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3 years ago

10

Use the Activity Series Chart

1 answer:

omeli [17]3 years ago

8 0

Answer:

Mg

Explanation:

Mg is above Al on the table meaning it is easier to oxidize.

Fe and Zn are below Al on the table showing they are not as reactive.

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Air, Sea water, alloy afe the examples of

ratelena [41]

Answer:

B. Mixture

because air,sea water and alloy are example of Mixture

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3 years ago

Which type of energy is associated with random motion of atoms and molecules in a sample of air?

aleksandrvk [35]

The answer is thermal energy

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3 years ago

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Can someone give me an example of balancing equations with a solution that is simple?

Monica [59]

Explanation:

$H _{2}O _{2(aq)} →H _{2}O _{(l)} + O _{2}(g) \\ solution : 2 \: and\: 2$

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3 years ago

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What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig

Mashutka [201]

Answer:

$P_H =2.86$

$c=1.4\times 10^{-4}$

Explanation:

first write the equilibrium equaion ,

$C_3H_6O_3$ ⇄ $C_3H_5O_3^{-} +H^{+}$

assuming degree of dissociation $\alpha$ =1/10;

and initial concentraion of $C_3H_6O_3$ =c;

At equlibrium ;

concentration of $C_3H_6O_3 = c-c\alpha$

$[C_3H_5O_3^{-} ]= c\alpha$

$[H^{+}] = c\alpha$

$K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}$

$\alpha$ is very small so $1-\alpha$ can be neglected

and equation is;

$K_a = {c\alpha \times \alpha}$

$[H^{+}] = c\alpha$ = $\frac{K_a}{\alpha}$

$P_H =- log[H^{+} ]$

$P_H =-logK_a + log\alpha$

$K_a =1.38\times10^{-4}$

$\alpha = \frac{1}{10}$

$P_H= 3.86-1$

$P_H =2.86$

composiion ;

$c=\frac{1}{\alpha} \times [H^{+}]$

$[H^{+}] =antilog(-P_H)$

$[H^{+} ] =0.0014$

$c=0.0014\times \frac{1}{10}$

$c=1.4\times 10^{-4}$

6 0

3 years ago

Mexico City has an elevation of 7400 ft above sea level. Its normal atmospheric pressure is 589 mm Hg. At what temperature does

IRISSAK [1]

Given that the volume and amount of water are kept constant,
P/T = constant
P₁/T₁ = P₂/T₂
Normal atmospheric pressure is 746 mmHg and normal boiling point of water is 100 °C.
746/100 = 589/T₂
T₂ = 79.0 °C

5 0

3 years ago