Answer:
- <em>Chemical equations are balanced </em><u>to comply with the law of conservation of mass.</u>
Explanation:
Law of conservation of mass states that matter cannot be either created or destroyed.
A skeleton chemical equation shows the reactants and products of a chemical reaction without taking into account the real proportion in which the reactants combine and the products are obtained.
An example of a skeleton reaction is the combustion of methane:
Such as that equation is shown, there are four atoms of hydrogen in the reactants but only 2 atoms of hydrogen in the products. Also, there are 2 atoms of oxygen in the reactants but three atoms of oxygen in the products. This seems to show that some atoms of hydrogen have been destroyed and some atoms of oxygen have been created. This is impossible as it is against the law of conservation of matter.
Then, to show a real situation, the chemical equation of combustion must be balanced, adjusting the coefficients. This is the balanced chemical equation:
Now you see that the number of atoms of each matter is conserved: the number of carbon atoms in each side is 1, the number of atoms of hydrogen in each side is 4, and the number of atoms of oxygen in each side is 4. Thus, by balancing the chemical equation, the law of conservation of mass is not violated.
I Believe the answer to your question is C.Protista and D.Animalia and B.Eubacteria
Answer:
4.8 grams
Explanation:
Use PV=nRT
P: 775 mmHg (divide by 760 mmHg to get atm) -> 1.02 atm
V: 3700 mL (divide by 1000 to get L) -> 3.7L
n: ?
R: (a constant) 0.0821L * atm/k *mol
T: 33 C (add 273 to get K) -> 306K
Move equation so n is on the side: PV/RT = n. Plug the numbers into the equation.

Then, convert moles to grams using the molar mass of O2 which is 32g/mol.
= 4.8g
The expansion diameters of the steel cylinder is 0.3mm
The expansion stress on the steel cylinder is 150MPa
<h3>Diameter of thin walled cylinder </h3>
t = Dp / 2s
low pressure cylinder of cast iron used for certain engine
t = Dp / 2500 + 0.3
wall thickness = 0.3 mm
p = internal pressure
d = inside diameter of cylinder
t = wall thikness
s = allowable tensile stress
<h3>The stress of the cylinder is</h3>
The pressure in a thin walled tube with diameter 0.3 m and thickness 0.001 m is 1000 kPa (10 bar).
The hoop stress can be calculated
σh = (1000 kPa) (0.3 m) / (2 (0.001 m))
= 150000 kPa
= 150 MPa
Hence , The diameter of the thin walled cylinder is 0.3mm
The stress of the thin walled cylinder is 150 MPa
Learn more about the thin walled cylinder on
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