Answer:
10.000 grams
Explanation:
For the first law of thermodynamics, the energy must be conserved, that means that the energy in form of heat (Q) must be equal to the sum of work (W) and internal energy(ΔU) :
Q = W + ΔU
ΔU depends on the temperature and W in the variation of pressure and volume. Q depends on the temperature, but also the mass. So, there is the same temperature, ΔU is equal for both reaction, if there is no work done, the heat must be equal for both of them. So the mass such be the same.
Answer:
Increasing the pressure on a reaction involving reacting gases increases the rate of reaction. Changing the pressure on a reaction which involves only solids or liquids has no effect on the rate.
Explanation:
The correct answer is the first option. The piece of evidence would suggest an unknown substance is an ionic compound is that <span>it conducts electricity when dissolved in water.</span><span> A</span><span>n ionic compound dissociates
into ions when in aqueous solution. These ions move freely in the solution and
can allow the flow of electricity into the solution.</span>
Answer:
63.546 u
Would be the molar mass of copper!
Answer:
Pb (NO₃)₂(aq) + 2KCl(aq) → 2KNO₃(aq) + PbCl₂(s)
Explanation:
In given chemical equation the aqueous lead (II) nitrate react with aqueous potassium chloride and form aqueous potassium nitrate and lead chloride.
Chemical equation:
Pb (NO₃)₂(aq) + KCl(aq) → KNO₃(aq) + PbCl₂(s)
Balanced chemical equation:
Pb (NO₃)₂(aq) + 2KCl(aq) → 2KNO₃(aq) + PbCl₂(s)
ionic equation:
Pb²⁺ (aq) + 2NO₃⁻ (aq) + 2K⁺(aq) + 2Cl⁻ (aq) → 2NO₃⁻(aq) + 2K⁺(aq) + PbCl₂(s)
Net ionic equation:
Pb²⁺ (aq) + 2Cl⁻ (aq) → PbCl₂(s)
The NO₃⁻(aq) and K⁺(aq) are spectator ions that's why these are not written in net ionic equation. The PbCl₂ can not be splitted into ions because it is present in solid form.
Spectator ions:
These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.