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Brut [27]
3 years ago
10

Typical "hard" water contains about 2.0 x 10–3 mol of Ca2+ per liter. Calculate the maximum concentration of fluoride ion that c

ould be present in hard water. Assume fluoride is the only anion present that will precipitate calcium ion.
Chemistry
1 answer:
malfutka [58]3 years ago
8 0

Answer:

[F^-]_{max}=4x10{-3}\frac{molF^-}{L}

Explanation:

Hello,

In this case, for the described situation, we infer that calcium reacts with fluoride ions to yield insoluble calcium fluoride as shown below:

Ca^{+2}(aq)+2F^-(aq)\rightleftharpoons CaF_2(s)

Which is typically an equilibrium reaction, since calcium fluoride is able to come back to the ions. In such a way, since the maximum amount is computed via stoichiometry, we can see a 1:2 mole ratio between the ions, therefore, the required maximum amount of fluoride ions in the "hard" water (assuming no other ions) turns out:

[F^-]_{max}=2.0x10^{-3}\frac{molCa^{2+}}{L}*\frac{2molF^-}{1molCa^{2+}}  \\

[F^-]_{max}=4x10{-3}\frac{molF^-}{L}

Best regards.

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Why homoannular diene in uv spectrometry have higher wavlenth than hetroannular diene?
Leno4ka [110]

Answer:

Homoanular dienes have a greater base value than heteroanular dienes

Explanation:

Woodward in 1945 gave a set of rules relating the wavelength of maximum absorption to the structure of a compound. These rules were modified by Fieser in 1959. These sets of rules describe the absorption of organic molecules in the UV region of the electromagnetic spectrum.

Each system of diene or triene has a given fixed value at which maximum absorption is expected to occur according to Woodward rules. This given fixed value is called the base or parent value. If the two double bonds are trans to each other, the diene is said to be transoid. If the two double bonds belong to different rings, the system is said to be heteroanular and the base value in each case is 215nm. If the double bonds are cis to each other (cisoid), or the two double bonds are in the same ring (homoanular), then the base value is 253nm.

Since λmax = base value + ∑ substituent contributions + ∑ other contributions, if the other contributions are not very significant, homoanular diene will have a greater λmax because of its larger base value compared to heteroanular diene. This correlates well with the fact that conjugated systems absorb at a longer wavelength.

4 0
3 years ago
Calculate the mass of vanadium(V) oxide (V2O5) that contains a million (1.0 *10^6) vanadium atoms. Be sure your answer has a uni
sweet [91]

Answer : The mass of vanadium(V) oxide will be 1.51\times 10^{-16}g

Explanation : Given,

Number of atoms of V_2O_5 = 1.0\times 10^{6}

Molar mass of V_2O_5 = 181.88 g/mole

In V_2O_5, there are 2 atoms of vanadium and 5 atoms of oxygen.

First we have to determine the moles of V_2O_5.

As, 2\times 6.022\times 10^{23} number of vanadium atom present in 1 moles of V_2O_5

So, 1.0\times 10^{6} number of vanadium atom present in \frac{1.0\times 10^{6}}{2\times 6.022\times 10^{23}}=8.3\times 10^{-19} moles of V_2O_5

Now we have to determine the mass of V_2O_5.

\text{Mass of }V_2O_5=\text{Moles of }V_2O_5\times \text{Molar mass of }V_2O_5

\text{Mass of }V_2O_5=(8.3\times 10^{-19}mole)\times (181.88g/mole)=1.51\times 10^{-16}g

Therefore, the mass of vanadium(V) oxide will be 1.51\times 10^{-16}g

4 0
4 years ago
Read 2 more answers
Josh heated a certain amount of blue copper sulfate crystals to get 2.1 g of white copper sulfate powder and 1.4 g of water. Wha
kolbaska11 [484]

Answer: The mass of blue copper sulfate is 3.5 g

Explanation:

Law of conservation of mass states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.

This also means that total mass on the reactant side must be equal to the total mass on the product side.

The chemical equation for the heating of copper sulfate crystals is:

Let the mass of blue copper sulfate be 'x' grams

We are given:

Mass of copper sulfate powder = 2.1 grams

Mass of water = 1.4 grams

Total mass on reactant side = x

Total mass on product side = (2.1 + 1.4) g

So, by applying law of conservation of mass, we get:

Hence, the mass of blue copper sulfate is 3.5 grams

7 0
3 years ago
Calculate the pH at the equivalence point for the titration of 0.110 M methylamine (CH3NH2) with 0.110 M HCl. The Kb of methylam
nikklg [1K]
The reaction between the reactants would be:

CH₃NH₂ + HCl ↔ CH₃NH₃⁺ + Cl⁻

Let the conjugate acid undergo hydrolysis. Then, apply the ICE approach.

             CH₃NH₃⁺ + H₂O → H₃O⁺ + CH₃NH₂
I                0.11                       0             0
C               -x                          +x           +x
E            0.11 - x                     x             x

Ka = [H₃O⁺][CH₃NH₂]/[CH₃NH₃⁺]

Since the given information is Kb, let's find Ka in terms of Kb.

Ka = Kw/Kb, where Kw = 10⁻¹⁴

So,
Ka = 10⁻¹⁴/5×10⁻⁴ = 2×10⁻¹¹ = [H₃O⁺][CH₃NH₂]/[CH₃NH₃⁺]
2×10⁻¹¹ = [x][x]/[0.11-x]
Solving for x,
x = 1.483×10⁻⁶ = [H₃O⁺]

Since pH = -log[H₃O⁺],
pH = -log(1.483×10⁻⁶)
<em>pH = 5.83</em>


4 0
3 years ago
Can someone help me with this question ^
lisov135 [29]

Answer:

There are 8 electrons in the last level. Because of the octet rule.

7 0
2 years ago
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