Answer : The balanced chemical equation is,
![2MnO_4^-(aq)+Br^-(aq)+H_2O(l)\rightarrow 2MnO_2(s)+BrO_3^-(aq)+2OH^-(aq)](https://tex.z-dn.net/?f=2MnO_4%5E-%28aq%29%2BBr%5E-%28aq%29%2BH_2O%28l%29%5Crightarrow%202MnO_2%28s%29%2BBrO_3%5E-%28aq%29%2B2OH%5E-%28aq%29)
Explanation :
Rules for the balanced chemical equation in basic solution are :
- First we have to write into the two half-reactions.
- Now balance the main atoms in the reaction.
- Now balance the hydrogen and oxygen atoms on both the sides of the reaction.
- If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the more number of oxygen are present.
- If the hydrogen atoms are not balanced on both the sides then adding hydroxide ion
at that side where the less number of hydrogen are present. - Now balance the charge.
The half reactions in the basic solution are :
Reduction :
......(1)
Oxidation :
.......(2)
Now multiply the equation (1) by 2 and then added both equation, we get the balanced redox reaction.
The balanced chemical equation in a basic solution will be,
![2MnO_4^-(aq)+Br^-(aq)+H_2O(l)\rightarrow 2MnO_2(s)+BrO_3^-(aq)+2OH^-(aq)](https://tex.z-dn.net/?f=2MnO_4%5E-%28aq%29%2BBr%5E-%28aq%29%2BH_2O%28l%29%5Crightarrow%202MnO_2%28s%29%2BBrO_3%5E-%28aq%29%2B2OH%5E-%28aq%29)
Answer:All six of the ions contain 10 electrons in the 1s, 2s, and 2p orbitals, but the nuclear charge varies from +7 (N) to +13 (Al)
Explanation:
hope this helped if it did may i have brainliest
Answer:
So first thing to do in these types of problems is write out your chemical reaction and balance it:
Mg + O2 --> MgO
Then you need to start thinking about moles of Magnesium for moles of Magnesium Oxide. Based on the above equation 1 mole of Magnesium is needed to make one mole of Magnesium Oxide.
To get moles of magnesium you need to take the grams you started with (.418) and convert to moles by dividing by molecular weight of Mg (24.305), this gives you .0172 moles of Mg.
The theoretical yield would be the assumption that 100% of the magnesium will be converted into Magnesium Oxide, so you would get, based on the first equation, .0172 mol of MgO. Multiplying this by the molecular weight of MgO (24.305+16) gives us .693 g of MgO.
The percent yield is what you actually got in the experiment, and for this you subtract off the total mass from the crucible mass, or 27.374 - 26.687, which gives .66 g of MgO obtained.
Percent yield is acutal/theoretical, .66/.693, or 95.24%.
I'll let you do the same for the second trial, and average percent yield is just an average of the two trials percent yield.
Hope this helps.
Answer: Thus the new volume of the gas is 530 ml
Explanation:
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
The combined gas equation is,
![\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}](https://tex.z-dn.net/?f=%5Cfrac%7BP_1V_1%7D%7BT_1%7D%3D%5Cfrac%7BP_2V_2%7D%7BT_2%7D)
where,
= initial pressure of gas = 740 torr
= final pressure of gas = 760 torr
= initial volume of gas = 500 ml
= final volume of gas = ?
= initial temperature of gas = ![25^oC=273+25=298K](https://tex.z-dn.net/?f=25%5EoC%3D273%2B25%3D298K)
= final temperature of gas = ![50^oC=273+50=323K](https://tex.z-dn.net/?f=50%5EoC%3D273%2B50%3D323K)
Now put all the given values in the above equation, we get:
![\frac{740\times 500}{298}=\frac{760\times V_2}{323}](https://tex.z-dn.net/?f=%5Cfrac%7B740%5Ctimes%20500%7D%7B298%7D%3D%5Cfrac%7B760%5Ctimes%20V_2%7D%7B323%7D)
![V_2=530ml](https://tex.z-dn.net/?f=V_2%3D530ml)
Thus the new volume of the gas is 530 ml