<h3>
Answer:</h3>
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
- Analyzing Reactions RxN
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] 2C + O₂ → 2CO₂
[Given] 0.25 moles O₂
[Solve] moles CO₂
<u>Step 2: Identify Conversions</u>
[RxN] 1 mol O₂ → 2 mol CO₂
<u>Step 3: Stoichiometry</u>
- [DA] Set up:
- [DA] Multiply/Divide [Cancel out units]:
107.9 should be it according to my calculations.Please mark as Brainliest!
Answer:
When mensuration volume of air within the flask at the primary temperature, a volume of 250mil was recorded, referred to as V1. The temperature of the air within the flask in boiling water was recorded as 99ᵒC, referred to as T2. so as to seek out the right calculations, 99ᵒC must be born-again to Kelvin by adding 273.
the primary temperature in Kelvin is 372K. the worth of V1/T1, may be found by swing 250/372. This involves a complete of zero.67. the quantity of the air within the flask of the second temperature was 177 mil, referred to as V2.
The temperature of the air within the cooled flask is 7ᵒC, referred to as T2. 7ᵒC must be born-again to Kelvin by adding 273 that involves a final total of 280K. the worth of V2/T2, found by swing 177/280 involves a complete of zero.63.
The close to equality in numbers may be attributed to Charles Law. Charles Law states that “as the temperature will increase, thus will the quantity of a gas sample once the pressure is command constant”.
The results of V1/T1 and V2/T2 were terribly on the brink of one another. this can be because of the very fact that this experiment was worn out as a closed system. In Charles Law, if there's a closed system the 2 ratios ought to have equal numbers. {this is|this is often|this may be} why it can be expected for the magnitude relation numbers to be terribly equal.
Answer:
The amount of mass and matter in all the transformations of the clay ball will remain the same or constant
Explanation:
From the law of conservation of mass we have, for an enclosed system to and from which there is no transfer of matter or energy, mass cannot be created nor destroyed, and remains constant at the given value, but the matter which make up the mass can be changed into different forms
Therefore, the clay ball can be transformed into different shapes and will still posses the same initial mass before the transformation, provided there are no transfer of matter or energy from the clay ball system.