Why does the light bulbs have to be part of the loop in order for it to be succesful?

1 answer:

6 0

The light bulbs ave to be a part of the loop in order for it to be successful because it have to have wires to a circuit breaker so you can see.

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Aluminum boils at 2467°C. Aluminum’s boiling point in Kelvin is 2194.<br> T or F

dusya [7]

That is false because aluminum melts at 2,470C

7 0

2 years ago

Describe the difference between an independent and dependent valuable?

kolbaska11 [484]

Answer:

A dependent valuable is a valuable whose variation depend on another variable usually the independent variable. An independent variable is a variable whose variation do not depend on another variable but the reseacher experimenting.

7 0

3 years ago

The activation energy of a certain uncatalyzed reaction is 64 kJ/mol. In the presence of a catalyst, the Ea is 55 kJ/mol. How ma

Ksivusya [100]

Answer:

About 5 times faster.

Explanation:

Hello,

In this case, since the Arrhenius equation is considered for both the catalyzed reaction (1) and the uncatalized reaction (2), one determines the relationship between them as follows:

$\frac{k_1}{k_2}=\frac{Aexp(-\frac{Ea_1}{RT} )}{Aexp(-\frac{Ea_2}{RT})} \\\frac{k_1}{k_2}=\frac{exp(-\frac{Ea_1}{RT} )}{exp(-\frac{Ea_2}{RT})}$

By replacing the corresponding values we obtain:

$\frac{k_1}{k_2}=\frac{exp(-\frac{55000J/mol}{8.314J/molK*673.15K} )}{exp(-\frac{64000J/mol}{8.314J/molK*673.15K} )} =4.8$

Such result means that the catalyzed reaction is about five times faster than the uncatalyzed reaction.

Best regards.

4 0

2 years ago

In the compound sodium methoxide (naoch3), there is ________ bonding

zvonat [6]

Maybe covalent bonding?

5 0

3 years ago

Two unknown molecular compounds were being studied. A solution containing 5.00 g of compound A in 100. g of water froze at a low

LenaWriter [7]

Answer:

Compound B has greater molar mass.

Explanation:

The depression in freezing point is given by ;

$\Delta T_f=i\times k_f\times m$ ..[1]

$m=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in kg}}$

Where:

i = van't Hoff factor

$k_f$ = Molal depression constant

m = molality of the solution

According to question , solution with 5.00 g of A in 100.0 grams of water froze at at lower temperature than solution with 5.00 g of B in 100.0 grams of water.

The depression in freezing point of solution with A solute: $\Delta T_{f,A}$