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LUCKY_DIMON [66]
3 years ago
7

How many anions are in 0.500 g of MgBr2

Chemistry
1 answer:
Oksana_A [137]3 years ago
4 0

<u>Given:</u>

Mass of MgBr2 = 0.500 g

<u>To determine:</u>

Number of anions in 0.500 g MgBr2

<u>Explanation:</u>

Molar mass of MgBr2 = 24 + 2 (80) = 184 g/mol

Moles of MgBr2 = 0.500 g/184 g.mol-1 = 0.00271 moles

Based on stoichiometry-

1 mole of MgBr2 has 1 mole of Mg2+ cations and 2 moles of Br- anions

Therefore, 0.00271 moles of MgBr2 will have: 2 * 0.00271 = 0.00542 moles of Br-

Now,

1 mole of Br- contains 6.023 * 10²³ anions

0.00542 moles of Br- contain: 0.00542 * 6.023*10²³ = 3.264*10²¹ anions

Ans: There are 3.264*10²¹ anions in 0.5 g of MgBr2


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1kg of water has greater internal energy compared to 1g of water because 1kg of water has more mass.

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A sample of gold has a mass of 100 grams and a volume of 5cm^3, calculate the density by dividing the mass by volume
Dafna11 [192]

Answer:

20cm^2

Explanation:

Here, Density= Mass/ Volume

=100/5

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7 0
3 years ago
If you have access to stock solutions of 1.00 M H3PO4, 1.00 M of HCl, and 1.00 M NaOH solution, (and distilled water of course),
garri49 [273]

Answer:

0.10L of 1.00M of H₃PO₄ and 0.1613L of 1.00M NaOH

Explanation:

The pKa's of phosphoric acid are:

H₃PO₄/H₂PO₄⁻ = 2.1

H₂PO₄⁻/HPO₄²⁻ = 7.2

HPO₄²⁻/PO₄³⁻ = 12.0

To make a buffer with pH 9.40 we need to convert all H₃PO₄ to H₂PO₄⁻ and an amount of H₂PO₄⁻ to HPO₄²⁻

To have a 50mM solution of phosphoures we need:

2L * (0.050mol / L) = 0.10 moles of H₃PO₄

0.10 mol * (1L / mol) = 0.10L of 1.00M of H3PO4

To convert the H₃PO₄ to H₂PO₄⁻ and to HPO₄²⁻ must be added NaOH, thus:

H₃PO₄ + NaOH → H₂PO₄⁻ + H₂O + Na⁺

H₂PO₄⁻ + NaOH → HPO₄²⁻ + H₂O + Na⁺

Using H-H equation we can find the amount of NaOH added:

pH = pKa + log [A⁻] / [HA] <em>(1)</em>

<em>Where [A-] is conjugate base, HPO₄²⁻ and [HA] is weak acid, H₂PO₄⁻</em>

<em>pH = 7.40</em>

<em>pKa = 7.20</em>

[A-] + [HA] = 0.10moles <em>(2)</em>

Replacing (2) in (1):

7.40 = 7.20 + log 0.10mol - [HA] / [HA]

0.2 = log 0.10mol - [HA] / [HA]

1.5849 = 0.10mol - [HA] / [HA]

1.5849 [HA] = 0.10mol - [HA]

2.5849[HA] = 0.10mol

[HA] = 0.0387 moles = H₂PO₄⁻ moles

That means moles of HPO₄²⁻ are 0.10mol - 0.0387moles = 0.0613 moles

The moles of NaOH needed to convert all H₃PO₄ in H₂PO₄⁻ are 0.10 moles

And moles needed to obtain 0.0613 moles of HPO₄²⁻ are 0.0613 moles

Total moles of NaOH are 0.1613moles * (1L / 1mol) = 0.1613L of 1.00M NaOH

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3 years ago
A solution with pH of 9 has [OH-] concentration of
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PH = -log([H+])
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Answer:

They form a covalent bond

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