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2 years ago

How many moles of water are in 1.23 x 10 to the 18th power water molecules

Chemistry

2 answers:

Ksivusya [100]2 years ago

6 0

Answer:

The answer is 2.04x10^-6 moles of water.

Explanation:

Avogadro's number is defined as the number of particles found in an amount of substance per mole. It is the factor that relates the moles of a substance to the mass of that substance. We will use Avogadro's number to calculate the number of moles of water. as follows:

Avogadro's number = 1 mol = 6.022 x 10^23 particles.

we will use the conversion factor to calculate the number of moles of water:

Moles of water = 1.23x10^18 particles x (1mol/6.022x10^23 particles) = 2.04x10^-6 moles

mamaluj [8]2 years ago

3 0

Divide the number of molecules you have by, 6.022 x 10^23. This will give you the moles of water, or the moles of anything, since there is always 6.022 x 10^23 molecules in 1 mole of substance.

1.23x10^24 atoms/6.022x10^23 atom/mole = 2.04 mole H20

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trans-2-Butene does not exhibit a signal in the double-bond region of the spectrum (1600–1850 cm-1); however, IR spectroscopy is

spayn [35]

Answer:

The other signal that would indicate the presence of a C= C bond appears close to 3100 $cm^{-1}$ .

Explanation:

Bands that appear above 3000 $cm^{-1}$ are often unsaturation diagnoses suggest. The band at 3000- 3100 $cm^{-1}$ is characteristics for C-H stretching frequencies and normally is overlaps with the ones for alkanes because it is a band of weak intensity.

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2 years ago

Chlorine gas was first prepared in 1774 by the oxidation of NaCl with MnO2:

irina1246 [14]

Answer:

0.5

Explanation:

2NaCl(s) + 2H2SO4(l) + MnO2(s) → Na2SO4(s) + MnSO4(s) + 2H2O(g) + Cl2(g)

Using ideal gas equation,

PV = nRT

28.7torr

Converting torr to atm,

= 0.0378atm

V = 0.597L

T = 27 °C

= 300 K

a) PV = nRT

(0.0378atm) * (0.597L) = n(0.0821) * (300k)

= 0.000915 mol

moles of water and chlorine = 0.000915 mol

From the above equation, the ratio of water to chlorine = 1 : 2

Therefore, mole of chlorine = 0.000915/2

= 0.000458 mol

mole fraction = moles of specie/moles of all the species present

= 0.000458/0.000915

= 0.5

5 0

2 years ago

Eddie is going to do an experiment to find out which freezes more easily—distilled water or salt water. In what order should he

mel-nik [20]

1. Make a Prediction

2. Fill both beakers with water

3. Dissolve salt in one of the beakers

4. Place both in the freezer and observe

5. Write a report

(Always make the prediction first! That's a hypothesis!)

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2 years ago

Read 2 more answers

A sample of ammonia gas at 75°c and 445 mm hg has a volume of 16.0 l. what volume will it occupy if the pressure rises to 1225 m

ioda

In this kind of exercises, you should use the "ideal gas" rules: PV = nRT
P should be in Pascal:
445mmHg = 59328Pa
1225mmHg = 163319Pa

V should be in cubic meter:
16L = 0.016 m3

R = $\frac{PV}{nT}$ = constant
$\frac{P1 V1}{n T}$ = $\frac{P2 V2}{n T}$
==> P1 * V1 = P2 * V2
V2 = $\frac{P1 V1}{P2}$ = $\frac{445 0.016}{1225}$
V2 = 0.00581 m3 = 5.81 L

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2 years ago

Describe the hybrid orbitals used by the central atom and the types of bonds formed in o3

Tems11 [23]

Hybridization in ozone, O3......

...O = O ........ 1 lone pair on central O, 2 lone pairs on terminal O 
../ 
O .................. 3 lone pairs on terminal O 

I didn't show the second of two resonance structures in which the single and double bonds are reversed. In reality, both bonds are identical have a bond order of 1.5 due to delocalized pi-bonding. 

The central atom exhibits sp2 hybridization since there is trigonal planar electron pair geometry. The notion of hybrid orbitals was "invented" by Linus Pauling in the 1930's as a way of explaining the geometry of molecules, primarily the geometry of carbon compounds. 

If the electron pair geometry is linear, the hybridization is sp. 
If the electron pair geometry is trigonal planar, the hybridization is sp2. 
If the electron pair geometry is tetrahedral, the hybridization is sp3. 

The notion that there is sp3d and sp3d2 because of d-orbital participation has been debunked. Chemists know today that there is no d-orbital involvement in hypervalent molecules regardless of what some out-of-date textbooks and some teachers' dusty old notes may say. Instead, the best explanation involves 3-center, 4-electron bonding.

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2 years ago