Answer: Fe<em>(aq)</em>+S<em>(aq)</em>=FeS<em>(s)</em>
Explanation: The Sodium and Bromine are spectator ions because they don't react with anything, you can see this by writing the ionic equation like so:
1.) Molecular formula (given): FeBr2 (aq)+Na2S (aq)= FeS(s)+2NaBr(aq)
Each dissolved FeBr2 breaks up into one Fe with a charge of 2+ and two Br with a negative charge. This gives you:
Fe(aq)+ 2Br(aq)+Na2S(aq)=FeS(s)+2NaBr
2.) Now repeat what was shown with the other compounds in the given molecular formula, and pay attention to the states that each ion is in (solid, liquid, aqueous, gas) because this will give you the ionic equation, which from there you can get rid of any ions that don't change amount or state.
3.) Ionic formula: Fe(aq)+ <u>2Br(aq)</u>+<u>2 Na(aq)</u>+S (aq)=FeS(s)+<u>2 Na(aq)+2Br(aq)</u>
4.)When you've derived a total ionic equation (above), you'll find that some ions appear on both sides of the equation in equal numbers. For example, in this case two Na cations and two Br anions appear on both sides of the total ionic equation. What does this mean? It means these ions don't participate in the chemical reaction. They're present before and after the reaction. Nothing happens to them. So those are removed and you're left with the net ionic: Fe(aq)+S(aq)=FeS(s)
Hope this helps :)
Answer:
The volume of NO₂ gas collected over water at 25.0 °C is 1.68 Liters.
Explanation:

Moles of copper = 
According to reaction, 1 mol of copper gives 2 moles of nitrogen dioxide gas.
Then 0.03613 moles of copper will give:
of nitrogen dioxide gas
Moles of nitrogen dioxide gas = n = 0.06326 mol
Pressure of the gas = P
P = Total pressure - vapor pressure of water
P = 726 mmHg - 23.8 mmHg = 702.2 mmHg
P = 0.924 atm (1 atm = 760 mmHg)
Temperature of the gas = T = 25.0°C =298.15 K
Volume of the gas = V


V = 1.68 L
The volume of NO₂ gas collected over water at 25.0 °C is 1.68 Liters.
Answer:
Explanation:
CH₃Br+NaOH⟶CH₃OH+NaBr
It is a single step bimolecular reaction so order of reaction is 2 , one for CH₃Br and one for NaOH .
rate of reaction = k x [CH₃Br] [ NaOH]
.008 = k x .12 x .12
k = .55555
when concentration of CH₃Br is doubled
rate of reaction = .555555 x [.24] [ .12 ]
= .016 M/s
when concentration of NaOH is halved
rate of reaction = .555555 x [.12] [ .06 ]
= .004 M/s
when concentration of both CH₃Br and Na OH is made 5 times
rate of reaction = .555555 x .6 x .6
= 0.2 M/s
Answer:
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