All categories

3 years ago

What is the main energy conversion that occurs when you plug in a lamp

Chemistry

2 answers:

4vir4ik [10]3 years ago

5 0

Answer:

Electric energy conservation which converts into heat and light

Explanation:

The transformation of energy from one form energy into another form, and throughout this conversion, the energy that we give to a device will comes out. However, the energy that is given may or may not coming out in the required form.

For example, we plug in a lump means we give an electrical energy into a lamp and the lamp produces light, but we also get heat energy from the lamp(it is the form of energy which is undesirable coming from a lamp).

MrRa [10]3 years ago

4 0

Elecrtical or Radiantal

You might be interested in

Dioxin is a by-product of various industrial chemical processes. It is suspected of causing cancer and birth defects in both hum

Nuetrik [128]

Answer:

The natural ligand of the cytosolic Aryl hydrocarbon receptor which is still not known.

Explanation:

Dioxin is formed during some industrial chemical processes. It is considered harmful because it causes cancer, weakens the immune system, results in nervous breakdown and also affects reproduction.

The mechanism of operation of Dioxins has not been fully understood. However, it is known that dioxins works through a receptor known as the Aryl hydrocarbon receptor. This receptor is useful in gene expression and also acts as a transcription factor.

4 0

3 years ago

Read 2 more answers

When a 2 kg log is burned, it produces 0.5 kg of ash. Explain why the mass of the ash is less than the mass of the log. In other

Lostsunrise [7]

Answer:

my mass turned into gasses and other substances besides ash

Explanation:

burning a log products smoke and ash. the smoke takes some mass and the ash takes the rest.

3 0

3 years ago

Determine the empirical formula. a 3.880g sample contains 0.691g of magnesium , 1.84 g of sulfur , and 1.365 g of oxygen .

Aliun [14]

Answer:

Mg S2 O3

Explanation:

.691 g of Mg is .284 mole

1.84 g of S is .5739 mole

1.365 g of O is .8531 mole you can see the ratio is ~ 1 :2 :3

Mg S2 O3

4 0

2 years ago

Determine Z and V for steam at 250°C and 1800 kPa by the following: (a) The truncated virial equation [Eq. (3.38)] with the foll

makvit [3.9K]

Answer:

Explanation:

Given that:

the temperature $T_1$ = 250 °C= ( 250+ 273.15 ) K = 523.15 K

Pressure = 1800 kPa

The truncated viral equation is expressed as:

$\frac{PV}{RT} = 1 + \frac{B}{V} + \frac{C}{V^2}$

where; B = - $152.5 \ cm^3 /mol$ C = -5800 $cm^6/mol^2$

R = 8.314 × 10³ cm³ kPa. K⁻¹.mol⁻¹

Plugging all our values; we have

$\frac{1800*V}{8.314*10^3*523.15} = 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

$4.138*10^{-4} \ V= 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

Multiplying through with V² ; we have

$4.138*10^4 \ V ^3 = V^2 - 152.5 V - 5800 = 0$

$4.138*10^4 \ V ^3 - V^2 + 152.5 V + 5800 = 0$

V = 2250.06 cm³ mol⁻¹

Z = $\frac{PV}{RT}$

Z = $\frac{1800*2250.06}{8.314*10^3*523.15}$

Z = 0.931

b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)].

The generalized Pitzer correlation is :

$T_c = 647.1 \ K \\ \\ P_c = 22055 \ kPa \\ \\ \omega = 0.345$

$T__{\gamma}} = \frac{T}{T_c}$

$T__{\gamma}} = \frac{523.15}{647.1}$

$T__{\gamma}} = 0.808$

$P__{\gamma}} = \frac{P}{P_c}$

$P__{\gamma}} = \frac{1800}{22055}$

$P__{\gamma}} = 0.0816$

$B_o = 0.083 - \frac{0.422}{T__{\gamma}}^{1.6}}$

$B_o = 0.083 - \frac{0.422}{0.808^{1.6}}$

$B_o = 0.51$

$B_1 = 0.139 - \frac{0.172}{T__{\gamma}}^{ \ 4.2}}$

$B_1 = -0.282$

The compressibility is calculated as:

$Z = 1+ (B_o + \omega B_1 ) \frac{P__{\gamma}}{T__{\gamma}}$

$Z = 1+ (-0.51 +(0.345* - 0.282) ) \frac{0.0816}{0.808}$

Z = 0.9386

$V= \frac{ZRT}{P}$

$V= \frac{0.9386*8.314*10^3*523.15}{1800}$

V = 2268.01 cm³ mol⁻¹

c) From the steam tables (App. E).

At $T_1 = 523.15 \ K \ and \ P = 1800 \ k Pa$

V = 0.1249 m³/ kg

M (molecular weight) = 18.015 gm/mol

V = 0.1249 × 10³ × 18.015

V = 2250.07 cm³/mol⁻¹

R = 729.77 J/kg.K

Z = $\frac{PV}{RT}$

Z = $\frac{1800*10^3 *0.1249}{729.77*523.15}$

Z = 0.588

3 0

3 years ago

C + 2H2 -> CH4

ankoles [38]

Answer:

C) 4

Explanation:

The reaction :

C (s) + 2H₂ (g) ⇒ CH₄ (g)

12g 4g 16g

Hence, based on this we can say that : 2 moles of hydrogen gas are needed to produce 16g of methane.

For 32g of methane

Number of moles of H₂ = 32/16 × 2
Number of moles of H₂ = 4

4 0

2 years ago

Read 2 more answers