The amount of heat energy needed to heat 200 g of water from 15 °C to its boiling point, and boil it, is

1 mole of any gas is equivalent to?

iragen [17]

Answer:

22.4 L at standard temperature and pressure.

6 0

2 years ago

A white crystalline salt conducts electricity when it is melted and when it dissolves in water.

sergeinik [125]

Answer: Option (a) is the correct answer.

Explanation:

Ionic salts are defined as the salts which tend to contain ionic bonds as there occurs transfer of electrons between its combining atoms.

So, when an ionic salt melts or it is dissolved in water then it will dissociate into its respective ions and as electricity is the flow of electrons or ions. Hence, this salt is then able to conduct electricity.

As covalent compounds are insoluble in water so, they do no dissociate into ions. Hence, they do not conduct electricity.

Similarly, metallic and network solids do not dissociate into ions either when melted or dissolved in water. Therefore, they also do not conduct electricity.

Thus, we can conclude that when a white crystalline salt conducts electricity when it is melted and when it dissolves in water then this bond is of ionic type.

7 0

2 years ago

Coke is an impure form of carbon that is often used in the in-dustrial production of metals from their oxides. if a sample of co

olasank [31]

The reaction equation is:
2CuO(s) + C(s) → 2Cu(s) + CO₂(g)

First, we determine the number of grams present in one ton of copper oxide. This is:
1 ton = 9.09 x 10⁵ g

We convert this into moles by dividing by the molecular mass of copper oxide, which is:
9.09 x 10⁵ / 79.5 = 11,434 moles

Each mole of carbon reduces two moles of copper oxide, so the moles of carbon required are:
11,434 / 2 = 5,717 moles of Carbon required

The mass of carbon is then:
5,717 x 12 = 68,604 grams

The mass of coke is:
68,604 / 0.95 = 72,214 g

The mass of coke required is 7.22 x 10⁴ grams

5 0

2 years ago

An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respective

katovenus [111]

Answer:

The new partial pressures after equilibrium is reestablished:

$PCl_3,p_1'=6.798 Torr$

$Cl_2,p_2'=26.398 Torr$

$PCl_5,p_3'=223.402 Torr$

Explanation:

$PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)$

At equilibrium before adding chlorine gas:

Partial pressure of the $PCl_3=p_1=13.2 Torr$

Partial pressure of the $Cl_2=p_2=13.2 Torr$

Partial pressure of the $PCl_5=p_3=217.0 Torr$

The expression of an equilibrium constant is given by :

$K_p=\frac{p_1}{p_1\times p_2}$

$=\frac{217.0 Torr}{13.2 Torr\times 13.2 Torr}=1.245$

At equilibrium after adding chlorine gas:

Partial pressure of the $PCl_3=p_1'=13.2 Torr$

Partial pressure of the $Cl_2=p_2'=?$

Partial pressure of the $PCl_5=p_3'=217.0 Torr$

Total pressure of the system = P = 263.0 Torr

$P=p_1'+p_2'+p_3'$

$263.0Torr=13.2 Torr+p_2'+217.0 Torr$

$p_2'=32.8 Torr$

$PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)$

At initail

(13.2) Torr (32.8) Torr (13.2) Torr

At equilbriumm

(13.2-x) Torr (32.8-x) Torr (217.0+x) Torr

$K_p=\frac{p_3'}{p_1'\times p_2'}$

$1.245=\frac{(217.0+x)}{(13.2-x)(32.8-x)}$

Solving for x;

x = 6.402 Torr

The new partial pressures after equilibrium is reestablished:

$p_1'=(13.2-x) Torr=(13.2-6.402) Torr=6.798 Torr$

$p_2'=(32.8-x) Torr=(32.8-6.402) Torr=26.398 Torr$

$p_3'=(217.0+x) Torr=(217+6.402) Torr=223.402 Torr$

6 0

2 years ago

How many moles of oxygen atoms are in 7.9E-1 moles of CO_2

Ilya [14]

Answer:

The number of moles of O atom in $(7.9\times10^{-1})$ mol of $CO_{2}$ = 1.6

Explanation:

1 molecule of $CO_{2}$ contains 2 atoms of O

So, $(6.023\times 10^{23})$ molecules of $CO_{2}$ contains $(2\times6.023\times10^{23})$ atoms of O.

We know that 1 mol of an atom/molecule/ion represents $6.023\times10^{23}$ numbers of atoms/molecules/ions respectively.

So, $(6.023\times 10^{23})$ molecules of $CO_{2}$ is equal to 1 mol of $CO_{2}$ .

$(2\times6.023\times10^{23})$ atoms of O is equal to 2 moles of O atom.

Hence, 1 mol of $CO_{2}$ contains 2 moles of O atom.

Therefore, $(7.9\times10^{-1})$ mol of $CO_{2}$ contains $(2\times7.9\times10^{-1})$ moles of O atom or 1.6 moles of O atom.

3 0

2 years ago