we are asked to multiply 5.4 mm by 6.02 mm
5.4 mm has 2 significant figures
6.02 mm has 3 significant figures
When multiplying these 2 numbers the answer should have the least number of significant figures
From the 2 numbers the least number of significant figures is 2 , therefore answer should be rounded off to 2 significant numbers
5.4 x 6.02 = 32.5 rounded off is 33
Answer is b. 33 m^2
The final volume of the gas is 73.359 mL
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Given :
A sample gas has an initial volume of 72.0 mL
The work done = 141.2 J
Pressure = 783 torr
The objective is to determine the final volume of the gas.
Since, the process does 141.2 J of work on its surroundings at a constant pressure of 783 Torr. Then, the pressure is external.
Converting the external pressure to atm; we have
External Pressure P
:
= 783 torr × 
= 1.03 atm
The work done W = 
The change in volume ΔV= 
ΔV = 
ΔV = 
ΔV = 0.001359 L
ΔV = 1.359 mL
The initial volume = 72.0 mL
The change in volume V is ΔV = V₂ - V₁
- V₂ = - ΔV - V₁
multiply both sides by (-), we have:
V₂ = ΔV + V₁
= 1.359 mL + 72.0 mL
= 73.359 mL
Therefore, the final volume of the gas is 73.359 mL .
Learn more about volume here:
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Explanation:
The reaction is as follows:
2Mg(s) + O2(g) ---> 2MgO(s)
and the researcher said that 32 g of MgO was produced.
Stoichiometry:
28 g Mg × (1 mol Mg/24.305 g Mg) = 1.15 mol Mg
15 g O2 × (1 mol O2/15.999 g O2) = 0.938 mol O2
1.15 mol Mg × (2 mol MgO/2 mol MgO) = 1.15 mol MgO
1.15 mol MgO × (40.3044 g MgO/1 mol MgO) = 46.6 g MgO
0.938 mol O2 × (2 mol MgO/1 mol O2) = 1.88 mol MgO
1.88 mol MgO × (40.3044 g MgO/1 mol MgO = 75.6 g MgO
Based on these numbers, the amount of product after the reaction is much less than expected so these results don't seem to support the law of conservation of matter.
Answer:
277.7 g of CO2
Explanation:
Equation of reaction
C13H18O2 + 11O2 ---> 13CO2 + 9H2O
From the equation of reaction
1 mole of ibuprofen produces 13 moles of CO2
Molar mass of ibuprofen is 206g
Molar mass of CO2 is 44g
13 moles of CO2 weighs 572g
Therefore, 100g of ibuprofen will produce (100×572)/206 of CO2
= 277.7g
