Answer:
30 Liters of 40% acid solution and 10 L of 60% acid solution is needed.
Explanation:
Let volume of the 40% acid solution be x.
Let volume of the 60% acid solution be y.
Volume of solution formed after mixing both solution = 40 L
x + y = 40 L..[1]
Volume of acid 40% solution = 40% of x= 0.4x
Volume of acid 60% solution = 60% of y= 0.6y
Volume of acid formed = 45% of 40 L = 
..[2]
Solving [1] and [2]
x = 30 L , y = 10 L
30 Liters of 40% acid solution and 10 L of 60% acid solution is needed.
Answer:
163.2g
Explanation:
First let us generate a balanced equation for the reaction. This is shown below:
4Al + 3O2 —> 2Al2O3
From the question given, were were told that 3.2moles of aluminium was exposed to 2.7moles of oxygen. Judging by this, oxygen is excess.
From the equation,
4moles of Al produced 2moles of Al2O3.
Therefore, 3.2moles of Al will produce = (3.2x2)/4 = 1.6mol of Al2O3.
Now, let us covert 1.6mol of Al2O3 to obtain the theoretical yield. This is illustrated below:
Mole of Al2O3 = 1.6mole
Molar Mass of Al2O3 = (27x2) + (16x3) = 54 + 48 =102g/mol
Mass of Al2O3 =?
Number of mole = Mass /Molar Mass
Mass = number of mole x molar Mass
Mass of Al2O3 = 1.6 x 102 = 163.2g
Therefore the theoretical of Al2O3 is 163.2g
The ocean hold 97 percent of earths water.
The correct answer is 30 seconds
Answer:

Explanation:
Hello!
In this case, since the equation we use to model the heat exchange into the calorimeter and compute the heat of reaction is:

We plug in the mass of water, temperature change and specific heat to obtain:

Now, this enthalpy of reaction corresponds to the combustion of propyne:

Whose enthalpy change involves the enthalpies of formation of propyne, carbon dioxide and water, considering that of propyne is the target:

However, the enthalpy of reaction should be expressed in kJ per moles of C3H4, so we divide by the appropriate moles in 7.00 g of this compound:

Now, we solve for the enthalpy of formation of C3H4 as shown below:

So we plug in to obtain (enthalpies of formation of CO2 and H2O are found on NIST data base):

Best regards!