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Aleksandr [31]
3 years ago
15

which of the following parts of a circuit would be considered a load ? Justify your answer by explaining your choices • light •

switch • motor • battery
Chemistry
1 answer:
QveST [7]3 years ago
6 0
The answer to the question given above is letter A. Light
 
Light  is considered a load of the parts of a circuit. <span>The load in a circuit can be any electrical device that converts electrical energy into other usable forms of energy such as a <span>light bulb.

>></span></span><span>Energy sources include batteries and generating stations
>>switch-</span><span>is used in electric circuits to allow the circuit to be turned on and off.</span>
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CH4 with pressure 1 atm and volume 10 liter at 27°C is passed into a reactor with 20% excess oxygen, how many moles of oxygen is
BaLLatris [955]

Answer : The moles of O_2 left in the products are 0.16 moles.

Explanation :

First we have to calculate the moles of CH_4.

Using ideal gas equation:

PV=nRT

where,

P = pressure of gas = 1 atm

V = volume of gas = 10 L

T = temperature of gas = 27^oC=273+27=300K

n = number of moles of gas = ?

R = gas constant = 0.0821 L.atm/mol.K

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

(1atm)\times (10L)=n\times (0.0821L.atm/mol.K)\times (300K)

n=0.406mole

Now we have to calculate the moles of O_2.

The balanced chemical reaction will be:

CH_4+2O_2\rightarrow CO_2+2H_2O

From the balanced reaction we conclude that,

As, 1 mole of CH_4 react with 2 moles of O_2

So, 0.406 mole of CH_4 react with 2\times 0.406=0.812 moles of O_2

Now we have to calculate the excess moles of O_2.

O_2 is 20 % excess. That means,

Excess moles of O_2 = \frac{(100 + 20)}{100} × Required moles of O_2

Excess moles of O_2 = 1.2 × Required moles of O_2

Excess moles of O_2 = 1.2 × 0.812 = 0.97 mole

Now we have to calculate the moles of O_2 left in the products.

Moles of O_2 left in the products = Excess moles of O_2 - Required moles of O_2

Moles of O_2 left in the products = 0.97 - 0.812 = 0.16 mole

Therefore, the moles of O_2 left in the products are 0.16 moles.

7 0
3 years ago
c) If a mixture of zinc powder and cobalt(II) oxide is heated, the following reaction occurs: Zn(s) + COO(s) → ZnO(s) + Co(s) (i
bazaltina [42]

Answer:

The metal which reduces the other compound is the one higher in reactivity. So in this case, it is.

Explanation:

5 0
3 years ago
A car with a mass of 1,500 kg is traveling at a speed of 30 m/s. What force must be applied to stop the car in 3 seconds ?
saw5 [17]

Yo sup??

we can solve this problem by applying Newton's 2nd law

F*t=Δp

p=momentum

pi=mu=1500*30

pf=mv=m*0=0

Therefore

F*3=1500*30

F=15000 N

Hope this helps.

5 0
3 years ago
Zoe is frying eggs in a pan. The eggs cook because of _____ of heat. Zoe feels hot as she stands near the stove mostly because o
erica [24]

Answer:

1) conduction

2) Radiation

Explanation:

Conduction is a mode of heat transfer by which heat energy is transferred through a material, the average position of the particles of the material remaining the same.

Radiation is a process of heat transfer by which heat is transferred from a hotter to a cooler point without any intervening medium.

The pan used to fry the egg is a conductor of heat hence heat can pass through it by conduction thereby enabling the eggs to cook.

Heat can travel without an intervening medium hence Zoe feels hot near the stove. This ability of the heat to travel without an intervening medium is called radiation.

6 0
3 years ago
Read 2 more answers
Gas has a volume of 247.3 ML and is at 100 Celsius and 745 Hg. If the mass of the gas is 0.347 g what is the molar mass of the v
Ahat [919]

Answer:

The molar mass of the vapor is 43.83 g/mol

Explanation:

Given volume of gas = V = 247.3 mL = 0.2473 L

Temperature = T = 100^{\circ}C = 373 K

Pressure of the gas = P = 745 mmHg  (1 atm = 760 mmHg)

P = \displaystyle \frac{745}{760} \textrm{ atm} = 0.9802 \textrm{ atm}

Mass of vapor = 0.347 g

Assuming molar mass of gas to be M g/mol

The ideal gas equation is shown below

\textrm{PV} =\textrm{nRT} \\\textrm{PV} = \displaystyle \frac{m}{M}\textrm{ RT } \\0.98026 \textrm{ atm}\times 0.2473 \textrm{ L} = \displaystyle \frac{3.47 \textrm{ g}}{M}\times 0.0821 \textrm{ L.atm.mol}^{-1}.K^{-1}\times 373\textrm{K} \\M = 43.834 \textrm{ g/mol}

The molar mass of the vapor comes out to be 43.834 g/mol

4 0
3 years ago
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