Pure water has a freezing point of 0 degrees centigrade but seawater has a lower freezing point because of the salt (NaCl)
Answer: The enthalpy of combustion, per mole, of butane is -2657.4 kJ
Explanation:
The balanced chemical reaction is,
The expression for enthalpy change is,
Putting the values we get :
2 moles of butane releases heat = 5314.8 kJ
1 mole of butane release heat = 
Thus enthalpy of combustion per mole of butane is -2657.4 kJ
Answer: option (1) an electron.
Justification:
1) The plum pudding model of the atom conceived by the scientist J.J. Thompson, described the atom as a solid sphere positively charged with the electrons (particles negatively charged) embedded.
2) The next model of the atom, developed by the scientist Ernest Rutherford, depicted the atom a mostly empty space with a small dense positively charged nucleous and the electrons surrounding it.
3) Then, Niels Bhor came out with the model of electrons in fixed orbits around the nucleous, just like the planets orbit the Sun. So, the path followed by the electrons were orbits.
4) The quantum model of the atom did not place the electrons in fixed orbits around the nucleous but in regions around the nucleous. Those regions were named orbitals. And they are regions were it is most probable to find the electron, since it is not possible to tell the exact position of an electron.
As per this model, the electron has a wave function associated. The scientist Schrodinger developed the wave equation which predicts the location of the electron as a probability.
The orbitals are those regions were it is most likely to find the electron. Those regions are thought as clouds of electrons.
I think the correct answers are X2Y and X3Y, X2Y5 and X3Y5, and X4Y2 and X3Y,
for the following reason:
If you look at the combining masses of X and Y in
each of the two compounds,
The first compound contains 0.25g of X combined with
0.75g of Y
so the ratio (by mass) of X to Y = 1 : 3
The second compound contains 0.33 g of X combined with
0.67 g of Y
so the ratio (by mass) of X to Y = 1 : 2
Now, you suppose to prepare each of these two
compounds, starting with the same fixed mass of element Y ( I will choose 12g
of Y for an easy calculation!)
The first compound will then contain 4g of X and 12g
of Y
The second compound will then contain 6g of X and
12g of Y
<span>The ratio which combined
the masses of X and the fixed mass (12g) of Y
= 4 : 6
<span>or 2 : 3 </span>
So, the ratio of MOLES of X which combined with the
fixed amount of Y in the two compounds is also = 2 : 3 </span>
The two compounds given with the plausible formula must therefore contain
the same ratio.