Answer:
The genetic material of a prokaryotic cell is found in the cytoplasm.
Explanation:
Answer:
a) 72 °F= 22.22 °C
b) 213.8 °C= 416.84°F
c) 180 °C= 453.15 °K
d) 315 °K= 107.33 °F
e) 1750 °F= 1227.594 °K
f) 0 °K= -459.67 °F
Explanation:
Para realizar el intercambio de unidades debes tener en cuenta las siguientes conversiones:
- Fahrenheit a Celsius:

- Celsius a Fahrenheit: °F= °C*1.8 + 32
- Celsius a Kelvin: °K= °C + 273.15
- Kelvin a Fahrenheit: F= (K -273.15)*1.8 + 32
- Fahrenheit a Kelvin:

Entonces se obtiene:
a) 72 °F=
=22.22 °C
b) 213.8 °C= 213.8*1.8 + 32= 416.84°F
c) 180 °C= 180°C + 273.15= 453.15 °K
d) 315 °K= (315 -273.15)*1.8 + 32= 107.33 °F
e) 1750 °F=
= 1227.594 °K
f) 0 °K= (0 -273.15)*1.8 + 32= -459.67 °F
Answer:
Molality = 0.0862 mole/kg
Explanation:
Molality = (number of moles of solute)/(mass of solvent in kg)
Number of moles of solute = (mass of Creatinine in the blood sample)/(Molar mass of Creatinine)
To obtain the mass of creatinine in 10 mL of blood. We're told that 1 mg of Creatinine is contained in 1 decilitre of blood.
1 decilitre = 100 mL
1 mg of Creatinine is contained in 100 mL of blood
x mg of Creatinine is contained in 10 mL of blood.
x = (1×10/100) = 0.1 mg = 0.0001 g
Molar mass of Creatinine (C₄H₇N₃O) = 113.12 g/mol
Number of moles of Creatinine in the 10 mL blood sample = (0.0001/113.12) = 0.000000884 moles
Mass of 10 mL of blood = density × volume = 1.025 × 10 = 10.25 mg = 0.01025 g = 0.00001025 kg
Molality of normal creatinine level in a 10.0-ml blood sample = (0.000000884/0.00001025)
Molality = 0.0862 moles of Creatinine per kg of blood.
Hope this Helps!!!
Via the half-life equation we have:

Where A is the amount initially and final amount after some time has passed, t is the time elapsed, h is the half life time and so:

Therefore there will be 2.8125 grams left after 140 seconds.