Acid palmitic acid has higher melting point, because it has two more methylene groups.
Acid palmitic acid has higher melting point, because it has two more methylene groups.
Giving it a greater surface area and therefore more intermolecular van der waals interact than the myristic acid.
stearic arid 
linoleic acid 
(two double bond)
Stearic acid has higher Melting point, because it does not have any Carbon-Carbon double bonds, whereas linoleic acid has two cis double bonds which prevent the molecules from packing closely together.
Oleic Acid and Linoleic acid.
-one double bond (cis)
Acid palmitic acid has higher melting point, because it has two more methylene groups.
For more such question on methylene group.
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The false positive from the response of hydrogen peroxide and the immunizing circle would be created by poor specificity. The recipe for specificity is TN/TN+FP. False-positive outcomes can be ascribed to meddling substances in nature where the strips are put away or utilized, for example, hydrogen peroxide (H2O2) or fade (hypochlorite).
Answer:
C. Glaciers
Explanation:
Over 68 percent of the fresh water on Earth is found in icecaps and glaciers, and just over 30 percent is found in ground water. Only about 0.3 percent of our fresh water is found in the surface water of lakes, rivers, and swamps.
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Answer:
a) 24.7 mol
b) 790 g
Explanation:
Step 1: Given data
- Volume of the chamber (V): 200. L
- Room temperature (T): 23 °C
- Pressure of the gas (P): 3.00 atm
Step 2: Convert "T" to Kelvin
We will use the following expression.
K = °C + 273.15
K = 23°C + 273.15 = 296 K
Step 3: Calculate the moles (n) of oxygen
We will use the ideal gas equation.
P × V = n × R × T
n = P × V/R × T
n = 3.00 atm × 200. L/(0.0821 atm.L/mol.K) × 296 K = 24.7 mol
Step 4: Calculate the mass (m) corresponding to 24.7 moles of oxygen
The molar mass (M) of oxygen ga sis 32.00 g/mol. We will calculate the mass of oxygen using the following expression.
m = n × M
m = 24.7 mol × 32.00 g/mol = 790 g
1) Chemical reaction:
2(CH₃COO)₃Fe + 3MgCrO₄ → Fe₂(CrO₄)₃ + 3(CH₃COO)₂Mg.
m((CH₃COO)₃Fe) = 15,0 g.
m(MgCrO₄) = 10,0 g.
n((CH₃COO)₃Fe) = m((CH₃COO)₃Fe) ÷ M((CH₃COO)₃Fe).
n((CH₃COO)₃Fe) = 15 g ÷ 233 g/mol.
n((CH₃COO)₃Fe) = 0,064 mol.
n(MgCrO₄) = m(MgCrO₄) ÷ M(MgCrO₄).
n(MgCrO₄) = 10 g ÷ 140,3 g/mol.
n(MgCrO₄) = 0,071 mol; limiting reagens.
From chemical reaction: n(MgCrO₄) : n((CH₃COO)₂Mg) = 3 : 3.
n((CH₃COO)₂Mg) = 0,071 mol.
m((CH₃COO)₂Mg) = 0,071 mol · 142,4 g/mol.
n((CH₃COO)₂Mg) = 10,11 g.
2) Chemical reaction:
2(CH₃COO)₃Fe + 3MgSO₄ → Fe₂(SO₄)₃ + 3(CH₃COO)₂Mg.
m((CH₃COO)₃Fe) = 15,0 g.
m(MgSO₄) = 15,0 g.
n((CH₃COO)₃Fe) = m((CH₃COO)₃Fe) ÷ M((CH₃COO)₃Fe).
n((CH₃COO)₃Fe) = 15 g ÷ 233 g/mol.
n((CH₃COO)₃Fe) = 0,064 mol; limiting ragens.
n(MgSO₄) = m(MgSO₄) ÷ M(MgSO₄).
n(MgSO₄) = 15 g ÷ 120,36 g/mol.
n(MgSO₄) = 0,125 mol; limiting reagens.
From chemical reaction: n(CH₃COO)₃Fe) : n((CH₃COO)₂Mg) = 2 : 3.
n((CH₃COO)₂Mg) = 0,064 mol · 3/2.
n((CH₃COO)₂Mg) = 0,096 mol.
m((CH₃COO)₂Mg) = 0,096 mol · 142,4 g/mol.
m((CH₃COO)₂Mg) = 13,7 g.
