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Jet001 [13]
2 years ago
4

An analytical chemist has determined by measurements that there are moles of magnesium in a sample of talc. how many moles of si

licon are in the sample
Chemistry
1 answer:
ladessa [460]2 years ago
4 0
The chemical formula for talc is:
Mg₃(SiO₂)₅(OH)₂

The formula shows us that there are:
3 moles of magnesium
5 moles of silicon
12 moles of oxygen
2 moles of hydrogen
For every mole of talc

Using this, we can determine the ratio between magnesium and silicon. This is:
3 moles Mg : 5 moles Si

Given a number of moles of magnesium, we may simply multiply by 5/3 to determine the number of moles of silicon present.
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If a certain gas occupies a volume of 13 L when the applied pressure is 6.5 atm , find the pressure when the gas occupies a volu
Sav [38]
We should apply Boyle's Law here given initial pressure, initial volume and final volume.

P1V1= P2V2
(6.5 atm) (13 L) = P2 (3.3 L)

Solve for P2 on your calculator and that should get you to the answer.


5 0
2 years ago
Read 2 more answers
Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown c
Alexus [3.1K]

Answer:

0.3229 M HBr(aq)

0.08436M H₂SO₄(aq)

Explanation:

<em>Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown concentration HBr(aq) required 18.45 mL of 0.3500 M NaOH(aq) to neutralize it, to the point where thymol blue indicator changed from pale yellow to very pale blue. Calculate the concentration (molarity) of Stu's HBr(aq) sample.</em>

<em />

Let's consider the balanced equation for the reaction between HBr(aq) and NaOH(aq).

NaOH(aq) + HBr(aq) ⇄ NaBr(aq) + H₂O(l)

When the neutralization is complete, all the HBr present reacts with NaOH in a 1:1 molar ratio.

18.45 \times 10^{-3} L NaOH.\frac{0.3500molNaOH}{1LNaOH} .\frac{1molHBr}{1molNaOH} .\frac{1}{20.00 \times 10^{-3} LHBr} =\frac{0.3229molHBr}{1LHBr} =0.3229M

<em>Kemmi Major also does a titration. She measures 25.00 mL of unknown concentration H₂SO₄(aq) and titrates it with 0.1000 M NaOH(aq). When she has added 42.18 mL of the base, her phenolphthalein indicator turns light pink. What is the concentration (molarity) of Kemmi's H₂SO₄(aq) sample?</em>

<em />

Let's consider the balanced equation for the reaction between H₂SO₄(aq) and NaOH(aq).

2 NaOH(aq) + H₂SO₄(aq) ⇄ Na₂SO₄(aq) + 2 H₂O(l)

When the neutralization is complete, all the H₂SO₄ present reacts with NaOH in a 1:2 molar ratio.

42.18 \times 10^{-3} LNaOH.\frac{0.1000molNaOH}{1LNaOH} .\frac{1molH_{2}SO_{4}}{2molNaOH} .\frac{1}{25.00\times 10^{-3}LH_{2}SO_{4}} =\frac{0.08436molH_{2}SO_{4}}{1LH_{2}SO_{4}} =0.08436M

6 0
2 years ago
How many hydrogen bonds can CH2O make to water
VladimirAG [237]
Hydrogen bonds are not like covalent bonds. They are nowhere near as strong and you can't think of them in terms of a definite number like a valence. Polar molecules interact with each other and hydrogen bonds are an example of this where the interaction is especially strong. In your example you could represent it like this: 

<span>H2C=O---------H-OH </span>

<span>But you should remember that the H2O molecule will be exchanging constantly with others in the solvation shell of the formaldehyde molecule and these in turn will be exchanging with other H2O molecules in the bulk solution. </span>

<span>Formaldehyde in aqueous solution is in equilibrium with its hydrate. </span>

<span>H2C=O + H2O <-----------------> H2C(OH)2</span>
5 0
2 years ago
The intrinsic solubility of sulfathiazole, an old antimicrobial, is 0.002 mol/l. it is a weak acid and its pka is 7.12. what is
lys-0071 [83]

The formula we will use is:

pH = 0.5 (pKa – log C)

where C is the concentration of sulfathiazole in molarity or mol / L

Since all values are given, we can compute directly for pH:

 

pH = 0.5 (7.12 – log 0.002)

<span>pH = 4.91</span>

4 0
2 years ago
Calculate the pka of hypochlorous acid. The ph of a 0.015 m solution of hypochlorous acid has a ph of 4.64.
o-na [289]

Answer:

  • pKa = 7.46

Explanation:

<u>1) Data:</u>

a) Hypochlorous acid = HClO

b) [HClO} = 0.015

c) pH = 4.64

d) pKa = ?

<u>2) Strategy:</u>

With the pH calculate [H₃O⁺], then use the equilibrium equation to calculate the equilibrium constant, Ka, and finally calculate pKa from the definition.

<u>3) Solution:</u>

a) pH

  • pH = - log [H₃O⁺]

  • 4.64 = - log [H₃O⁺]

  • [H_3O^+]= 10^{-4.64} = 2.29.10^{-5}

b) Equilibrium equation: HClO (aq) ⇄ ClO⁻ (aq) + H₃O⁺ (aq)

c) Equilibrium constant: Ka =  [ClO⁻] [H₃O⁺] / [HClO]

d) From the stoichiometry: [CLO⁻] = [H₃O⁺] = 2.29 × 10 ⁻⁵ M

e) By substitution: Ka = (2.29 × 10 ⁻⁵ M)² / 0.015M = 3.50 × 10⁻⁸ M

f) By definition: pKa = - log Ka = - log (3.50 × 10 ⁻⁸) = 7.46

5 0
2 years ago
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