Answer: Pink R-groups contain primarily C and H, which have approximately equal electronegativities and lack O; most green R-groups also contain a highlyelectronegative O atom.Green R-groups contain primarily C and H, which have approximately equal electronegativities and lack O; most pink R-groups contain a highlyelectronegative O atom
Explanation: I've been researching it for a while now
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2AgNO3+K2CrO4⇒Ag2CrO4(s)+2KNO3
Hence by mixing 0.0024M AgNO3 and 0.004M
K2CrO4, we will have Ag2CrO4 which is precipitated out and leave us with
0.0024M KN03 which is mixed with (0.004-0.0024/2)M, it can be 0.0028M, of K2Cr04
The specialized type of cell in human eyes which help to detect color, a layer at the back of the eyeball containing cells that are sensitive to light and that trigger nerve impulses that pass via the optic nerve to the brain, that is called a retina.
Answer:
pHe = 3.2 × 10⁻³ atm
pNe = 2.5 × 10⁻³ atm
P = 5.7 × 10⁻³ atm
Explanation:
Given data
Volume = 1.00 L
Temperature = 25°C + 273 = 298 K
mHe = 0.52 mg = 0.52 × 10⁻³ g
mNe = 2.05 mg = 2.05 × 10⁻³ g
The molar mass of He is 4.00 g/mol. The moles of He are:
0.52 × 10⁻³ g × (1 mol / 4.00 g) = 1.3 × 10⁻⁴ mol
We can find the partial pressure of He using the ideal gas equation.
P × V = n × R × T
P × 1.00 L = 1.3 × 10⁻⁴ mol × (0.082 atm.L/mol.K) × 298 K
P = 3.2 × 10⁻³ atm
The molar mass of Ne is 20.18 g/mol. The moles of Ne are:
2.05 × 10⁻³ g × (1 mol / 20.18 g) = 1.02 × 10⁻⁴ mol
We can find the partial pressure of Ne using the ideal gas equation.
P × V = n × R × T
P × 1.00 L = 1.02 × 10⁻⁴ mol × (0.082 atm.L/mol.K) × 298 K
P = 2.5 × 10⁻³ atm
The total pressure is the sum of the partial pressures.
P = 3.2 × 10⁻³ atm + 2.5 × 10⁻³ atm = 5.7 × 10⁻³ atm
