Answer:
Explanation:
Uncundcuhecuheduendudnueduhdeundeuednusjiswj8
Control group: 50 dogs continuing their normal diet
Experiments group: 50 dogs chosen to eat the new food
Independent variable: dog food
Dependent variable: the dogs’ weight
Answer:
11.31 g.
Explanation:
Molarity is defined as the no. of moles of a solute per 1.0 L of the solution.
M = (no. of moles of solute)/(V of the solution (L)).
<em>∴ M = (mass/molar mass)of NaCl/(V of the solution (L)).</em>
<em></em>
<em>∴ mass of NaCl remained after evaporation of water = (M)(V of the solution (L))(molar mass)</em> = (0.45 M)(0.43 L)(58.44 g/mol) = <em>11.31 g.</em>
Hello!
The process in which hydrogen nuclei combine with each other to form helium and other heavier elements is called Nuclear Fusion.
Nuclear Fusion happens when the repulsion between the two nuclei is overcome by the Strong Interaction Force, spending and releasing high amounts of energy.
The two nuclei need to be close enough for Nuclear Fusion to happen, and the energy needed to achieve that is given by the extremely hot temperatures that exist within stars (about 15 000 000 Kelvin). That's why this kind of reactions are common in stars.
Have a nice day!
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!