Radium-221 has a half-life of 30 s. how long will it take for 95% of a sample to decay
this would be 2.22 seconds
Mizar is a quadruple or quaternary star system.
When there is more than two stars, it is called a star system or multiple star system or stellar system.
Depending on the number of stars, they are also called a triple, quadruple or quintuple star system or trinary, ternary, quaternary, quintenary, sextuple, sextenary, septuple or septenary star system.
Hydrogen should be your final answer.
Answer:
162 kJ
Explanation:
The reaction given by the problem is:
- NBr₃(g) + 3 H₂O(g) → 3 HOBr(g) + NH₃(g) ∆H = +81 kJ
If we turn it around, we have:
- 3 HOBr(g) + NH₃(g) → NBr₃(g) + 3 H₂O(g) ∆H = -81 kJ
If we think now of HOBr and NH₃ as our reactants, then now we need to find out <u>which one will be the </u><em><u>limiting reactant</u></em> when we have 9 moles of HOBr and 2 moles of NH₃:
- When we have 1 mol NH₃, we need 3 mol HOBr. So when we have 2 moles NH₃, we need 6 moles HOBr. We have more than 6 moles HOBr so that's our <em>reactant in excess</em>, thus NH₃ is our limiting reactant.
-81 kJ is our energy change when there's one mol of NH₃ reacting, so we <u>multiply that value by two when there's two moles of NH₃ reacting</u>. The answer is 81*2 = 162 kJ.
Answer:- 10 L of ethane.
Solution:- The given balanced equation is:
From this equation, ethane and oxygen react in 2:7 mol ratio, the ratio of volumes would also be same if they are at same temperature and pressure.
Since 14 L of each gas are taken, the oxygen will be the limiting reactant and ethane will be the excess reactant. Let's calculate the volume of ethane used:
=
From above calculations, 4 L of ethane are used. So, excess volume of ethane left after the completion of reaction = 14 L - 4 L = 10 L
Hence, 10 L of ethane will be remaining.