Answer:
Sample A is a mixture
Sample B is a mixture
Explanation:
For sample A, we are told that the originally yellow solid was dissolved and we obtained an orange powder at the bottom of the beaker. Subsequently, only about 30.0 g of solid was recovered out of the 50.0g of solid dissolved. This implies that the solid is not pure and must be a mixture. The other components of the mixture must have remained in solution accounting for the loss in mass of solid obtained.
For sample B, we are told that boiling started at 66.2°C and continued until 76.0°C. The implication of this is that B must be a mixture since it boils over a range of temperatures. Pure substances have a sharp boiling point.
Answer:
Reaction 1 is balanced but 2 is not balanced , the balance equation are :
1. 
2.
Explanation:
Balanced Equations : These are the equation which follows the law of conservation of mass .
The total number of atoms present in reactant is equal to total number of atoms present in product.
1. 
This is acid - base type reaction where
act as Acid
act as weak base
Reactant :
,
Number of atoms of :
C = 2 (
) + 1 (
)
= 2 + 1
= 3
H = 4(
) + 1 (
)
= 4 + 1
5
O = 2(
) + 3 (
)
= 5
Na = 1 (
)
= 1
Product :
,
, 
Number of atoms :
C = 1(
) + 2(
)
= 1 + 2
= 3
H = 2(
) + 3(
)
= 2 + 3
= 5
O = 1(
) + 2(
)
+2(
= 1 + 2 + 2
= 5
Na = 1(
= 1
Number of Na =1 , C = 3 , H= 5 and O =5 in both reactant and product , so it is a balanced reaction
2.
This is double displacement reaction .
Check the balancing in both reactant and products should be :
Na = 2
H = 2
Ca = 1
C = 2
O = 6
Cl = 2
pH=4.625
The classification of this sample of saliva : acid
<h3>Further explanation</h3>
The water equilibrium constant (Kw) is the product of concentration
the ions:
Kw = [H₃O⁺] [OH⁻]
Kw value at 25° C = 10⁻¹⁴
It is known [OH-] = 4.22 x 10⁻¹⁰ M
then the concentration of H₃O⁺:
![\tt 10^{-14}=4.22\times 10^{-10}\times [H_3O^+]\\\\(H_3O^+]=\dfrac{10^{-14}}{4.22\times 10^{-10}}=2.37\times 10^{-5}](https://tex.z-dn.net/?f=%5Ctt%2010%5E%7B-14%7D%3D4.22%5Ctimes%2010%5E%7B-10%7D%5Ctimes%20%5BH_3O%5E%2B%5D%5C%5C%5C%5C%28H_3O%5E%2B%5D%3D%5Cdfrac%7B10%5E%7B-14%7D%7D%7B4.22%5Ctimes%2010%5E%7B-10%7D%7D%3D2.37%5Ctimes%2010%5E%7B-5%7D)
pH=-log[H₃O⁺]
Saliva⇒acid(pH<7)
The answer is (2) higher vapor pressure and weaker intermolecular forces. Propanone has a lower boiling point, so it is more volatile than water. Propanone's vapor pressure is, therefore, higher than that of water at 50 degrees Celsius. Propanone is more volatile due to the fact that the intermolecular forces that hold its molecuels together are not as strong as those that hold together molecules of water. Since the IMFs are weaker, it takes less thermal energy to break individual molecules free of each other.
Answer:

Explanation:
Data:
50/50 ethylene glycol (EG):water
V = 4.70 gal
ρ(EG) = 1.11 g/mL
ρ(water) = 0.988 g/mL
Calculations:
The formula for the boiling point elevation ΔTb is

i is the van’t Hoff factor — the number of moles of particles you get from 1 mol of solute. For EG, i = 1.
1. Moles of EG

2. Kilograms of water

3. Molal concentration of EG

4. Increase in boiling point

5. Boiling point