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3 years ago

Write conversion factors between moles of each constituent element and moles of the compound for c12h22o11

Chemistry

1 answer:

olga nikolaevna [1]3 years ago

3 0

The conversion factor from compound to element is actually just based on the number of elements in the compound itself.

From the compound C12H22O11, we can say that there are 12 C, 22 H and 11 O, therefore the conversion factors are:

1 mol C12H22O11=12 moles of C

1 mol C12H22O11=22 moles of H

1 mol C12H22O11= 11 moles of O

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Find the mass of 4.5 moles of H3PO4

Contact [7]

Hey there!

H₃PO₄

Find molar mass.

H: 3 x 1.008 = 3.024

P: 1 x 30.97 = 30.97

O: 4 x 16 = 64

---------------------------------

97.994 grams

The mass of 1 mole of H₃PO₄ is 97.994 grams.

We have 4.5 moles.

97.994 x 4.5 = 440

The mass of 4.5 moles of H₃PO₄ is 440 grams.

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2 years ago

What is wrong with this model of the atom.

sp2606 [1]

Answer:

A. The neutrons and electrons are in the wrong place.

Explanation:

The atom's nucleus contains both protons and neutrons, whilst the electrons are arranged in shells around the nucleus.

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2 years ago

The ionization of pure water forms _____.

Evgesh-ka [11]

Answer: The ionization of pure water forms <u><em>hydroxide and hydronium ions.</em></u>

Explanation:

Ionization is a reaction in the pure water in which water breaks down into its constituting ions that hydronium ion and hydroxide ions.

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3 years ago

Part A

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Answer:

$n_{Cl_2}=0.3molCl_2$

Explanation:

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In this case, according to the given chemical reaction whereas the sodium chloride is in a 2:1 mole ratio with chlorine, the required moles of the later are computed as shown below:

$n_{Cl_2}=0.6molNaCl*\frac{1molCl_2}{2molNaCl}$

So we cancel out the moles of NaCl to obtain:

$n_{Cl_2}=0.3molCl_2$

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2 years ago

A sample of sodium-24 with an activity of 14 mCi is used to study the rate of blood flow in the circulatory system. If sodium-24

SOVA2 [1]

Answer:

See explanation

Explanation:

From;

0.693/t1/2 = 2.303/t log (Ao/A)

Where;

t1/2 = half life of the sodium-24

t = time taken

Ao = initial activity of sodium-24

A= activity of sodium-24 at time t

0.693/15 = 2.303/15 log (14/A)

0.0462 = 0.1535 log (14/A)

0.0462/0.1535 = log (14/A)

log (14/A) = 0.0462/0.1535

14/A = Antilog(0.3)

14/A= 1.995

A = 14/1.995

A = 7.0 mCi

0.693/15 = 2.303/30 log (14/A)

0.0462 =0.0768 log(14/A)

0.0462/0.0768 =log (14/A)

(14/A) =Antilog (0.6)

A = 14/Antilog (0.6)

A = 3.5 mCi

0.693/15 = 2.303/45 log (14/A)

0.0462= 0.0512 log (14/A)

log (14/A) = 0.0462/0.0512

log (14/A) = 0.9

(14/A) = Antilog (0.9)

A= 14/Antilog (0.9)

A = 14/7.9

A = 1.77 mCi

2.5 days = 2.5 * 24 hours = 60 hours

0.693/15 = 2.303/60 log (14/A)

0.0462 = 0.03838 log (14/A)

log (14/A) = 0.0462/0.03838

(14/A) = Antilog(1.2)

A= 14/Antilog(1.2)

A = 14/15.8

A = 0.886 mCi

Note that activity (A) decreases as time increases.

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3 years ago