For the first blank, that is the endoplasmic reticulum
For the second, it is lysosome
For the third blank, it is the cell membrane
For the fourth, sorry I don’t know this one
For the fifth, that is the vacuole
For the sixth, that is mitochondrion
For the seventh, that is Golgi body
And lastly the eighth, it is the nucleus
Sorry I did not know what the fourth was but everything else is good.
a. the ratio of mass to charge of an electron
Explanation:
The experiment permitted the direct measurement of the ratio of mass to charge of an electron.
- The charge to mass ratio of an electron was determine by accelerating a beam of cathode rays in magnetic and electric fields.
- No matter the gas used in the tube or the nature of the material of the electrodes, the rays were found to have constant charge to mass ratio of 1.76 x 10¹¹coulombkg⁻¹.
learn more:
Subatomic particles brainly.com/question/2757829
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Answer:
Final temperature = 
Explanation:
Given that,
Heat added, Q = 250 J
Mass, m = 30 g
Initial temperature, T₁ = 22°C
The Specific heat of Cu= 0.387 J/g °C
We know that, heat added due to the change in temperature is given by :
Put all the values,
So, the final temperature is equal to .
Answer:
b) add 130 g of NaCH₃CO₂ to 100 mL of H₂O at 80 °C while stirring until all the solid dissolves, then let the solution cool to room temperature.
Explanation:
The solubility of NaCH₃CO₂ in water is ~1.23 g/mL. This means that at room temperature, we can dissolve 1.23 g of solute in 1 mL of water (solvent).
<em>What would be the best method for preparing a supersaturated NaCH₃CO₂ solution?</em>
<em>a) add 130 g of NaCH₃CO₂ to 100 mL of H₂O at room temperature while stirring until all the solid dissolves.</em> NO. At room temperature, in 100 mL of H₂O can only be dissolved 123 g of solute. If we add 130 g of solute, 123 g will dissolve and the rest (7 g) will precipitate. The resulting solution will be saturated.
<em>b) add 130 g of NaCH₃CO₂ to 100 mL of H₂O at 80 °C while stirring until all the solid dissolves, then let the solution cool to room temperature. </em>YES. The solubility of NaCH₃CO₂ at 80 °C is ~1.50g/mL. If we add 130 g of solute at 80 °C and let it slowly cool (and without any perturbation), the resulting solution at room temperature will be supersaturated.
<em>c) add 1.23 g of NaCH₃CO₂ to 200 mL of H₂O at 80 °C while stirring until all the solid dissolves, then let the solution cool to room temperature.</em> NO. If we add 1.23 g of solute to 200 mL of water, the resulting solution will have a concentration of 1.23 g/200 mL = 0.00615 g/mL, which represents an unsaturated solution.
