It is an ideal gas therefore we can use the ideal gas equation to solve the problem. The ideal gas equation is expressed as PV = nRT. First, we solve the amount of the gas in moles using the said equation and the first conditions.
(2.0 atm) (5.0 x 10^3 cm^3) = n (82.0575 atm.cm^3/mol.K)(215 K)
n=0.5668 mol
Using the second conditions given, we obtain the new pressure.
P (4.0 x 10^3) = 0.5668 x <span>82.0575 x 265
P= 3.08 atm</span>
<h3>Answer:</h3><h2>Chemical properties</h2><h3>Explanation:</h3>
By its very definition, a chemical property is one which is exhibited as a result of a chemical reaction. This may happen during or after the reaction. This is because in a chemical reaction there is a transformation in the physical composition of the components and this directly affects its chemical properties.
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Answer: 1000x
Explanation:
I hope this helped!
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- Zack Slocum
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<h3><u>Answer</u>;</h3>
C3H4O
<h3><u>Explanation;</u></h3>
Empirical formula is the simplest formula of a compound;
Molar mass CO2 = 44.01
Mass of CO2 produced = 2.053 g
Mass of carbon in original sample = 12.01/44.01 × 2.053
= 0.5603g
Molar mass H2O = 18
Mass of H in original sample = 2/18 ×0.5601
= 0.0622 g
Thus; original sample contained 0.5603g C and 0.0622g H. The balance of the sample was O
Mass of O = 0.8715 - (0.5603 + 0.0622) = 0.249g
The mole ratio of C:H:O will be;
Moles C = 0.5603/12 = 0.0467
Moles H = 0.0622
Moles O = 0.249/16 = 0.01556
C:H:O = 0.0467:0.0622:0.01556
Divide through by 0.01556:
C:H:O = 3:4:1
Empirical formula is thus C3H4O
The organelle that is primarily used is glucose
