Answer:
D. 44.2 g O₂
General Formulas and Concepts:
<u>Gas Laws</u>
- STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at <em>1 atm, 273 K</em>
<u>Stoichiometry</u>
- Dimensional Analysis
- Mole Ratio
Explanation:
<u>Step 1: Define</u>
<em>Identify given.</em>
61.9 L O₂ at STP
<u>Step 2: Convert</u>
We know that the oxygen gas is at STP. Therefore, we can set up and solve for how many <em>moles</em> of O₂ is present:

Recall the Periodic Table (Refer to attachments). Oxygen's atomic mass is roughly 16.00 grams per mole (g/mol). We can use a mole ratio to convert from <em>moles</em> to <em>grams</em>:

Now we deal with sig figs. From the original problem, we are given 3 significant figures. Round your answer to the <u>exact</u> same number of sig figs:

∴ our answer is letter choice D.
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Topic: AP Chemistry
Unit: Stoichiometry
I hate to tell you this but this isnt tinder hunny
The new pressure will be 7.65 atm
<h3>General gas law</h3>
The problem is solved using the general gas equation:
P1V1/T1 = P2V2/T2
In this case, P1 = 3.4 atm, V1 = 1500 mL, T1 = 25
, V2 = 2000 mL, and T2 = 75 
What we are looking for is P2.
Thus, P2 = P1V1T2/T1V2
= 3.4 x 1500 x 75/25 x 2000 = 382500/50000 = 7.65 atm
More on general gas laws can be found here: brainly.com/question/2542293
#SPJ1
Answer:
Explanation:i joule is equal to 0.238902957619 calories so 1251 joules is equal to 298.87 calories divided by 25.0 degrees centigrade is equal to 11.95 calories divided by the 35.2 gram sample weight to get the calories per gram per degree centigrade would come to 0.3396 calories/gram degree centigrade. Presumably this, if correct, could be used to obtain the metal in question by consulting a chart or table with specific heats of various metals because they should always be the same specific heat for each metal.
Answer:
The answer is TRUE!!!
Explanation:
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