The conversion factor 100mL/1 L would be used to change what

Chemistry

1 answer:

ValentinkaMS [17]3 years ago

8 0

Answer:

Liters to milliliters.

Explanation:

Conversion factors are used to convert from one unit to another. The unit you are converting from goes in the denominator, and the unit you are converting to goes in the numerator. Since liters is in the denominator, you are converting from liters to another unit. Since milliliters is in the numerator, you are converting to that unit. Therefore, you would be converting from liters to milliliters.

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A chemist studies the reaction below. 2NO(g) + Cl2(g) 2NOCl(g) He performs three experiments using different concentrations and

Dmitry_Shevchenko [17]

Answer:

1. $Rate =k [NO]^{2}[Cl_{2}]$

2. $k= 0.42 \frac{L^{2}}{mol^{2}*s}$

Explanation:

$Rate =k [NO]^{m}[Cl_{2}]^{n}$

$Rate1 = k[0.4]^{m}[0.3]^{n}=0.02\\Rate 2=k [0.8]^{m}[0.3}]^{n}=0.08\\\\\frac{Rate1}{Rate2}=\frac{0.02}{0.08} =\frac{k[0.4]^{m}[0.3]^{n}}{k[0.8]^{m}[0.3]^n}} \\\\\frac{1}{4} =(\frac{1}{2} )^{m},\\m=2$

$Rate3 =k [0.8]^{m}[0.6]^{n}=0.16\\Rate 2= k[0.8]^{m}[0.3}]^{n}=0.08\\\\\frac{Rate3}{Rate2}=\frac{0.16}{0.08} =\frac{k[0.8]^{m}[0.6]^{n}}{k[0.8]^{m}[0.3]^n}} \\\\\frac{2}{1} =(\frac{2}{1} )^{n},\\n=1$

$Rate =k [NO]^{2}[Cl_{2}]^{1}$

$Rate =k [NO]^{2}[Cl_{2}]^{1}\\Rate 1=k [0.4]^{2}[0.3]^{1} =0.02\\k*0.16*0.3=0.02\\k=\frac{0.02}{0.16*0.3}=\frac{1}{8*(\frac{3}{10} )}=\frac{5}{12} = 0.42 \frac{L^{2}}{mol^{2}*s}$