Answer:
Secondary research
Explanation:
Secondary research involves using already existing data, like what can be found in in books, articles, and electronic databases.
A is wrong. Primary research is the research that you do to obtain new information, as in experiments or in interviews and surveys.
B is wrong. Qualitative research is the gathering of non-numerical data.
C is wrong. Summative research is done at the end of a project to determine its success. It can also point the way to future research projects.
Answer:
Helps eliminate waste products such as urea, uric acid ammonia, and other products via urine.
It helps maintain the osmotic level of blood and plasma.
It helps maintain the electrolyte balance in the body.
And it also helps in the metabolism of those drugs that do not get metabolized in the liver.
Explanation:
<em>"The excretory system is a passive biological system that removes excess, unnecessary materials from the body fluids of an organism, so as to help maintain internal chemical homeostasis and prevent damage to the body. The dual function of excretory systems is the elimination of the waste products of metabolism and to drain the body of used up and broken down components in a liquid and gaseous state"</em>
A proton<span> is one of the most </span>important<span> types of subatomic particles. </span>Protons <span>combine with electrons and neutrons to </span>make<span> atoms. </span>Protons<span> are nearly the same size as neutrons and are much larger than electrons. ... </span>Protons<span> are composed of two up quarks and one down quark.......Hope this helps can you mark me brainliest?</span>
Answer:
Theoretical yield = 3.52 g
Percent yield =65.34%
Explanation:
Given data:
Mass of HgO = 46.8 g
Theoretical yield of O₂ = ?
Percent yield of O₂ = ?
Actual yield of O₂ = 2.30 g
Solution:
Chemical equation:
2HgO → 2Hg + O₂
Number of moles of HgO = mass/ molar mass
Number of moles of HgO = 46.8 g / 216.6 g/mol
Number of moles of HgO = 0.22 mol
Now we will compare the moles of HgO with oxygen.
HgO : O₂
2 : 1
0.22 : 1/2×0.22 = 0.11 mol
Theoretical yield:
Mass of oxygen = number of moles × molar mass
Mass of oxygen = 0.11 mol × 32 g/mol
Mass of oxygen = 3.52 g
Percent yield :
Percent yield = actual yield / theoretical yield × 100
Percent yield = 2.30 g/ 3.52 g × 100
Percent yield =65.34%
Answer:
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