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kumpel [21]
3 years ago
11

Water (H2O) is not found on the periodic table because water is it an _.

Chemistry
2 answers:
tamaranim1 [39]3 years ago
6 0
Compound, not element. water is made up of hydrogen and 2 oxygen, which are elements but compounds are 2 and more elements
omeli [17]3 years ago
3 0
It does not consist of a single element 

Hope I helped! ( Smiles )

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Distilled water density 1.0 g/cm^3 propane density 0.494 g/cm^3 salt water density 1.025 g/cm^3 liquid gold density 17.31g/cm^3
I am Lyosha [343]
What are you asking?
6 0
3 years ago
2. Which of the following elements does not lose an electron easily? (a) Na (6) F (c) Mg (d) Al​
vaieri [72.5K]

Answer:

Answer: (b) F

Explanation:

Sodium has 1, magnesium has 2 and Aluminium has 3 electrons in its outermost shell whereas Fluorine has 7 electrons in its outermost shell hence Fluorine does not lose electrons easily.

The electronic configuration of fluorine is 2,7.

Fluorine is the ninth element with a total of 9 electrons.

The first two electrons will go in the 1s orbital.

The next 2 electrons for F go in the 2s orbital.

The remaining five electrons will go in the 2p orbital. Therefore the F electron configuration will be 1s22s22p5.

3 0
2 years ago
What is the equilibrium expression for the reatcion bellow<br><br> 2SO3(g)&lt;=&gt;O2(g)+2SO2(g)
Luda [366]

Answer:

[O2(g)][SO2(g)]^2/[SO3(g)]^2

8 0
3 years ago
Please help on 8 thank youuuuuuuuuu
enyata [817]

Answer:

b)5l x 10kg  c)10kg + 9l   (Not sure for the last 1)

6 0
3 years ago
Look at sample problem 19.10 in the 8th ed Silberberg book. Write the Ksp expression. Find the concentrations of the ions you ne
777dan777 [17]

Answer:

Explanation:

From the information given:

CaF_2 \to Ca^{2+} + 2F^-

Ksp = 3.2 \times 10^{-11}

no of moles of Ca^{2+} = 0.01 L × 0.0010 mol/L

no of moles of Ca^{2+} = 1 \times 10^{-5} \ mol

no of moles of F^- = 0.01 L × 0.00010 mol/L

no of moles of F^- = 1 \times 10^{-6}\ mol

Total volume = 0.02 L

[Ca^{2+}}] = \dfrac{1\times10^{-5} \ mol}{0.02 \ L} \\ \\  \\  \[[Ca^{2+}}] = 0.0005 \ mol/L

[F^{-}] = \dfrac{(1\times 10^{-6} \ mol)}{0.02 \ L}

[F^{-}] = 5 \times 10^{-5}  \ mol/L

Q = [Ca^{2+}][F^-]^2 \\ \\ Q = 0.0005 \times (5\times 10^{-5})^2 \\ \\ Q = 1.25 \times 10^{-12}

Since Q<ksp, then there will no be any precipitation of CaF2

3 0
3 years ago
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