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3 years ago

Substances that are readily combustible or may cause fire through friction

Chemistry

1 answer:

Dvinal [7]3 years ago

3 0

Answer :

Flammable substances

Explanation :

<em>Flammable substances</em> will catch fire and continue to burn when they contact an ignition source like a spark or a flame.

For example, <em>methanol</em> is a flammable liquid.

A flammable solid may also catch fire through friction. <em>Matches</em> are flammable solids.

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Different isotopes of the same element emit light at slightly different wavelengths. A wavelength in the emission spectrum of a

Inessa [10]

Different isotopes of the same element emit light at slightly different wavelengths, the minimum number of slits is mathematically given as

N=1820slits

<h3>What minimum number of slits is required to resolve these two wavelengths in second-order?</h3>

Generally, the equation for the wave is mathematically given as

$d\ sin\ (\theta\ m) \ = \ m\ \lambda$

Where the chromatic resolving power (R) is defined by

$R\ =\ \lambda\ / \ d \ \lambda$

R = nN,

Therefore

$\lambda_1 \ = \ (656.45)(1 \ * \ 10^{-9})/1mm$

$\lambda_1= 656.45*10^{-9}$

and

$\lambda_2= (656.27)(1*10^{-9})/1mm$

$\\\\\lambda_2= 656.27*10^{-9}m$

In conclusion, the minimum number of slits is required to resolve these two wavelengths in second-order

$N\ =\ \dfrac{\lambda}{m\ d\ T\ }\\\\$

Therefore

$N\ =\ \dfrac{656.45 \ * \ 10^{-9}}{2\ * \ (0.18*10^{-9})}$

N=1820slits

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7 0

2 years ago

A trial of this decomposition experiment, using different quantities of reactants than those listed in the question above produc

Paul [167]

Answer : The volume of $O_2(g)$ produced at standard conditions of temperature and pressure is 0.2422 L

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of $O_2$ gas = (740-22.4) torr = 717.6 torr

$P_2$ = final pressure of $O_2$ gas at STP= 760 torr

$V_1$ = initial volume of $O_2$ gas = 280 mL

$V_2$ = final volume of $O_2$ gas at STP = ?

$T_1$ = initial temperature of $O_2$ gas = $25^oC=273+25=298K$

$T_2$ = final temperature of $O_2$ gas = $0^oC=273+0=273K$

Now put all the given values in the above equation, we get:

$\frac{717.6torr\times 280mL}{298K}=\frac{760torr\times V_2}{273K}$

$V_2=242.2mL=0.2422L$

Therefore, the volume of $O_2(g)$ produced at standard conditions of temperature and pressure is 0.2422 L

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3 years ago

How many molecules of ethane ( C2H6) are present in 0.197g of C2H6?

gayaneshka [121]

C2H6/C2H6=1

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3 years ago

C10H22 + O2 -- CO2 + H2O balance

Tasya [4]

Answer:

1,31÷2 =10,11

Explanation:

c10h22+31÷2o2=10co2+11h2o

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3 years ago

What is the ph of a 0.027 M KOH solution?

Monica [59]

First you calculate the concentration of [OH⁻] in <span>solution :

POH = - log [ OH</span>⁻]

POH = - log [ 0.027 ]

POH = 1.56

PH + POH = 14

PH + 1.56 = 14

PH = 14 - 1.56

PH = 12.44

hope this helps!

8 0

3 years ago