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Ira Lisetskai [31]
4 years ago
6

he balanced net ionic equation for precipitation of CaCO 3 when aqueous solutions of Na 2CO 3 and CaCl 2 are mixed is ________.

2Na (aq) CO32- (aq) Na2CO3 (aq) 2Na (aq) 2Cl- (aq) 2NaCl (aq) Na (aq) Cl- (aq) NaCl (aq) Ca (aq) CO32- (aq) CaCO3 (s) Na2CO3 (aq) CaCl2 (aq) 2NaCl (aq) CaCO3 (s)
Chemistry
1 answer:
ale4655 [162]4 years ago
6 0

Answer:

Net ionic equation: Ca^{2+}(aq.)+CO_{3}^{2-}(aq.)\rightarrow CaCO_{3}(s)

Explanation:

Reaction between Na_{2}CO_{3} and CaCl_{2} is an example of double decomposition reaction where each cations and anions of each salt exchange their partners during reaction to form CaCO_{3} and NaCl

CaCO_{3} is an insoluble salt. Hence precipitation of CaCO_{3} has been observed.

Molecular equation: Na_{2}CO_{3}(aq.)+CaCl_{2}(aq.)\rightarrow 2NaCl(aq.)+CaCO_{3}(s)

Total ionic equation:

2Na^{+}(aq.)+CO_{3}^{2-}(aq.)+Ca^{2+}(aq.)+2Cl^{-}(aq.)\rightarrow CaCO_{3}(s)+2Na^{+}(aq.)+2Cl^{-}(aq.)

Here Na^{+} and Na^{+} are spectator ions. Therefore they are eliminated from both side of total ionic equation to get net ionic equation.

Net ionic equation: Ca^{2+}(aq.)+CO_{3}^{2-}(aq.)\rightarrow CaCO_{3}(s)

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What is the amount, in moles, of each elemental sample? a. 11.8 g Ar b. 3.55 g Zn c. 26.1 g Ta d. 0.211 g Li
spayn [35]

Answer:

A. 0.295 mole

B. 0.055 mole

C.0.144 mole

D. 0.03 mole

Explanation:

To find the amount in moles, we simply use a mathematical relation that connects mass, atomic mass and number of moles.

Number of moles = mass/atomic mass

A. Atomic mass of Argon is 40

n = 11.8/40 = 0.295 mole

B. Atomic mass of zinc is 65

n = 3.55/65 = 0.055 mole

C. Atomic mass of Tantalum is 181

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8 0
3 years ago
10) How many grams are there in 1.00 x 10^24 molecules of BC13?
lana66690 [7]

194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.

Explanation:

In order to convert the given number of molecules of BCl₃ to grams, first we have to convert the molecules to moles.

It is known that 1 moles of any element has 6.022×10²³ molecules.

Then 1 molecule will have \frac{1}{6.022*10^{23} } moles.

So 1*10^{24} molecules = \frac{1*10^{24} }{6.022*10^{23} } =1.66 moles

Thus, 1.66 moles are included in BCl₃.

Then in order to convert it from moles to grams, we have to multiply it with the molecular mass of the compound.

As it is known as 1 mole contains molecular mass of the compound.

As the molecular mass of BCl₃ will be

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Mass of boron is 10.811 g and the mass of chlorine is 35.453 g.

Molar mass of BCl₃ = 10.811+(3×35.453)=117.17 g.

grams of BCl_{3}=Molar mass of BCl_{3}*Number of moles

grams of BCl_{3}=117.17*1.66=194.5 g

So, 194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.

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