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Greeley [361]
3 years ago
8

Does Hornfels have biological origins

Chemistry
1 answer:
anzhelika [568]3 years ago
3 0

Answer:

Hornfels most commonly form in the aureole of granitic intrusions in the upper or middle crust. Hornfels formed from contact metamorphism by volcanic activity very close to the surface can produce unusual and distinctive minerals.

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Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the f
julia-pushkina [17]

Explanation:

As per Brønsted-Lowry concept of acids and bases, chemical species which donate proton are called Brønsted-Lowry acids.

The chemical species which accept proton are called Brønsted-Lowry base.

(a) HNO_3 + H_2O \rightarrow H_3O^+ + NO_3^-

HNO_3 is Bronsted lowry acid and NO_3^- is its conjugate base.

H_2O is Bronsted lowry base and H_3O^+ is its conjugate acid.

(b)

CN^- + H_2O \rightarrow HCN + OH^-

CN^- is Bronsted lowry base and HCN is its conjugate acid.

H_2O is Bronsted lowry acid and OH^- is its conjugate base.

(c)

H_2SO_4 + Cl^- \rightarrow HCl + HSO_4^-

H_2SO_4 is Bronsted lowry acid and HSO_4^- is its conjugate base.

Cl^- is Bronsted lowry base and HCl is its conjugate acid.

(d)

HSO_4^-+OH^- \rightarrow SO_4^{2-}+H_2O

HSO_4^- is Bronsted lowry acid and SO_4^{2-} is its conjugate base.

OH^- is Bronsted lowry base and H_2O is its conjugate acid.

(e)

O_{2-}+H_2O \rightarrow 2OH^-

O_{2-} is Bronsted lowry base and OH- is its conjugate acid.

H_2O is Bronsted lowry acid and OH- is its conjugate base.

6 0
3 years ago
How many moles is 2.2 x 1023 atoms of Fe?
kow [346]

Answer:

it's 3.77 moles

Explanation:

Pls give brainliest :)

4 0
3 years ago
HELP
Jobisdone [24]

Answer:

hello,

first one is 25.15

second is 301.55

Explanation:

honestly if you look up on the internet there is a converter to get you your answers

5 0
3 years ago
Read 2 more answers
Hydrogen sulfide burns form sulfur dioxide:
Helga [31]

Answer: 404.04 kJ.

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

moles of H_2S

\text{Number of moles}=\frac{26.7g}{34.1g/mol}=0.78moles

2H_2S(g)+3O_2(g)\rightarrow 2SO_2(g)+2H_2O(g)    \Delta H=-1036kJ

According to stoichiometry :

2 moles of H_2S on burning produces = 1036 kJ

Thus 0.78 moles of H_2S on burning produces =\frac{1036kJ}{2}\times 0.78=404.04

Thus the enthalpy change when burning 26.7 g of hydrogen sulfide is 404.04 kJ.

7 0
3 years ago
What was technicalagy like in the past
V125BC [204]
Most of it was analog
8 0
3 years ago
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