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loris [4]
3 years ago
9

Three things about the body systems?

Chemistry
1 answer:
makvit [3.9K]3 years ago
5 0

Answer:

Three things about our body's systems:

All systems have a method of self-regulation or exogenous regulation by other systems.

All systems have a balance in their functions.

All the systems of our organism are intertwined with each other thus giving general vitality.

Explanation:

Best known systems:

Renal, respiratory, circulatory, cardiac, nervous, immune, blood, muscular systems.

All of them include the participation of one or more organs

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Which of there is an indication that a chemical reaction has occurred?
Komok [63]
Answer is (b) , because a chemical change happened
6 0
3 years ago
Suppose a student started with 142.0 mg of trans-cinnamic acid, 412 mg of pyridinium tribromide, and 2.30 mL of glacial acetic a
nirvana33 [79]

Answer: Theoretical Yield = 0.2952 g

               Percentage Yield = 75.3%

Explanation:

Calculation of limiting reactant:

n-trans-cinnamic acid moles = (142mg/1000) / 148.16 = 9.584*10⁻⁴ mol

pyridium tribromide moles = (412mg/1000) / 319.82= 1.288*10⁻³ mol

  • n-trans-cinnamic acid is the limiting reactant

The molar ratio according to the equation mentioned is equals to 1:1

The brominated product moles is also = 9.584*10⁻⁴ mol

Theoretical yield = (9.584*10⁻⁴ mol) * (Mr of brominated product)

                             =  (9.584*10⁻⁴ mol) * (307.97) = 0.2952 g

Percentage Yield is : Actual Yield / Theoretical Yield = 0.2223/0.2952

                                                                                           = 75.3%

4 0
3 years ago
combustion analysis of a hydrocarbon produced 33.01g CO2 and 13.51g H2O. Calculate the empirical formula for the hydrocarbon
masya89 [10]

Answer:

\rm CH_2.

Explanation:

Carbon and hydrogen are the only two elements in a hydrocarbon. When a hydrocarbon combusts completely in excess oxygen, the products would be \rm CO_2 and \rm H_2O. The \rm C and \rm H would come from the hydrocarbon, while the \rm O atoms would come from oxygen.

Look up the relative atomic mass of these three elements on a modern periodic table:

  • \rm C: 12.011.
  • \rm H: 1.008.
  • \rm O: \rm 15.999.

Calculate the molar mass of \rm CO_2 and \rm H_2O:

M(\mathrm{CO_2}) = 12.011 + 2 \times 15.999 = 44.009\; \rm g \cdot mol^{-1}.

M(\mathrm{H_2O}) = 2 \times 1.008 + 15.999 = 18.015\; \rm g \cdot mol^{-1}

Calculate the number of moles of \rm CO_2 molecules in 33.01\; \rm g of \rm CO_2\!:

\displaystyle n(\mathrm{CO_2}) = \frac{m(\mathrm{CO_2})}{M(\mathrm{CO2})} = \frac{33.01\; \rm g}{44.009\; \rm g\cdot mol^{-1}} \approx 0.7501\; \rm mol.

Similarly, calculate the number of moles of \rm H_2O molecules in 13.51\; \rm g of \rm H_2O\!:

\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = \frac{13.51\; \rm g}{18.015\; \rm g\cdot mol^{-1}} \approx 0.7499\; \rm mol.

Note that there is one carbon atom in every \rm CO_2 molecule. Approximately0.7501\; \rm mol of \rm CO_2\! molecules would correspond to the same number of \rm C atoms. That is: n(\mathrm{C}) \approx 0.7501\; \rm mol.

On the other hand, there are two hydrogen atoms in every \rm H_2O molecule. approximately 0.7499\; \rm mol of \rm H_2O molecules would correspond to twice as many \rm H\! atoms. That is: n(\mathrm{H}) \approx 2 \times 0.7499 \; \rm mol\approx 1.500\; \rm mol.

The ratio between the two is: n(\mathrm{C}): n(\mathrm{H}) \approx 1:2.

The empirical formula of a compound gives the smallest whole-number ratio between the elements. For this hydrocarbon, the empirical formula would be \rm CH_2.

6 0
3 years ago
What percentage of known species are invertebrates?
ehidna [41]

95 percent of species known are invertebrates

6 0
3 years ago
A chemical company makes two brands of antifreeze. the first brand is 70% pure antifreeze, and the second brand is 95% pure anti
madreJ [45]

Answer is: 56 gallons of 70% antifreeze and 84 gallons of 95% antifreeze.


ω₁ = 70% ÷ 100% = 0.7; 70% pure antifreeze.

ω₂ = 95% ÷ 100% = 0.95.

ω₃<span> = 85% ÷ 100% = 0.85.
V</span>₁ = ?; volume of 70% antifreeze.

V₂ = ?; volume of 95% antifreeze.<span>
V</span>₃ = V₁ + V₂<span>.
V</span>₃ = 140 gal.

V₁ = 140 gal - V₂<span>.
ω</span>₁ · V₁ + ω₂ ·V₂ = ω₃ · V₃.

0.70 · (140 gal - V₂) + 0.95 · V₂ = 0.85 · 140 gal.

98 gal - 0.7V₂ + 0.95V₂ = 119 gal.

0.25V₂ = 21 gal.

V₂ = 21 gal ÷ 0.25.

V₂ = 84 gal.

V₁ = 140 gal - 84 gal.

V₁ = 56 gal.

4 0
3 years ago
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