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Andrei [34K]
3 years ago
11

Which best explains the high surface tension of water?

Chemistry
2 answers:
ololo11 [35]3 years ago
7 0
Your answer would be D
andrezito [222]3 years ago
5 0
D its hydrogen bonding interactions

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Sodium ethanoate ? --> solid sodium ethanoate
timurjin [86]
I’m pretty sure it craters sodium acetate
7 0
3 years ago
CH=C-CH=CH– CH = CH, what is the name of this molecule?​
Kobotan [32]

Answer:

pent-3-ene-1-yne

Explanation:

1 2 3 4 5

CH ≡ C - CH = CH - CH3

IUPAC name : Pent-3-ene-1-yne

3 0
2 years ago
Deterioration of buildings, bridges, and other structures through the rusting of iron costs millions of dollars a day. The actua
GrogVix [38]

The value of ∆H when 0.250kg of iron rusts is -1.846 × 10³kJ.

The rust forms when 4.85X10³ kJ of heat is released is 888.916 g.

<h3>Chemical reaction:</h3>

4 Fe + 3O2 ------ 2Fe2O3

∆H = -1.65×10³kJ

A) Given,

mass of iron = 0.250kg = 250 g

<h3>Calculation of number of moles</h3>

moles = given mass/ molar mass

= 250/ 55.85 g/mol.

= 4.476 mol

As we know that,

For the rusting of 4 moles of Fe, ∆H = -1.65×10³kJ

For the rusting of 4.476 moles of Fe ∆H required can be calculated as

-1.65×10³kJ × 4.476 mol/ 4mol

∆H required = -1.846 × 10³kJ

Now,

when 2 mol of Fe2O3 formed, ∆H = - 1.65×10³kJ

It can be said that,

-1.65×10³kJ energy released when 2 mol of Fe2O3 formed

So, -4.6 × 10³kJ energy released when 2 mol of Fe2O3 formed

= 2 × -4.6 × 10³kJ / -1.65×10³kJ

= 5.57 mol of Fe2O3 formed

Now,

mass of Fe2O3 formed = 5.57 mol × 159.59 g/mol

= 888.916 g

Thus, we calculated that the rust forms when 4.85X10³ kJ of heat is released is 888.916 g. and the value of ∆H when 0.250kg of iron rusts is -1.846 × 10³kJ.

learn more about ∆H:

brainly.com/question/24170335

#SPJ4

DISCLAIMER:

The given question is incomplete. Below is the complete question

QUESTION:

Deterioration of buildings, bridges, and other structures through the rusting of iron costs millions of dollars a day. The actual process requires water, but a simplified equation is 4Fe(s) + 3O₂(g) → 2Fe₂O₃(s) ΔH = -1.65×10³kJ

a) What is the ∆H when 0.250kg iron rusts.

(b) How much rust forms when 4.85X10³ kJ of heat is released?

7 0
1 year ago
4Fe + 30₂ ⇒ Fe₂0₃
DaniilM [7]

Answer:

A

The nuber of each one should be same

5 0
1 year ago
A 35 MGD surface Water treatement plant proposes to to use alum Coagulation for turbidity reduction using 44 mg/ L alum. the raw
CaHeK987 [17]
C) is the water alkainty adequate. I.e
5 0
3 years ago
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