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4 years ago

10

Most warm-water ocean currents move away from the equator and toward the poles because warm water is driven by __ from _

_ regions to _______ regions.

1 answer:

aksik [14]4 years ago

6 0

convection; hot; cold

Hope This Work

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In a hydrogen fuel cell, hydrogen gas and oxygen gas are combined to form water. write the balanced chemical equation describing

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3 years ago

Read 2 more answers

Be sure to answer all parts. Write the balanced equations corresponding to the following rate expressions: a) rate = − 1 3 Δ[CH4

Alinara [238K]

Answer : The balanced equations will be:

(a) $3CH_4+2H_2O+CO_2\rightarrow 4CH_3OH$

(b) $2N_2O_5\rightarrow 2N_2+5O_2$

(c) $2H_2+2CO_2+O_2\rightarrow 2H_2CO_3$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

Now we have to determine the balanced equations corresponding to the following rate expressions.

(a) $Rate=-\frac{1}{3}\frac{d[CH_4]}{dt}=-\frac{1}{2}\frac{d[H_2O]}{dt}=-\frac{d[CO_2]}{dt}=+\frac{1}{4}\frac{d[CH_3OH]}{dt}$

The balanced equations will be:

$3CH_4+2H_2O+CO_2\rightarrow 4CH_3OH$

(b) $Rate=-\frac{1}{2}\frac{d[N_2O_5]}{dt}=+\frac{1}{2}\frac{d[N_2]}{dt}=+\frac{1}{5}\frac{d[O_2]}{dt}$

The balanced equations will be:

$2N_2O_5\rightarrow 2N_2+5O_2$

(c) $Rate=-\frac{1}{2}\frac{d[H_2]}{dt}=-\frac{1}{2}\frac{d[CO_2]}{dt}=-\frac{d[O_2]}{dt}=+\frac{1}{2}\frac{d[H_2CO_3]}{dt}$

The balanced equations will be:

$2H_2+2CO_2+O_2\rightarrow 2H_2CO_3$

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3 years ago