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2 years ago

Which word equation best describes the following formula equation: 4Al + 302 + 2Al2O3?

Chemistry

2 answers:

liberstina [14]2 years ago

6 0

Answer:

Explanation:

The equation is not correctly written, so let's start with that.

4Al + 3O2 ====> 2Al2O3?

Totally correct, including the balance numbers, would be

4moles aluminum + 3 moles Oxygen ===> 2 moles of Aluminum Oxide

Anything else is incomplete.

o-na [289]2 years ago

5 0

Answer:

aluminum reacts with oxygen to produce aluminum oxide.

I hope this helps

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

8 0

2 years ago

Use the drop-down menus to identify the experimental group with additional nutrients and the control group without additional nu

Paladinen [302]

Answer:

A. 1 to 16

B. 17 to 32

Explanation:

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2 years ago

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Determine the percent water in MgSO4*7H20?

Semmy [17]

Answer:

$\% H_2O=51.2\%$

Explanation:

Hello!

In this case, since the percent water is computed by dividing the amount of water by the total mass of the hydrate; we infer we first need the molar mass of water and that of the hydrate as shown below:

$MM_{MgSO_4* 7H_2O}=120.36 g/mol+7*18.02g/mol\\\\MM_{MgSO_4* 7H_2O}=246.5g/mol$

Thus, the percent water is:

$\% H_2O=\frac{7*MM_{H_2O}}{MM_{MgSO_4* 7H_2O}} *100\%\\\\$

So we plug in to obtain:

$\% H_2O=\frac{7*18.02}{246.5} *100\%\\\\\% H_2O=51.2\%$

Best regards!

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2 years ago

A(n) _____ is an organic compound that contains a nitrogen atom bonded to one, two, or three carbon atoms.

Katyanochek1 [597]

Tertiary Amine

The reason Why it is Tertiary Amine is because it has three carbons and one nitrogen.

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3 years ago

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a vehicle of known mass accelerates along a straight path. according to newtons second law of motion, what caused this accelerat

postnew [5]

According to Newton's Second law of motion, t<span>he force acting on an object is equal to the mass of that object times its acceleration. The acceleration of the vehicle is caused by the force that is applied and the mass it posses. Hope this answers the question.</span>

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3 years ago

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