I just looked it up and I found this:
Polar substances are likely to dissolve in polar solvents
Hope this helps!
Answer: 1. 1.59 x 10^23 Particles
2. 4.79 x 10^21 particles
3. 2.67 x 10^25 particles
4. 2.12 x 10^23 Particles
Explanation: 1mole of any substance contains 6.02x10^23 Particles
1. 0.264 mol of silver will contain = 0.264 x 6.02x10^23 = 1.59 x 10^23 Particles
2. 7.95 x 10^-3 mol sodium chloride will contain = 7.95 x 10^-3 X 6.02x10^23 = 4.79 x 10^21 particles
3. 44.4 mol carbon dioxide will contain = 44.4 x 6.02x10^23 = 2.67 x 10^25 particles
4. 0.352 mol nitrogen will contain = 0.352 x 6.02x10^23 = 2.12 x 10^23 Particles
Answer:
Explanation:
1. find the molar mass (amu) of each element and add them to get the whole molar mass.
2. divide the 1 element molar mass with the whole molar mass
3. multiple by 100 and that gives you the % composition.
<h2><u><em>56-57: NaCl</em></u></h2>
1. Na(22.99amu) + Cl (35.453amu)=58.443
2(Na): 
= .393
2(Cl): 
= .607
3(Na): .393 * 100=39.3%
3(Cl): .607 * 100= 60.7%
<h2><u>58-60 </u>
<u /></h2>
1. K: (39.098)(2)=78.196
_ C: (12.011)(1)= 12.011
_O: (15.99)(3) = 47.997
78.196+12.011+47.997= 138.204
2:K: 
= .566 <u>Step </u>3: (.566)(100)= 56.6%
2: C: 
= .087 <u>Step 3</u>: (.087)(100)= 8.7%
2: O: 
= .347 <u>Step 3</u>: (.347)(100) = 34.7%
<h2>
61-62 
</h2>
1. Fe (55.845)(3)= 167.535
_ O (15.999)(4) = 63.996
167.535+63.996=231.531
2: Fe: 
= .724 Step 3: (.724)(100)= 72.4%
2: O : 
= .276 Step 3: (.276)(100) = 27.6%
<h2>63-65
</h2>
1.
C(12.011*3)=36.033
H(1.008*5)=5.04 + (1.008*3)=3.024 so its 8.064
O(15.999*3)=47.997
add them: 92.094
2: C: 
= .391 Step 3: (.391)(100) = 39.1%
2: H: 
= .088 step 3: (.088)(100) = 8.8%
2: O: 
= .521 step 3: (.521)(100) = 52.1%
Answer:
Option C. +150KJ
Explanation:
Data obtained from the question include:
Heat of reactant (Hr) = 200KJ
Heat of product (Hp) = 350KJ
Change in enthalphy (ΔH) =..?
The enthalphy of the reaction can be obtained as follow:
Change in enthalphy (ΔH) = Heat of reactant (Hp) – Heat of reactant (Hr)
ΔH = Hp – Hr
ΔH = 350 – 200
ΔH = +150KJ
Therefore, the enthalphy for the reaction above is +150KJ
The unit cell of Aluminum is a face-centered cubic structure. This lattice unit cell consists of 4 atoms per unit cell. The volume of the unit cell is a³ where a is the length of the side of the unit cell. In terms of radius r, volume is equal to:
V = (4r/√2)³
V = (4(0.143×10⁻⁹ m)/√2)³
V = 6.62×10⁻²⁹ m³
