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3 years ago

What energy does a heated element make

Chemistry

1 answer:

bekas [8.4K]3 years ago

8 0

Answer:

electrical energy

Explanation:

A heating element converts electrical energy into heat through the process of resistive (otherwise known as Joule heating). The electric current passing through the element encounters resistance, which produces heat.

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CS2 (s) + 3 O2 (g) → CO2 (g) + 2 SO2 (g)

Natasha_Volkova [10]

Answer:

2.067 L ≅ 2.07 L.

Explanation:

The balanced equation for the mentioned reaction is:

CS₂(g) + 3O₂(g) → CO₂(g) + 2SO₂(g),

It is clear that 1.0 mole of CS₂ react with 3.0 mole of O₂ to produce 1.0 mole of CO₂ and 2.0 moles of SO₂.

At STP, 3.6 L of H₂ reacts with (?? L) of oxygen gas:

It is known that at STP: every 1.0 mol of any gas occupies 22.4 L.

using cross multiplication:

1.0 mol of O₂ represents → 22.4 L.

??? mol of O₂ represents → 3.1 L.

∴ 3.1 L of O₂ represents = (1.0 mol)(3.1 L)/(22.4 L) = 0.1384 mol.

To find the no. of moles of SO₂ produced from 3.1 liters (0.1384 mol) of hydrogen:

Using cross multiplication:

3.0 mol of O₂ produce → 2.0 mol of SO₂, from stichiometry.

0.1384 mol of O₂ produce → ??? mol of SO₂.

∴ The no. of moles of SO₂ = (2.0 mol)(0.1384 mol)/(3.0 mol) = 0.09227 mol.

Again, using cross multiplication:

1.0 mol of SO₂ represents → 22.4 L, at STP.

0.09227 mol of SO₂ represents → ??? L.

∴ The no. of liters of SO₂ will be produced = (0.09227 mol)(22.4 L)/(1.0 mol) = 2.067 L ≅ 2.07 L.

8 0

3 years ago

When the equation Eg(s) + DnGO4(aq) ---> Dn(s) + Eg2(GO4)3(aq) is balanced, what is the coefficient in

Artemon [7]

B) 2

You would first balance GO4 by adding a coefficient of 3 in front of DnGO4 in the reactants. Then you’d balance the 3DnGO4 by adding a coefficient of 3 in front of Dn in the products. Finally you’d balance Eg by adding a coefficient of 2 in front of Eg to balance with the Eg2(GO4)3 in the products.

5 0

3 years ago

an engineer wishes to design a container that will hold 12.0 mol of ethane at a pressure no greater than 5.00x10*2 kPa and a tem

OleMash [197]

Answer:

The minimum volume of the container is 0.0649 cubic meters, which is the same as 64.9 liters.

Explanation:

Assume that ethane behaves as an ideal gas under these conditions.

By the ideal gas law,

$P\cdot V = n\cdot R\cdot T$ ,

$\displaystyle V = \frac{n\cdot R\cdot T}{P}$ .

where

$P$ is the pressure of the gas,
$V$ is the volume of the gas,
$n$ is the number of moles of particles in this gas,
$R$ is the ideal gas constant, and
$T$ is the absolute temperature of the gas (in degrees Kelvins.)

The numerical value of $R$ will be $8.314$ if $P$ , $V$ , and $T$ are in SI units. Convert these values to SI units:

$P =\rm 5.00\times 10^{2}\;kPa = 5.00\times 10^{2}\times 10^{3}\; Pa = 5.00\times 10^{5}\; Pa$ ;
$V$ shall be in cubic meters, $\rm m^{3}$ ;
$T = \rm 52.0 \textdegree C = (52.0 + 273.15)\; K = 325.15\; K$ .

Apply the ideal gas law:

$\displaystyle \begin{aligned}V &= \frac{n\cdot R\cdot T}{P}\\ &= \frac{12.0\times 8.314\times 325.15}{5.00\times 10^{5}}\\ &= \rm 0.0649\; m^{3} \\ &= \rm (0.0649\times 10^{3})\; L \\ &=\rm 64.9\; L\end{aligned}$ .

4 0

3 years ago

What is the cell potential for the reaction mg(s+fe2+(aq?mg2+(aq+fe(s at 77 ?c when [fe2+]= 3.40 m and [mg2+]= 0.210 m . express

Galina-37 [17]

First, you need to calculate the standard cell potential using standard reduction potential from a textbook or online. Since Mg becomes Mg+2, magnesium is being oxidized because it is losing electrons, you need to flip its potential

Fe+2 + 2e- --> Fe potential= -0.44
Mg+2 + 2e- --> Mg potential= -2.37

Cell potential= (-0.44) + (+2.37)= 1.93 V

Now, you need to use Nernst formula to get the answer. I have attached a PDF with the work.

Download pdf

8 0

3 years ago

Suppose you need to prepare 141.9 mL of a 0.223 M aqueous solution of NaCl. What mass of NaCl do you need to use to make the sol

Afina-wow [57]

Answer:

1.811 g

Explanation:

The computation of the mass need to use to make the solution is shown below:

We know that molarity is

$Molarity = \frac{Number\ of\ moles}{Volume\ in\ L}$