Archaea<span> and </span>bacteria are tiny, single-cell organisms which cannot be seen by the naked human eye. They are microbes and both prokaryotes. This means that they do not have a nucleus and lack membrane-bound organelles.
Protists on the other hand are small eukaryotes. They can<span> be unicellular or multicellular, but most of them are unicellular which is characteristic they share with the bacteria and archaea.
</span><span>- They transcribe sections of their DNA code into RNA, then translate this RNA into protein using complex structures called ribosomes.
</span><span>- Have cell membranes made of chemicals called phospholipids.
</span>- Have similar biochemical processes, for example the process glycolysis (the process <span>used to break down glucose).</span>
Answer:
i have an answer but i can only show you because my teacher helped my on it and wrote it down for me to remember! hope this helps!!!
Explanation:
A 1.00 g sample ofn-hexane (C6H14) under-goes complete combustion with excess O2ina bomb calorimeter. The temperature of the1502 g of water surrounding the bomb risesfrom 22.64◦C to 29.30◦C. The heat capacityof the hardware component of the calorimeter(everything that is not water) is 4042 J/◦C.What is ΔUfor the combustion ofn-C6H14?One mole ofn-C6H14is 86.1 g.The specificheat of water is 4.184 J/g·◦C.1.-9.96×103kJ/mol2.-7.40×104kJ/mol3.-1.15×104kJ/mol4.-4.52×103kJ/mol5.-5.92×103kJ/molcorrectExplanation:mC6H8= 1.00 gmwater= 1502 gSH = 4.184 J/g·◦CHC = 4042 J/◦CΔT= 29.30◦C-22.64◦C = 6.66◦CThe increase in the water temperature is29.30◦C-22.64◦C = 6.66◦C. The amount ofheat responsible for this increase in tempera-ture for 1502 g of water isq= (6.66◦C)parenleftbigg4.184Jg·◦Cparenrightbigg(1502 g)= 41854 J = 41.85 kJThe amount of heat responsible for the warm-ing of the calorimeter isq= (6.66◦C)(4042 J/◦C)= 26920 J = 26.92 kJ
We are given with a 2.5 M stock solution of acetic acid and we are required to calculate the volume of the solution needed to prepare 100 milliliters of 0.5 M acetic acid solution. To solve this, we acquire the formula <span>Mconcentrated*Vconcentrated = Mdilute*Vdilute. That is 2.5 M*x=0.5M*100 ml where x is the volume of 2.5 M needed. x is equal to 20 ml. So we need 20 ml of 2.5 M solution and dilute to 100 ml using water as diluent.</span>
Answer:
39.3%
Explanation:
CaF2 + H2SO4 --> CaSO4 + 2HF
We must first determine the limiting reactant, the limiting reactant is the reactant that yields the least number of moles of products. The question explicitly says that H2SO4 is in excess so CaF2 is the limiting reactant hence:
For CaF2;
Number of moles reacted= mass/molar mass
Molar mass of CaF2= 78.07 g/mol
Number of moles reacted= 11g/78.07 g/mol = 0.14 moles of Calcium flouride
Since 1 mole of calcium fluoride yields two moles of 2 moles hydrogen fluoride
0.14 moles of calcium fluoride will yield 0.14×2= 0.28 moles of hydrogen fluoride
Mass of hydrogen fluoride formed (theoretical yield) = number of moles× molar mass
Molar mass of hydrogen fluoride= 20.01 g/mol
Mass of HF= 0.28 moles × 20.01 g/mol= 5.6 g ( theoretical yield of HF)
Actual yield of HF was given in the question as 2.2g
% yield of HF= actual yield/ theoretical yield ×100
%yield of HF= 2.2/5.6 ×100
% yield of HF= 39.3%
Ok so this all boils down to Newton's second law
Force = mass X acceleration
so the force = 10 X 3 = 30N
