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KonstantinChe [14]
3 years ago
12

What is the difference between double salt and complex salt​

Chemistry
1 answer:
Ksivusya [100]3 years ago
5 0

Answer:

The main different of double salt and complex salt is that a double salt is a combination of two salt compounds whereas a complex salt is a molecular structure that is composed of one or more complex ions.

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What is the ratio of 32oz of fuel to oil for 25 to 1.
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Answer:

<u>25 to 1 fuel mix 1 gallon</u>

Explanation:

Add two of the 2.5 ounce containers to 1 gallon of gas. For a 50:1 ratio add 1 container 2.5 ounces to 1 gallon of gas. 50=parts of fuel 1=parts of oil . For 25:1 and 2 of the 2.5 oz containers to 1 gallon of gas.

5 0
2 years ago
Just a curious question to the Chemists,
Vikki [24]

Answer:

well the world would be shocked

Explanation:

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8 0
3 years ago
Which particle changes the mass of the isotope the most? which particle changes the mass of the isotope the most? electron captu
MAXImum [283]
This question is absurd
7 0
3 years ago
Read 2 more answers
Will a precipitate of magnesium fluoride form when 300. mL of 1.1 × 10 –3 M MgCl 2 are added to 500. mL of 1.2 × 10 –3 M NaF? [K
Tju [1.3M]

Answer:

No precipitate is formed.

Explanation:

Hello,

In this case, given the dissociation reaction of magnesium fluoride:

MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-

And the undergoing chemical reaction:

MgCl_2+2NaF\rightarrow MgF_2+2NaCl

We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:

n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2

Next, the moles of magnesium chloride consumed by the sodium fluoride:

n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2

Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:

n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2

Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:

[Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M

[F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M

Thereby, the reaction quotient is:

Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}

In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.

Regards.

6 0
3 years ago
A sample of a pure compound that weighs 59.8 g contains 27.6 g Sb (antimony) and 32.2 g F (fluorine). What is the percent compos
Rama09 [41]

Answer:

53.85%

Explanation:

Data obtained from the question include:

Mass of antimony (Sb) = 27.6g

Mass of Fluorine (F) = 32.2g

Mass of compound = 59.8g

Percentage composition of fluorine (F) =..?

The percentage composition of fluorine can be obtained as follow:

Percentage composition of fluorine = mass of fluorine/mass of compound x 100

Percentage composition of fluorine = 32.2/59.8 x 100

= 53.85%

Therefore, the percentage composition of fluorine in the compound is 53.85%

3 0
3 years ago
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