Answer:
Ethyl alcohol is soluble in water because <span>ethyl alcohol exhibits dipole-dipole and h-bonding interactions with water.
Explanation:
Ethyl alcohol and water are miscible in each other because both are polar in nature and "Like dissolves Like".
The bond between oxygen and hydrogen atoms, both in alcohol and water are polar in nature and results in intermolecular hydrogen bond interactions between them as hydrogen bonding results when hydrogen atom in one molecule directly attached to highly electronegative atoms like fluorine, oxygen and nitrogen forms interaction with higly electronegative atom of neighbor atom.</span>
The empirical formula of the following compounds 0.903 g of phosphorus combined with 6.99 g of bromine.
<h3>What is empirical formula?</h3>
The simplest whole number ratio of atoms in a compound is the empirical formula of a chemical compound in chemistry. Sulfur monoxide's empirical formula, SO, and disulfur dioxide's empirical formula, S2O2, are two straightforward examples of this idea. As a result, both the sulfur and oxygen compounds sulfur monoxide and disulfur dioxide have the same empirical formula.
<h3>
How to find the empirical formula?</h3>
Convert the given masses of phosphorus and bromine into moles by multiplying the reciprocal of their molar masses. The molar masses of phosphorus and bromine are 30.97 and 79.90 g/mol, respectively.
Moles phosphorus = 0.903 g phosphorus 
= 0.0293 mol
Moles bromine 6.99 g bromine
=0.0875 mol
The preliminary formula for compound is P0.0293Bro.0875. Divide all the subscripts by the subscript with the smallest value which is 0.0293. The empirical formula is P1.00Br2.99 ≈ P₁Br3 or PBr3
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Answer: The correct answer is -297 kJ.
Explanation:
To solve this problem, we want to modify each of the equations given to get the equation at the bottom of the photo. To do this, we realize that we need SO2 on the right side of the equation (as a product). This lets us know that we must reverse the first equation. This gives us:
2SO3 —> O2 + 2SO2 (196 kJ)
Remember that we take the opposite of the enthalpy change (reverse the sign) when we reverse the equation.
Now, both equations have double the coefficients that we would like (for example, there is 2S in the second equation when we need only S). This means we should multiply each equation (and their enthalpy changes) by 1/2. This gives us:
SO3 —>1/2O2 + SO2 (98 kJ)
S + 3/2O2 —> SO3 (-395 kJ)
Now, we add the two equations together. Notice that the SO3 in the reactants in the first equation and the SO3 in the products of the second equation cancel. Also note that O2 is present on both sides of the equation, so we must subtract 3/2 - 1/2, giving us a net 1O2 on the left side of the equation.
S + O2 —> SO2
Now, we must add the enthalpies together to get our final answer.
-395 kJ + 98 kJ = -297 kJ
Hope this helps!
I would say CuSO4 or Copper Sulfate, as option 1 is Methane and would create a fire so heat, and the last one is Sugar which doesn't conduct electricity. And C6H6 I believe is not soluble in water.<span />
Answer:
The first ionization energy for K is less than Ca because Ca has a larger effective nuclear charge.
