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ANEK [815]
10 months ago
12

what volume (in ml) of 1.508 m h3po4 (aq) would be required to react completely with 11.73 ml of 1.006 m ca(oh)2? (aq)

Chemistry
1 answer:
alexdok [17]10 months ago
6 0

The amount of 4.86 ml needed to completely react with 11.73 ml of 1.006 m ca(oh)2 is 4.86 ml.

given data to  be calculated:-

Ca3(PO4)2 (s) + 6H2O can be used to calculate this: 3Ca(OH)2 (aq) + 2H3PO4 (aq) (l)

ca moles (OH)

2  present 11.00 moles of H3PO4 are required; 7.33 moles of H3PO4 in a volume of ml are required; 7.33/1.508 = 4.86.

<h3>How do moles react to an increase in volume?</h3>

The direction that creates more moles of gas benefits from an increase in volume, and since there are more moles of products in this case, the reaction will shift to the right and produce more moles of products.

To know more about mole visit:-

brainly.com/question/26416088

#SPJ4

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6) (a) Calculate the absorbance of the solution if its concentration is 0.0278 M and its molar extinction coefficient is 35.9 L/
Anvisha [2.4K]

Answer:

6) (a) 0.499; (b) 31.7 %

7) 0.15

Explanation:

6) (a) Absorbance

Beer's Law is

A = \epsilon cl\\A = \text{35.9 L&\cdot$mol$^{-1}$cm$^{-1}$} $\times$ 0.0278 mol$\cdot$L$^{-1} \times $ 0.5 cm = \mathbf{0.499}

(b) Percent transmission

A = \log {\left (\dfrac{1}{T}}\right)}\\\\\%T = 100T\\\\T = \dfrac{\%T}{100}\\\\\dfrac{1}{T} = \dfrac{100 }{\%T}\\\\A = \log \left(\dfrac{100 }{\%T} \right ) = 2 - \log \%T\\\\0.499 = 2 - \log \%T\\\\\log \%T = 2 - 0.499 = 1.501\\\\\%T = 10^{1.501} = \mathbf{31.7}

7) Absorbance

A = \log \left (\dfrac{I_{0}}{I} \right ) = \log \left (\dfrac{I_{0}}{0.70I_{0}} \right ) = \log \left (\dfrac{1}{0.70} \right ) = -\log(0.70) = \mathbf{0.15}}

8 0
3 years ago
What kind of bond occurs between two<br> elements sharing three pairs of electrons?
Zigmanuir [339]
Covalent bond i believe
6 0
2 years ago
_____ and _______ are necessary for rusting.​
Anastaziya [24]

Answer :-

<h3>Water and Air are necessary for rusting .</h3>

3 0
2 years ago
Read 2 more answers
Why does the melting of the polar ice caps facilitate the absorption of CO2 by ocean waters?​
Korvikt [17]

hi brainly user! ૮₍ ˃ ⤙ ˂ ₎ა

⊱┈────────────────────────⊰

Answer:

The melting of the caps increases the concentration of CO2 in the water, making its absorption faster.

⊱┈────────────────────────⊰

Explanation:

<h2><u>CO2 absorption</u><u> </u></h2>

The polar caps of our planet have in their composition several elements, among them the CO2 that is absorbed by the atmosphere. Cold waters, which are present in the Arctic, have an easier time absorbing CO2 compared to other waters.

When glaciers melt, the CO2 that is present in the mixture is dissolved in the ocean, increasing its concentration. The cold waters that came from the ice caps increase the absorption of CO2.

Learn more about natural methods remove CO2 from the atmosphere in:

brainly.com/question/14323197

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4 0
1 year ago
The diameter of the He He atom is approximately 0.10 nm nm . Calculate the density of the He atom in g/cm 3 g/cm3 (assuming that
Sladkaya [172]

Answer:

Density of the He atom = 12.69 g/cm³

Explanation:

From the information given:

Since 1 mole of an atom = 6.022x 10²³ atoms)

1 atom of He = 1  \ atom \times  (\dfrac{1  \ mole}{  6.022 \times  10^{23}  \ atoms}) \times ( \dfrac{4.003 \ grams}{  1  \ mole})

=6.647 \times  10^{-24} \  grams

The volume can be determined as  folows:

since the diameter of the He atom is approximately 0.10 nm

the radius of the He = \dfrac{0.10}{2} = 0.05 nm

Converting it into cm, we have:

0.05 nm \times  \dfrac{10^{-9} \  meters}{ 1  nm} \times \dfrac{ 1 cm }{10^{-2} \ meters}

=5 \times  10^{-9}  \ cm

Assuming that it is a sphere, the volume of a sphere is

= \dfrac{4}{3}\pi r^3

= \dfrac{4}{3}\pi  \times (5\times 10^{-9})^3

= 5.236 \times 10^{-25} \ cm^3

Finally, the density can be calcuated by using the formula :

Density = \dfrac{mass}{volume}

D =  \dfrac{6.647 \times  10^{-24} \  grams }{ 5.236 \times 10^{-25} \  cm^3}

D = 12.69 g/cm³

Density of the He atom = 12.69 g/cm³

4 0
2 years ago
Read 2 more answers
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