Multiple Choice 1. Protons are located in the nucleus of the atom. A proton has a) No charge b) A negative charge C A positive c
2 answers:
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Answer:
The volume of the gas at a pressure of 65.0 kPa would be 363 mL
Explanation:
Boyle's Law is a gas law that relates the pressure and volume of a certain amount of gas, without temperature variation, that is, at constant temperature.
Boyle's law states that the pressure of a gas in a closed container is inversely proportional to the volume of the container, when the temperature is constant. In other words, the product P · V remains constant at the same temperature:
P*V=k
Being P1 and V1 the pressure and volume in state 1 and P2 and V2 the pressure and volume in state 2 are fulfilled:
P1*V1=P2*V2
In this case:
- P1= 45 kPa= 45,000 Pa (being 1 kPa=1,000 Pa)
- V1= 525 mL= 0.525 L (being 1 L=1,000 mL)
- P2= 65 kPa= 65,000 Pa
- V2= ?
Replacing:
45,000 Pa* 0.525 L= 65,000 Pa*V2
Solving:
V2=0.363 L=363 mL
<u><em>The volume of the gas at a pressure of 65.0 kPa would be 363 mL</em></u>
Answer: " 145,000 g " .
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Explanation:
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To solve: Let's use a technique known as "dimensional analysis"
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* ( 537 mol) = ? g ;
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<u>Note</u>: "g" stands for "grams" ;
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The units of "mol" ["mole(s)"] on the "left-hand side of the equation"
→ "cancel out" ; {since: "mol/mol" = 1 "} ;
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And we have:
(287.34 * 537) g ;
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Using a calculator, we do the multiplication to get:
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154301.58 g ; We round this to <u>3 (three) significant figures</u>;
since: (287.34 * 537) g ; the limiting number with the "least number of significant figure" is "537" ; which has 3 (three) significant figures;
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As such:
We have 154301.58 g ;
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We round this to 145,000 g ; since:
"154" are the first 3 (three) significant figures in our value:
" 154301.58" ; and the next significant figure, "3" ; is less than "5" ;
If the next significant figure were "0, 1, 2, 3, or 4" ; we would round down to: 145,000 g; such as in our case. If the next significant figure were "5, 6, 7, 8, or 9" , we would round up to: 146,000 g .
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The correct answer is: " 145,000 g ."