Explanation:
30 lb is 480 ounces
34 mi/second is 54.718 kilometre/ second
455 lb/ gal is 54521.024 grams / litre
50 cl is 500 millilitres
55nm is 5.5 × 10^-6 centimetre
Answer:
Increasing the surface area of a reactant increases the frequency of collisions and increases the reaction rate. Several smaller particles have more surface area than one large particle. The more surface area that is available for particles to collide, the faster the reaction will occur.
Explanation:
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Answer:
The molecular formula of the compound is
. The molecular formula is obtained by the following expression shown below

Explanation:
Given molecular mass of the compound is 176 g/mol
Given empirical formula is
Atomic mass of carbon, hydrogen and oxygen are 12 u , 1 u and 16 u respectively.
Empirical formula mass of the compound = 


Molecular formula = 4
Molecular formula is 
Answer:
3) NaCl.
Explanation:
<em>∵ ΔTf = iKf.m</em>
where, <em>i</em> is the van 't Hoff factor.
<em>Kf </em>is the molal depression freezing constant.
<em>m</em> is the molality of the solute.
<em>The van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. </em>
<em></em>
- For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.
<em>So, for sugar: i = 1.</em>
<em>∴ ΔTf for sugar = iKf.m = (1)(Kf)(2.0 m) = 2 Kf.</em>
<em></em>
- For most ionic compounds dissolved in water, the van 't Hoff factor is equal to the number of discrete ions in a formula unit of the substance.
For NaCl, it is electrolyte compound which dissociates to Na⁺ and Cl⁻.
<em>So, i for NaCl = 2.</em>
<em>∴ ΔTf for NaCl = iKf.m = (2)(Kf)(1.0 m) = 2 Kf.</em>
<em></em>
<em>So, the right choice is: 3) NaCl.</em>
<em></em>
Answer:
643g of methane will there be in the room
Explanation:
To solve this question we must, as first, find the volume of methane after 1h = 3600s. With the volume we can find the moles of methane using PV = nRT -<em>Assuming STP-</em>. With the moles and the molar mass of methane (16g/mol) we can find the mass of methane gas after 1 hour as follows:
<em>Volume Methane:</em>
3600s * (0.25L / s) = 900L Methane
<em>Moles methane:</em>
PV = nRT; PV / RT = n
<em>Where P = 1atm at STP, V is volume = 900L; R is gas constant = 0.082atmL/molK; T is absolute temperature = 273.15K at sTP</em>
Replacing:
PV / RT = n
1atm*900L / 0.082atmL/molK*273.15 = n
n = 40.18mol methane
<em>Mass methane:</em>
40.18 moles * (16g/mol) =
<h3>643g of methane will there be in the room</h3>