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miss Akunina [59]
3 years ago
11

What are the natural components of water?

Chemistry
2 answers:
11Alexandr11 [23.1K]3 years ago
8 0
Two Hydrogen and 1 Oxygen Atoms
DerKrebs [107]3 years ago
7 0

two hydrogen atoms and one oxygen atom. hope this helped.

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Describe the Rutherford model... and and what are its limitation...<br><br>going offline forever ✌️​
SashulF [63]

Answer:

Limitations of Rutherford Atomic Model

Although the Rutherford atomic model was based on experimental observations it failed to explain certain things. Rutherford proposed that the electrons revolve around the nucleus in fixed paths called orbits. ... Ultimately the electrons would collapse in the nucleus.

8 0
3 years ago
A 111.6 gram sample of iron (MW=55.8) was heated from 0 degrees C to 20 degrees C. It absorbed 1004 Joules of energy. What is th
qwelly [4]
Use the equation q=ncΔT.
q= heat absorbed our released (in this case 1004J)
n= number of moles of sample ( in this case 2.08 mol)
c=molar heat capacity 
ΔT=change in temperature (in this case 20°C)
You have to rewrite the equation for c.
c=q/nΔT
c=1004J/(2.08mol x 20°C)
c=24.1 J/mol°C
 
I hope this helps
7 0
3 years ago
You have to prepare some 2 M solutions, with 10 g of solute in each. What volume of solution will you prepare, for each solute b
alexdok [17]

Answer:

             0.045 L  or 45 mL

Explanation:

Moles = Mass/M.Mass

Moles = 10 g / 109.94 g/mol

Moles = 0.09 moles

Also,

Molarity = Moles / Vol in L

Or,

Vol in L = Moles / Molarity

Vol in L = 0.09 mol / 2 mol/L

Vol in L = 0.045 L

8 0
2 years ago
the radius of magnesium atom is 0.160nm. the radius of a magnesium ion is 7.2x 10^-11m. explain the difference in size between t
Allushta [10]

Answer:

during reaction magnesium lises ions.

Explanation:

magnesium reacts by losing two ions which makes it smaller in size.

8 0
3 years ago
An ideal gas in a sealed container has an initial volume of 2.80 L. At constant pressure, it is cooled to 18.00 °C, where its
Aleks04 [339]

Answer:

T_1=-91.18\°C

Explanation:

Hello there!

In this case, given the T-V variation, we understand it is possible to apply the Charles' law as shown below:

\frac{T_1}{V_1}= \frac{T_2}{V_2}

Thus, since we are interested in the initial temperature, we can solve for T1, plug in the volumes and use T2 in kelvins:

T_1= \frac{T_2V_1}{V_2}\\\\T_1= \frac{(18.00+273.15)K(1.75L)}{(2.80L)}\\\\T_1=182K-273.15\\\\T_1=-91.18\°C

Best regards!

6 0
3 years ago
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