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Kruka [31]
3 years ago
12

Any body tell me Iron salt and water produced what ​

Chemistry
1 answer:
kolezko [41]3 years ago
7 0

iron salts react with water to form hydrated iron ( III ) oxide which is basically rust

hope that helps :)

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How much CaCO3 would have to be decomposed to produce 247 g of CaO
vovangra [49]

441 g CaCO₃ would have to be decomposed to produce 247 g of CaO

<h3>Further explanation</h3>

Reaction

Decomposition of CaCO₃

CaCO₃ ⇒ CaO + CO₂

mass CaO = 247 g

mol of CaO(MW=56 g/mol) :

\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{247}{56}\\\\mol=4.41

From equation, mol ratio CaCO₃ : CaO = 1 : 1, so mol CaO :

\tt \dfrac{1}{1}\times 4.41=4.41

mass CaCO₃(MW=100 g/mol) :

\tt mass=mol\times MW\\\\mass=4.41\times 100\\\\mass=441~g

6 0
3 years ago
What is the main function of the circulatory system?
likoan [24]
B- is the answer


Hope this helps


Have a nice day~
4 0
3 years ago
Read 2 more answers
What is the molarity of a solution that has 2.50 moles of NaOH dissolved in 0.500 L of solution?
const2013 [10]

Answer: 5 is the molarity

Explanation:

The molarity formula is moles over liters and that in your case is 2.50 moles divided by .500 L which results in 5 which is your answear hope this helped god bless

5 0
3 years ago
Read 2 more answers
Select all the correct locations.
NikAS [45]

。☆✼★ ━━━━━━━━━━━━━━  ☾  

The land breeze arrow is correct

The rest are incorrect.

Warm air should 'rise' by the land

This air then moves towards the sea and cools down

The cool air would then sink

Have A Nice Day ❤    

Stay Brainly! ヅ    

- Ally ✧    

。☆✼★ ━━━━━━━━━━━━━━  ☾

5 0
3 years ago
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. A bright violet line occurs at 435.8 nm in the emission spectrum of mercury vapor. What amount of energy, in joules, must be r
Tom [10]

Answer :  The energy released by an electron in a mercury atom to produce a photon of this light must be, 4.56\times 10^{-19}J

Explanation : Given,

Wavelength = 435.8nm=435.8\times 10^{-9}m

conversion used : 1nm=10^{-9}m

Formula used :

E=h\times \nu

As, \nu=\frac{c}{\lambda}

So, E=h\times \frac{c}{\lambda}

where,

\nu = frequency

h = Planck's constant = 6.626\times 10^{-34}Js

\lambda = wavelength = 435.8\times 10^{-9}m

c = speed of light = 3\times 10^8m/s

Now put all the given values in the above formula, we get:

E=(6.626\times 10^{-34}Js)\times \frac{(3\times 10^{8}m/s)}{(435.8\times 10^{-9}m)}

E=4.56\times 10^{-19}J

Therefore, the energy released by an electron in a mercury atom to produce a photon of this light must be, 4.56\times 10^{-19}J

3 0
3 years ago
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