Answer:
0.12693 mg/L
Explanation:
First we <u>calculate the concentration of compound X in the standard prior to dilution</u>:
- 10.751 mg / 100 mL = 0.10751 mg/mL
Then we <u>calculate the concentration of compound X in the standard after dilution</u>:
- 0.10751 mg/mL * 5 mL / 25 mL = 0.021502 mg/L
Now we calculate the<u> concentration of compound X in the sample</u>, using the <em>known concentration of standard and the given areas</em>:
- 2582 * 0.021502 mg/L ÷ 4374 = 0.012693 mg/L
Finally we <u>calculate the concentration of X in the sample prior to dilution</u>:
- 0.012693 mg/L * 50 mL / 5 mL = 0.12693 mg/L
The simplified solubility of glucose at 30°C is 1.25 g/g of water. Considering that the density of water at 30°C is 1 g/mL, the equivalent mass of 400 mL of water is also 400g.
The concentration of the solution in water is,
550 g/400g of water = 1.375 g glucose / g of water
Since the concentration is higher compared to the solubility of glucose at the specified temperature, it can be said that the solution is SATURATED.
A. 1.01 is the right answer
Since
The formula is Pv= nRT
P=1 atm
V= 22.4 L
N= x
r= 0.0821
t = 273 k (bc it’s standard temperature)
So (1)(22.4)=(x)(0.0821)(273)
X= 1.001
Answer:
a. The atom will go from a two-dimensional configuration to a three dimensional configuration.
d. The bond angle will increase.
f. The number of unhybridized p orbitals will decrease.
Explanation:
Sp2 is the atomic bond in which orbitals mixes with only two orbitals. These orbitals form three sp2. When two carbon atoms are overlapped they form sigma bond by overlapping of sp2 bonds. Sp3 bond is created when there is one lone molecule available for combination. When the bonding is updated from sp2 to sp3 then unhybridized orbitals will decrease causing the bond angle to increase.
Mass of methanol (CH3OH) = 1.922 g
Change in Temperature (t) = 4.20°C
Heat capacity of the bomb plus water = 10.4 KJ/oC
The heat absorbed by the bomb and water is equal to the product of the heat capacity and the temperature change.
Let’s assume that no heat is lost to the surroundings. First, let’s calculate the heat changes in the calorimeter. This is calculated using the formula shown below:
qcal = Ccalt
Where, qcal = heat of reaction
Ccal = heat capacity of calorimeter
t = change in temperature of the sample
Now, let’s calculate qcal:
qcal = (10.4 kJ/°C)(4.20°C)
= 43.68 kJ
Always qsys = qcal + qrxn = 0,
qrxn = -43.68 kJ
The heat change of the reaction is - 43.68 kJ which is the heat released by the combustion of 1.922 g of CH3OH. Therefore, the conversion factor is:
