When an element losses its electron its called a cation. When an element accepted that electron it called anion. This is called an ionic bond.
Answer:
no he just repeated the steps and made more of the same cleaner my guy
Explanation:
Answer:
1. ΔE = 0 J
2. ΔH = 0 J
3. q = 3.2 × 10³ J
4. w = -3.2 × 10³ J
Explanation:
The change in the internal energy (ΔE) and the change in the enthalpy (ΔH) are functions of the temperature. If the temperature is constant, ΔE = 0 and ΔH = 0.
The gas initially occupies a volume V₁ = 20.0 L at P₁ = 3.2 atm. When the pressure changes to P₂ = 1.6 atm, we can find the volume V₂ using Boyle's law.
P₁ × V₁ = P₂ × V₂
3.2 atm × 20.0 L = 1.6 atm × V₂
V₂ = 40 L
The work (w) can be calculated using the following expression.
w = - P . ΔV
where,
P is the external pressure for which the process happened
ΔV is the change in the volume
w = -1.6 atm × (40L - 20.0L) = -32 atm.L × (101.325 J/1atm.L) = -3.2 × 10³ J
The change in the internal energy is:
ΔE = q + w
0 = q + w
q = - w = 3.2 × 10³ J
If he was 30.8% too low, it means that he was at 69.2% of the boiling point needed. So 50o C is 69.2% of total.
In order to know what 100% is, you can divide the number by it's percentage and then multiply it by a hundred.
So: 50/30.8=1.623
1.623*100=162.3
So the correct boiling point of the liquid he was working with in the lab is 162.3 oC
Answer: 19.4 mL Ba(OH)2
Explanation:
H2(g) + Cl2(g) --> 2HCl(aq) (make sure this equation is balanced first)
At STP, 1 mol gas = 22.4 L gas. Use this conversion factor to convert the 100. mL of Cl2 to moles.
0.100 L Cl2 • (1 mol / 22.4 L) = 0.00446 mol Cl2
Use the mole ratio of 2 mol HCl for every 1 mol Cl2 to find moles of HCl produced.
0.00446 mol Cl2 • (2 mol HCl / 1 mol Cl2) = 0.00892 mol HCl
HCl is a strong acid and Ba(OH)2 is a strong base so both will completely ionize to release H+ and OH- respectively. You need 0.00892 mol OH- to neutralize all of the HCl. Note that one mole of Ba(OH)2 contains 2 moles of OH-.
0.00892 mol OH- • (1 mol Ba(OH)2 / 2 mol OH-) • (1 L Ba(OH)2 / 0.230 M Ba(OH)2) = 0.0194 L = 19.4 mL Ba(OH)2
