All categories

3 years ago

A. 1720 kJ B. 125.6 kJ C. 3440 kJ D. 4730 kJ

Chemistry

1 answer:

Feliz [49]3 years ago

7 0

Answer:

Q = 3440Kj

Explanation:

Given data:

Mass of gold = 2kg

Latent heat of vaporization = 1720 Kj/Kg

Energy required to vaporize 2kg gold = ?

Solution:

Equation

Q= mLvap

It is given that heat required to vaporize the one kilogram gold is 1720 Kj thus, for 2 kg

by putting values,

Q= 2kg × 1720 Kj/Kg

Q = 3440Kj

You might be interested in

Which of the following reactions is incorrectly labeled?

mars1129 [50]

Answer:

The option a termed as precipitation reaction is incorrectly labelled. 

Explanation:

The chemical reactions are classified based on the reactants used and products formed in a reaction. They are decomposition reaction, single displacement reaction, double displacement reaction, acid-base neutralisation reaction, precipitation reaction, combustion reaction, redox reaction and organic reaction.

Among these, the given options are labelled as precipitation and combustion reaction. The one which is labelled as combustion reaction is correct because combustion reactions occur in the presence of oxygen only and the products of combustion reaction should include water, oxygen or carbon and heat.

The other option which is labelled as precipitation reaction is incorrect because precipitation reaction occurs when an ionic substance will come out of a solution due to heating it or stirring it making the solubility of the ionic substance in a solution zero such that it will come out as solid and form a layer at the bottom of the solution.

But in this case all the products are in aqueous state, there is absence of any ionic substance in solid state, so the option which is labelled as precipitation reaction is incorrectly labelled.

3 0

3 years ago

A sample of methane occupies a volume of 370.0 ml at 25 oC and exerts a pressure of 1020 mm Hg. If the volume is allowed to expa

vagabundo [1.1K]

Answer:

744.9 mmHg ≅ 745 mmHg

Explanation:

The base to solve this, is the Ideal Gases Law. The mentioned formula is:

P . V = n . R . T

To compare two situations, we can propose:

For the first situation P₁ . V₁ = n₁. R . T₁

For the second situation P₂ . V₂ = n₂ . R . T₂

As the sample has the same moles and R is a constant value, we can avoid them so: (P₁ . V₁) / T₁ = (P₂ . V₂) / T₂

We need to make Tº unit conversion:

25ºC + 273 = 298K

We replace data → (370 mL . 1020 mmHg) / 298K = (P . 510 mL) / 300 K

(377400 mL.mmHg / 298K) . 300 K = P . 510 mL

379932.8 mL . mmHg = P . 510 mL

(379932.8 mL . mmHg) / 510 mL = P → 744.9 mmHg

7 0

3 years ago

Pls help me on this I have 5 minutes left to submit

pantera1 [17]

Heterogenous mixture mean unsimilar substances .

Chicken noodle soyp.
salt water
propane.

An element is a substance made of only one type of atoms

Example-Al,S

7 0

2 years ago

Read 2 more answers

What is the connection of ascorbic acid when one tablet is dissolved in 200cm cubed of water ?

BartSMP [9]

Answer:

Option-D: 2.3 × 10⁻³ mol/dm³

Explanation:

Calculate moles of ascorbic acid,

Moles = Mass / M.Mass

Moles = 0.080 g / 176 g/mol

Moles = 0.00045 mole

Also,

Molarity = Moles / Vol. in dm³

Molarity = 0.00045 mol / 0.20 dm³

Molarity = 0.00227 mol.dm⁻¹ or 2.3 × 10⁻³ mol/dm³

5 0

2 years ago

We want to calculate the concentrations of all species in a 0.58 M Na 2 SO 3 (sodium sulfite) solution. The ionization constants

NeX [460]

Explanation:

Reaction equation is as follows.

$Na_{2}SO_{3}(s) \rightarrow 2Na^{+}(aq) + SO^{2-}_{3}(aq)$

Here, 1 mole of $Na_{2}SO_{3}$ produces 2 moles of cations.

$[Na^{+}] = 2[Na_{2}SO_{3}] = 2 \times 0.58$

= 1.16 M

$[SO^{2-}_{3}] = [Na_{2}SO_{3}]$ = 0.58 M

The sulphite anion will act as a base and react with $H_{2}O$ to form $HSO^{-}_{3}$ and $OH^{-}$ .

As, $K_{b} = \frac{K_{w}}{K_{a_{2}}}$

= $\frac{10^{-14}}{6.3 \times 10^{-8}}$

= $1.59 \times 10^{-7}$

According to the ICE table for the given reaction,

$SO^{2-}_{3} + H_{2}O \rightleftharpoons HSO^{-}_{3} + OH^{-}$

Initial: 0.58 0 0

Change: -x +x +x

Equilibrium: 0.58 - x x x

So,

$K_{b} = \frac{[HSO^{-}_{3}][OH^{-}]}{[SO^{2-}_{3}]}$

$1.59 \times 10^{-7} = \frac{x^{2}}{0.58 - x}$

$x^{2} = 1.59 \times 10^{-7} \times (0.58 - x)$

x = 0.0003 M

So, x = $[HSO^{-}_{3}] = [OH^{-}]$ = 0.0003 M

$[SO^{2-}_{3}]$ = 0.58 - 0.0003

= 0.579 M

Now, we will use $[HSO^{-}_{3}]$ = 0.0003 M

The reaction will be as follows.

$HSO^{2-}_{3} + H_{2}O \rightleftharpoons H_{2}SO_{3} + OH^{-}$

Initial: 0.0003

Equilibrium: 0.0003 - x x x

$K_{b} = \frac{x^{2}}{0.0003 - x}$

$K_{b} = \frac{K_{w}}{K_{a_{1}}}$

= $\frac{10^{-14}}{1.4 \times 10^{-2}}$

= $7.14 \times 10^{-13}$

Therefore, $7.14 \times 10^{-13} = \frac{x^{2}}{0.0003 - x}$

As, x <<<< 0.0003. So, we can neglect x.

Therefore, $x^{2} = 7.14 \times 10^{-13} \times 0.0003$

= $0.00214 \times 10^{-13}$

x = $0.0146 \times 10^{-6}$

x = $[OH^{-}] = [H_{2}SO_{3}]$ = $1.46 \times 10^{-8}$

$[H^{+}] = \frac{10^{-14}}{[OH^{-}]}$

= $\frac{10^{-14}}{0.0003}$

= $3.33 \times 10^{-11}$ M

Thus, we can conclude that the concentration of spectator ion is $3.33 \times 10^{-11}$ M.

5 0

3 years ago