Answer:
A = Molarity = 0.22 M
B = Molarity = 0.36 M
Explanation:
Given data:
For first solution:
number of moles = 0.550 mol
Volume of solution = 2.50 L
Molarity = ?
Molarity:
Formula:
Molarity = number of moles of solute / volume of solution in L.
Molarity = 0.550 mol / 2.50 L
Molarity = 0.22 M
For second solution:
Mass of NaCl = 15.7 g
Volume of solution = 709 mL or 709/1000 = 0.709 L
Molarity = ?
Solution:
Number of moles = mass / molar mass
Number of moles = 14.7 g/ 58.44 g/mol
Number of moles = 0.252 mol
Molarity:
Molarity = number of moles of solute / volume of solution in L.
Molarity = 0.252 mol / 0.709 L
Molarity = 0.36 M
Answer:
The balanced equation is:
2 HNO3 + Mg ---> Mg(NO3)2 + H2
From the equation, we can see that we need twice the moles of HNO3 than the moles of Mg
Moles of Mg:
Molar mass of Mg = 24 g/mol
Moles = Given mass / Molar Mass
Moles of Mg = 4.47 / 24 = 0.18 moles (approx)
Hence, 2(moles of Mg) = 0.36 moles of HNO3 will be consumed
Number of moles of HNO3 after the reaction is finished is the number of unreacted moles of HNO3
Unreacted moles of HNO3 = Total Moles - Moles consumed
Unreacted moles of HNO3 = 0.64 moles (approx)
Since we approximated the value of moles of Mg, the value of remaining moles of HNO3 will also be approximate
From the given options, we can see that 0.632 moles is the closest value to our answer
Therefore, 0.632 moles will remain after the reaction
Answer:

Explanation:
Hello!
In this case, according to the following chemical reaction:

It means that we need to compute the moles of hydrogen and oxygen that are reacting, via the ideal gas equation as we know the volume, pressure and temperature:

Thus, the yielded moles of water are computed by firstly identifying the limiting reactant:

Thus, the fewest moles of water are 0.0609 mol so the limiting reactant is oxygen; in such a way, by using the ideal gas equation once again, we compute the pressure of water:

Best regards!
Answer:
5 seconds
Explanation:
Speed x Time. So t=ds. t=51=5.
Answer:
ans is (2) 2,4- hexadiene