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2 years ago

-2(3b-2)=8b-10 this is a problem that i am stuck on. math enough said

2 answers:

8 0

You first start by distributing the -2 in front of the parenthesis. You get -6b+4=8b-10. Then I would add the 6b to 8b so you would have 4=14b-10. Then add the 10 to the 4 and you get 14=14b. Then to get b by itself you want to divide by 14 so you end up with 1=b

adoni [48]2 years ago

3 0

The answer is B=1

-2(3b-2)=8b-10
-6b+4=8b-10
-14b+4=-10
-14b=-14
B=1

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HOCl dissociative into which 2 ions

trasher [3.6K]

Answer:

H+ and OCl-

Since this is an acid, the H+ will come off and what is left OCl- is the other part.

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3 years ago

For the reaction A+B+C→D+E, the initial reaction rate was measured for various initial concentrations of reactants. The followin

erastovalidia [21]

Answer : The initial rate for a reaction will be $3.4\times 10^{-3}Ms^{-1}$

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$A+B+C\rightarrow D+E$

Rate law expression for the reaction:

$\text{Rate}=k[A]^a[B]^b[C]^c$

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

$6.0\times 10^{-5}=k(0.20)^a(0.20)^b(0.20)^c$ ....(1)

Expression for rate law for second observation:

$1.8\times 10^{-4}=k(0.20)^a(0.20)^b(0.60)^c$ ....(2)

Expression for rate law for third observation:

$2.4\times 10^{-4}=k(0.40)^a(0.20)^b(0.20)^c$ ....(3)

Expression for rate law for fourth observation:

$2.4\times 10^{-4}=k(0.40)^a(0.40)^b(0.20)^c$ ....(4)

Dividing 1 from 2, we get:

$\frac{1.8\times 10^{-4}}{6.0\times 10^{-5}}=\frac{k(0.20)^a(0.20)^b(0.60)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\3=3^c\\c=1$

Dividing 1 from 3, we get:

$\frac{2.4\times 10^{-4}}{6.0\times 10^{-5}}=\frac{k(0.40)^a(0.20)^b(0.20)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\4=2^a\\a=2$

Dividing 3 from 4, we get:

$\frac{2.4\times 10^{-4}}{2.4\times 10^{-4}}=\frac{k(0.40)^a(0.40)^b(0.20)^c}{k(0.40)^a(0.20)^b(0.20)^c}\\\\1=2^b\\b=0$

Thus, the rate law becomes:

$\text{Rate}=k[A]^2[B]^0[C]^1$

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

$6.0\times 10^{-5}=k(0.20)^2(0.20)^0(0.20)^1$

$k=7.5\times 10^{-3}M^{-2}s^{-1}$

Now we have to calculate the initial rate for a reaction that starts with 0.75 M of reagent A and 0.90 M of reagents B and C.

$\text{Rate}=k[A]^2[B]^0[C]^1$

$\text{Rate}=(7.5\times 10^{-3})\times (0.75)^2(0.90)^0(0.90)^1$

$\text{Rate}=3.4\times 10^{-3}Ms^{-1}$

Therefore, the initial rate for a reaction will be $3.4\times 10^{-3}Ms^{-1}$

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2 years ago

What is your streak and what fact is it?

kobusy [5.1K]

Answer:

unknown fact I can't message anyone

6 0

2 years ago

Read 2 more answers

How much energy is required to melt 10. 0 g of ice at 0. 0°C, warm it to 100. 0°c and completely vaporize the sample?

natta225 [31]

The 7160 cal energy is required to melt 10. 0 g of ice at 0. 0°C, warm it to 100. 0°C and completely vaporize the sample.

Calculation,

Given data,

Mass of the ice = 10 g

Temperature of ice = 0. 0°C

The ice at 0. 0°C is to be converted into water at 0. 0°C

Heat required at this stage = mas of the ice ×latent heat of fusion of ice

Heat required at this stage = 10 g×80 = 800 cal

The temperature of the water is to be increased from 0. 0°C to 100. 0°C

Heat required for this = mass of the ice×rise in temperature×specific heat of water

Heat required for this = 10 g×100× 1 = 1000 cal

This water at 100. 0°C is to be converted into vapor.

Heat required for this = Mass of water× latent heat

Heat required for this = 10g ×536 =5360 cal

Total energy or heat required = sum of all heat = 800 +1000+ 5360 = 7160 cal

to learn more about energy

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1 year ago

Debido a que propiedad un cuerpo flota en el agua y otro no

Nata [24]

Answer:

cuerpo flota o se hunde es su densidad con respecto a la del líquido en que se sumerge. En cambio, una nuez (densidad = 0.5) flotara en ambos líquidos mientras que una piedra (densidad = 2) se hundirá.

Explanation:

8 0

3 years ago