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3 years ago

In which way does electricity flow in a DC circuit? *

Chemistry

1 answer:

horrorfan [7]3 years ago

4 0

Answer:

From the negative terminal of the battery to the positive one

Explanation:

DC is just dirrect current that has e- flowing from the negative terminal to the positive.

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10 m3 of carbon dioxide is originally at a temperature of 50 °C and pressure of 10 kPa. Determine the new density and volume of

Dimas [21]

Answer : The new density and new volume of carbon dioxide gas is 0.2281 g/L and $7.2m^3$ respectively.

Explanation :

First we have to calculate the new or final volume of carbon dioxide gas.

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 10 kPa

$P_2$ = final pressure of gas = 15 kPa

$V_1$ = initial volume of gas = $10m^3$

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = $50^oC=273+50=323K$

$T_2$ = final temperature of gas = $75^oC=273+75=348K$

Now put all the given values in the above equation, we get:

$\frac{10kPa\times 10m^3}{323K}=\frac{15kPa\times V_2}{348K}$

$V_2=7.2m^3$

The new volume of carbon dioxide gas is $7.2m^3$

Now we have to calculate the new density of carbon dioxide gas.

$PV=nRT\\\\PV=\frac{m}{M}RT\\\\P=\frac{m}{V}\frac{RT}{M}\\\\P=\rho \frac{RT}{M}\\\\\rho=\frac{PM}{RT}$

Formula for new density will be:

$\rho_2=\frac{P_2M}{RT_2}$

where,

$P_2$ = new pressure of gas = 15 kPa

$T_2$ = new temperature of gas = $75^oC=273+75=348K$

M = molar mass of carbon dioxide gas = 44 g/mole

R = gas constant = 8.314 L.kPa/mol.K

$\rho$ = new density

Now put all the given values in the above equation, we get:

$\rho_2=\frac{(15kPa)\times (44g/mole)}{(8.314L.kPa/mol.K)\times (348K)}$

$\rho_2=0.2281g/L$

The new density of carbon dioxide gas is 0.2281 g/L

5 0

3 years ago

Calculate the mass, in grams, of 1.2000 mol Mg₃N₂

fredd [130]

So what do you want me to do
Explanation

7 0

3 years ago

Boyles Law P1V1 = P2V2

arsen [322]

Answer:

A. The balloons will increase to twice their original volume.

Explanation:

Boyle's law states that the pressure exerted on a gas is inversely proportional to the volume occupied by the gas at constant temperature. That is:

P ∝ 1/V

P = k/V

PV = k (constant)

P = pressure, V = volume.

$P_1V_1=P_2V_2$

Let the initial pressure of the balloon be P, i.e. $P_1=P$ , initial volume be V, i.e. $V_1=V$ . The pressure is then halved, i.e. $P_2=\frac{P}{2}$

$P_1V_1=P_2V_2\\\\P*V=\frac{P}{2} *V_2\\\\V_2=\frac{2*P*V}{P}\\\\V_2=2V$

Therefore the balloon volume will increase to twice their original volume.

3 0

3 years ago

All of the following require breaking and forming chemical bonds except

skad [1K]

As far as I know, the answer is B (breaking a rock)

3 0

3 years ago

Complete the reaction, which is part of the electron transport chain. The abbreviation FMN represents flavin mononucleotide. Use

Margarita [4]

Answer:

1. NADH + H⁺ + FMN + Q ⟶ NAD⁺ + FMN + QH₂

2. The reactant that is reduced is Q

3. The charge on iron on the right side is +2, Fe²⁺

Explanation:

NADH + H⁺ + FMN + Q ⟶ NAD⁺ + FMN + QH₂

The reaction above is catalysed by NADH:ubiquinone oxidoreductase (complex 1), which transfers a hydride ion from NADH to FMN, from which two electrons pass through a series of of Fe-S centers to the iron-sulfur protein N-2. Electron transfer from N-2 to Ubiquinone forms QH₂

The species in a reaction which gains hydrogen irons is reduced, Therefore, the reactant that is reduced is Q, ubiquinone to form QH₂, ubiquinol.

To determine the oxidation number of iron on the right side of the reaction below,

QH2 + 2cyt c ( Fe3+) ⟶ Q + 2cyt c(Fex) + 2H^+

Sum of charges on the left side = Sum of charges on the right side

Sum of charges on the left side = 2 *+3 = +6

Therefore 2 * x + 2= 6

2x = 6 -2 = 4

x = 4/

x = 2

Therefore the charge on iron on the right side is +2, Fe²⁺

4 0

3 years ago