figs. as based on temperature change.)
Explanation:
Heat Transfer Equation of pure condensed phase substance => Q = mcΔT
Mixed phase (s ⇄ l melting/freezing, or l ⇄ g boiling/condensation) heat transfer equation => Q = m∙ΔHₓ; ΔHₓ = phase transition constant
Since this is a pure condensed phase (or, single phase) form of lead (Pb°(s)) and not melting/freezing or boiling/condensation, one should use
Q = m·c·ΔT
m = mass of lead = 35.0g
c = specific heat of lead = 0.16J/g°C
ΔT = Temp change = 74°C - 25°C = 49°C
Q = (35.0g)(0.16J/g·°C )(49°C) = 274.4 Joules ≅ 270 Joules (2 sig. figs. as based on temperature change.)
Answer:
Explanation:
To write the chemical symbol for an atom or cation with 22 electrons, we must understand some few things about an atom.
An atom is made up of positively charged paticles called protons which are located in the nucleus. The neutrons do not have any charges and are they reside together in the nucleus with the protons.
Electrons are negatively charged particles which orbits the nucleus.
Titanium in its neutral state is the 22nd element on the periodic table and it has 22 electrons.
Vanadium is the the 23rd element on the periodic table. If it loses an electron, it becomes positively charged and it is a cation.
Chromium is the 24th element on the periodic table. A loss of two electrons makes the net electrons remaining on it 22.
The symbols are:
₂₂Ti
₂₃V¹⁺
₂₄Cr²⁺
Answer:Reaccionan 54gr de Nitrato de plata al 39 % de pureza, con 72gr de ácido clorhídrico al 83% de pureza, en un proceso donde se obtienen 93 gr de cloruro de - 154… ... 93 gr de cloruro de plata. El otro producto es el ácido nítrico: Calcular el porcentaje de rendimiento de la reacción y balancearlo.
<span>294400 cal
The heating of the water will have 3 phases
1. Melting of the ice, the temperature will remain constant at 0 degrees C
2. Heating of water to boiling, the temperature will rise
3. Boiling of water, temperature will remain constant at 100 degrees C
So, let's see how many cal are needed for each phase.
We start with 320 g of ice and 100 g of liquid, both at 0 degrees C. We can ignore the liquid and focus on the ice only. To convert from the solid to the liquid, we need to add the heat of fusion for each gram. So multiply the amount of ice we have by the heat of fusion.
80 cal/g * 320 g = 25600 cal
Now we have 320 g of ice that's been melted into water and the 100 g of water we started with, resulting in 320 + 100 = 420 g of water at 0 degrees C. We need to heat that water to 100 degrees C
420 * 100 = 42000 cal
Finally, we have 420 g of water at the boiling point. We now need to pump in an additional 540 cal/g to boil it all away.
420 g * 540 cal/g = 226800 cal
So the total number of cal used is
25600 cal + 42000 cal + 226800 cal = 294400 cal</span>
