Answer:
30.33L
Explanation:
Using Boyle's law which states that the volume of a given mass of gas is inversely proportional to the pressure, provided temperature remains constant and Charles law states that the volume of a given mass of gas is directly proportional to the temperature provided the pressure remains constant
P1V1/T1 = P2V2/ T2
P1 = 78atm, V1 = 21L , T1 = 900K
P2 = 45atm, V2 = ? , T2 =750K
78× 21 / 900 = 45×V2 / 750
1638/900 = 45 V2 / 750
1638×750 = 900×45V2
1228500 = 40500V2
Divide both sides by 40500
1228500÷40500= V2
V2 = 30.33L
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The molarity of KOH is 0.1055 M
<u><em> calculation</em></u>
Step 1: write the equation for reaction between H₂C₂O₄.2H₂O and KOH
H₂C₂O₄.2H₂O + 2 KOH → K₂C₂O₄ +4 H₂O
step 2: find the moles of H₂C₂O₄.2H₂O
moles = mass÷ molar mass
from periodic table the molar mass H₂C₂O₄.2H₂O= (1 x2) +(12 x2) +(16 x4) + 2(18)=126 g/mol
= 0.2000 g ÷ 126 g/mol =0.00159 moles
step 3: use the mole ratio to calculate the moles of KOH
H₂C₂O₄.2H₂O : KOH is 1:2
therefore the moles of KOH =0.00159 x 2 = 0.00318 moles
step 4: find molarity of KOH
molarity = moles/volume in liters
volume in liters = 30.12/1000=0.03012 L
molarity is therefore = 0.00318/0.03012 =0.1055 M
9ml will be given for the case of dosage calculation order: 3 mg available: 2 mg per 6 ml
Conversion factors are necessary for dosage calculation, such as when translating from pounds to kilograms or liters to milliliters. This approach, which is straightforward in design, enables physicians to deal with different units of measurement and convert factors to arrive at the solution.
dosage calculation techniques serve as a second or third check on the accuracy of the previous computation techniques. Dimensional Analysis, Ratio Proportion, and Formula or Desired Over Have Method are the three main approaches for dosage calculation. dosage calculations are frequently prescribed and labeled based on their weight or, for solutions, their strength, which is the amount of weight dissolved or suspended in a given volume.
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Answer:
0.03g/mL
Explanation:
Given parameters include:
Five μL of a 10-to-1 dilution of a sample; This implies the Volume of dilute sample is given as 5 μL
Dilution factor = 10-to-1
The absorbance at 595 nm was 0.78
Mass of the diluted sample = 0.015 mg
We need to first determine the concentration of the diluted sample which is required in calculating the protein concentration of the original solution.
So, to determine the concentration of the diluted sample, we have:
concentration of diluted sample = 
= 
(where ∝ was use in place of μ in the expressed fraction)
= 0.003 mg/μL
The dilution of the sample is from 10-to-1 indicating that the original concentration is ten times higher; as such the protein concentration of the original solution can be calculated as:
protein concentration of the original solution = 10 × concentration of the diluted sample.
= 10 × 0.003 mg/μL
= 0.03 mg/μL
= 0.03g/mL
Hence, the protein concentration of the original solution is known to be 0.03g/mL
