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3 years ago

Bi2(CO3)3 Name of this compound

Chemistry

1 answer:

Keith_Richards [23]3 years ago

5 0

Answer:

The name of this compound is :

Bi2(CO3)3 = Bismuth Carbonate

Explanation:

The name of the compound is derived from the name of the elements present in it.

The rule followed while naming the compound are:

1. The first element (always the cation) is named as such .

2. The second element (The anion) end with "-ate , -ide ," etc

3. NO prefix is added while naming the first element.

For example : Bi2 can't be named as Dibismuth

Na2 = Can't be named as disodium

Hence the compound :

Bi2(CO3)3 contain two element : Bi and CO3. Here , Bi = cation (named as such) and CO3 = anion (named according to rules)

Bi = Bismuth

CO3 = carbonate

Bi2(CO3)3 = Bismuth Carbonate

The molecular mass of this compound is :

Molecular mass = 2 (mass of Bi) + 3(mass of C) + 6(mass of O)

= 2 (208.98)+3(12.01)+6(15.99)

= 597.987 u

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3 years ago

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.

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3 years ago

Describe the number and type of each component of the molecule methane

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Answer:

Methane is a chemical compound with the molecular formula CH4. It is the main component in natural gas. Methane is considered the simplest of alkanes, compounds that consist only of hydrogen (H) and carbon (C) elements. Methane is an odorless, colorless, tasteless gas that is lighter than air.

Explanation:

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3 years ago

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denis-greek [22]

Answer:

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Explanation:

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3 years ago

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Romashka-Z-Leto [24]

Answer:

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Explanation:

The mole ratio of nitrogen to hydrogen is 1:3; while that one of hydrogen to the products (ammonia) is 3:2

Thus if 3 moles of hydrogen gas produce 2 moles of ammonia gas

7.7 moles of hydrogen will produce:

(7.7moles×2)/3

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since the mass of an atom of nitrogen is 14 while that of hydrogen atom is 1.

Therefore 77/15 moles will have a mass of

77/15 moles × 17=87.27 grams

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3 years ago