45.0g
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Explanation:
Step 1:
Data obtained from the question. This include the following:
Initial pressure (P1) = 1atm
Initial temperature (T1) = 0°C = 0°C + 273 = 273K
Final temperature (T2) = 280°C = 280°C + 273 = 553K
Final pressure (P2) =...?
Step 2:
Determination of the new pressure of the gas.
Since the volume of the gas is constant, the following equation:
P1/T1 = P2/T2
will be used to obtain the pressure. This is illustrated below:
P1/T1 = P2/T2
1/273 = P2 / 553
Cross multiply
273x P2 = 553
Divide both side by 273
P2 = 553/273
P2 = 2.03atm
Therefore, the new pressure of the gas will be 2.03atm
Answer: The ion formed after the reduction of bromine is 
Explanation:
The electronic configuration of Sodium (Na) = ![[Ne]3s^1](https://tex.z-dn.net/?f=%5BNe%5D3s%5E1)
The electronic configuration of Bromine (Br) = ![[Ar]3d^{10}4s^24p^5](https://tex.z-dn.net/?f=%5BAr%5D3d%5E%7B10%7D4s%5E24p%5E5)
From the above configurations, Sodium ion will loose 1 electron in order to gain stable electronic configuration and that electron is accepted by the Bromine atom because it is 1 electron short of the stable electronic configuration.
(oxidation reaction)
(Reduction reaction)
Bromine atom is reduced to form
Reduction reactions are the reactions in which the element gain electrons.
Oxidation reactions are the reactions in which the element looses its electrons.
Answer:
See balanced equations below
Explanation:
1. Mg(s) +2 HCL (aq) →MgCl₂ (aq) +H₂(g)
This is a single replacement reaction, involving an acid with a metal
2. 2Al(s) + 3H₂SO₄ (aq)→Al₂(SO₄)₃(aq) + 3H₂
3. 3 Zn (s) + 2H₃PO₄(aq)→ Zn₃(PO₄)₂ (aq) + 3H₂ (g)
4. 2Al(s) + 6HCL (aq)→2AlCl₃(aq) +3H₂ (g)
B.
1. 2KOH(aq) + MgCl₂→Mg(OH)₂ (aq) + 2KCl (aq)
2. 3NaOH (aq)+ Al(NO₃)₃ (aq)→Al(OH)₃(s) + 3 NaNO₃(aq) ---this is a precipitation reaction
3. BaBr₂(aq) + H₂SO₄→BaSO₄ (s) + 2Br⁻(aq)
4. Na₂S + 2HCl → 2NaCl (aq) + H₂S (g)
5. 3CaCl₂ +2K₃PO₄→ Ca₃(PO₄)₂+6KCl
6.Ba(NO₃)₂ + (NH₄)₂CO₃→ 2(NH₄)⁺(aq) +BaCO₃(s)
