Weather can be predicted only as probable, not definite because the daily weather conditions depends on winds and storms.
It is common observation that weather forecasts are often given as a probability and never in definite terms.
This is because, the weather condition at anytime depends on the temperature of the earth's atmosphere which causes air masses to move leading to wind.
The temperature of the earth's atmosphere changes frequently hence weather conditions also change frequently.
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The molarity is the number of moles in 1 L of the solution.
The mass of NH₃ given - 2.35 g
Molar mass of NH₃ - 17 g/mol
The number of NH₃ moles in 2.35 g - 2.35 g / 17 g/mol = 0.138 mol
The number of moles in 0.05 L solution - 0.138 mol
Therefore number of moles in 1 L - 0.138 mol / 0.05 L x 1L = 2.76 mol
Therefore molarity of NH₃ - 2.76 M
Answer:
Density: Physical Property
Flammability: Chemical Property
Solubility In Water: Physical Property
Reactivity With Water: Chemical Property
Melting Pot: Physical Property
Color: Physical Property
Odor: Physical Property
Explanation:
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Answer: He did not discuss about any of these.
Explanation: Dalton proposed some of the postulates for his atomic theory. They are:
1) Matter is made up of atoms which are not divisible.
2) Atoms of different elements combine in a fixed ratio to form compounds.
3) The atomic properties of given element are same including mass. This states that all the atoms of an element have same mass but the atoms of different elements have different masses.
4) No atoms are either created or destroyed during a chemical reaction.
5) Atoms of an element are identical in mass, size and other chemical and physical properties.
As it is visible from the postulates, he only discussed only about the atoms but not subatomic particles or isotopes.
Answer:
0.12693 mg/L
Explanation:
First we <u>calculate the concentration of compound X in the standard prior to dilution</u>:
- 10.751 mg / 100 mL = 0.10751 mg/mL
Then we <u>calculate the concentration of compound X in the standard after dilution</u>:
- 0.10751 mg/mL * 5 mL / 25 mL = 0.021502 mg/L
Now we calculate the<u> concentration of compound X in the sample</u>, using the <em>known concentration of standard and the given areas</em>:
- 2582 * 0.021502 mg/L ÷ 4374 = 0.012693 mg/L
Finally we <u>calculate the concentration of X in the sample prior to dilution</u>:
- 0.012693 mg/L * 50 mL / 5 mL = 0.12693 mg/L