All categories

3 years ago

13. What is the symbol for Enthalpy change?

Chemistry

1 answer:

OverLord2011 [107]3 years ago

5 0

Answer:

Explanation:

13 ) symbol of enthalpy change = Δ H .

14 ) enthalpy change is nothing but heat absorbed or evolved .

During fusion enthalpy change

Δ H .= m Lf , m is mass and Ls is latent heat of fusion

During evaporation, enthalpy change

Δ H .= m Lv , m is mass and Lv is latent heat of evaporation

during temperature rise , enthalpy change

Δ H = m c Δ T

In case of gas , enthalpy change can be calculated by the following relation

Δ H = Δ E + W , Δ E is change in internal energy , W is work done by gas.

15 ) When enthalpy change is negative , that means heat is released to the environment .So reaction is called exothermic .

when heat is absorbed enthalpy change is positive . Reaction is endothermic.

You might be interested in

What uses lines of bubbles in a superheated liquid to track particles?

Dmitry [639]

Is bubble chamber one of your choices? Bubble chamber sounds like a good fit for the question.

7 0

3 years ago

The standard unit (SI) for mass is

telo118 [61]

The kilogram is the si

6 0

4 years ago

Help!!!!

V125BC [204]

Answer:D and C sorry if im wrong

Explanation:

8 0

3 years ago

Read 2 more answers

How many protons, neutrons, and electrons make up an ion of

yKpoI14uk [10]

Answer:

protons=15

electron=15

neutron=16

6 0

3 years ago

An iron(iii) sulfate hydrate is 18.4% water. What is the formula of the hydrate? What is the name of the hydrate?

aleksley [76]

Answer:- Formula of the hydrate is $Fe_2(SO_4)_3.5H_2O$ and it's name is Iron(III)sulfate pentahydrate.

Solution:- As per the given information, there is 18.4% water in the hydrate. If we assume the mass of the hydrate as 100 grams then there would be 18.4 grams of water and 81.6 grams of Iron(III)sulfate present in the hydrate.

Molar mass for Iron(III)sulfate is 399.88 gram per mol and the molar mass for water is 18.02 gram per mol.

We will calculate the moles of Iron(III)sulfate and water present in the compound on dividing their grams by their molar masses as:

$81.6gFe_2(SO_4)_3(\frac{1mol}{399.88g})$

= $0.204molFe_2(SO_4)_3$

$18.4gH_2O(\frac{1mol}{18.02g})$

= $1.02molH_2O$

Now, the next step is to calculate the mol ratio and for this we divide the moles of each by the least one of them means whose moles are less. Here, the moles of Iron(III)sulfate are less than moles of water. So, we divide the moles of each by 0.204.

$Fe_2(SO_4)_3=\frac{0.204}{0.204}$ = 1

$H_2O=\frac{1.02}{0.204}$ = 5

There is 1:5 mol ratio between Iron(III)sulfate and water. So, the formula of the hydrate is $Fe_2(SO_4)_3.5H_2O$ and the name of the hydrate is Iron(III)sulfate pentahydrate.

3 0

3 years ago