Recent content by a7med2009

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    What Are the Equilibrium Concentrations and Kc for 2NO(g) + Br2(g) ⇌ 2NOBr(g)?

    2NO(g) + Br2(g) <----___> http://www.freeimagehosting.net/uploads/b72c8010cb.jpg 2NOBr(g), 2.5 5.0 1.0 Since the total change in conc of [NOBr] is given as +0.5 so x[ extant of reaction]=0.50M (2mole NO to 2mole NOBr) Now using the general expression At eqm...
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    What Are the Equilibrium Concentrations and Kc for 2NO(g) + Br2(g) ⇌ 2NOBr(g)?

    I got the first part p=0.5 ,Q=0.25, R=4.75 , s=1.5 but for Kc is it Kc=[NOBr]^2 /[NO]^2 [Br2] or Kc=[NO]^2 [Br2]/[NOBr]^2
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    What Are the Equilibrium Concentrations and Kc for 2NO(g) + Br2(g) ⇌ 2NOBr(g)?

    Given 2NO(g) + Br2(g) <--------_____> 2NOBr(g), Concentration (M) [NO] [Br2] [NOBr] Initial 2.5 5.0 1.0 Change P Q...
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    What horizontal force is required to accelerate the block

    ohh 0_o it's ΣF=F-fk=ma F=ma+fk =(4.0)*(5)+(0.25)(39.2) =20+9.8 =29.8 N
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    What horizontal force is required to accelerate the block

    ΣF=F-fk=mg F=mg-fk =(4.0)*(9.8)-(0.25)(39.2) =29.4 N
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    What horizontal force is required to accelerate the block

    ma=4.0kg*5.0m/s^2 =20 N Fk= 0.25 * 20 = 5 NIs it correct?
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    What horizontal force is required to accelerate the block

    ΣF=ma Fk=mu * N
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    What horizontal force is required to accelerate the block

    A block of mass 4.0kg rests on a horizontal surface. What horizontal force is required to accelerate the block at 5.0m/s2 if the coefficient of kinetic friction is 0.25?
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