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rewona [7]
3 years ago
13

A student dissolved 5.00 g of Co(NO3)2 in enough water to make 100. mL of stock solution. He took 4.00 mL of the stock solution

and then diluted it with water to give 275. mL of a final solution. How many grams of NO3- ion are there in the final solution?
Chemistry
1 answer:
skad [1K]3 years ago
5 0

Answer:

0.136g

Explanation:

A student dissolved 5.00 g of Co(NO3)2 in enough water to make 100. mL of stock solution. He took 4.00 mL of the stock solution and then diluted it with water to give 275. mL of a final solution. How many grams of NO3- ion are there in the final solution?

Co(NO_3)_2(aq)\rightarrow Co^{2+}(aq)+2NO_3^{-}(aq)

Initial mole of Co(NO3)2  =\frac{mass}{molar mass}

=\frac{5.00}{182.94} \\\\=0.02733mol

Mole of Co(NO3)2 in final solution

=\frac{4.00}{100}\times 0.02733\\\\=0.04\times 0.02733\\\\= 0.001093mol

Mole of  NO3- in final solution = 2 x Mole of Co(NO3)2

=2\times 0.001093\\\\=0.002186mol

Mass of  NO3- in final solution is mole x Molar mass of NO3

=0.002186\times62.01\\\\=0.136g

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The only way to determine the products of a reaction is to carry out the reaction. All chemical reactions can be classified as o
myrzilka [38]

Answer:

All the statements are correct but "all chemical reactions can be classified as one of the five general types".

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So the wrong statement is:

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4 years ago
Menthol is a flavoring agent extracted from peppermint oil. It contains C, H, and O. In one combustion analysis, 10.00 mg of the
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Answer:

C₁₀H₂₀O

Explanation:

The molecular formula must be C_{x}H_yOz. The combustion reaction will occur between the fuel and oxygen gas:

C_{x}H_yOz + O₂ → CO₂ + H₂O

For Lavoisier law, the mass of the reactants must be equal to the mass of the products (mass conservation):

10.0 + mO₂ = 28.16 + 11.53

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Supposing that all the oxygen and the menthol were consumed, let's calculate the number of moles of the compounds, knowing, for the Periodic Table, that:

MC = 12 g/mol, MO = 16 g/mol, MH = 1 g/mol

MCO₂ = 12 + 2x16 = 44 g/mol

MH₂O = 2x1 + 16 = 18 g/mol

MO₂ = 2x16 = 32 g/mol

n = mass (g)/molar mass

nCO₂ = 0.02816/44 = 6.4x10⁻⁴ mol

nH₂O = 0.01153/18 = 6.4x10⁻⁴ mol

nO₂ = 0.02969/32 = 9.3x10⁻⁴ mol

The molar number is proportional in the molecule, so, in CO₂, the number of C is 6.4x10⁻⁴ mol, and of O is 1.28x10⁻³. All the carbon of the methol will be in CO₂, and all the H will be in H₂O. The number of moles of O will be the difference of moles in H₂O and the O₂, then:

nC = 6.4x10⁻⁴ mol

nH = 2x6.4x10⁻⁴ = 1.28x10⁻³ mol

nO = (2x6.4x10⁻⁴ + 6.4x10⁻⁴) - (2x9.3x10⁻⁴) = 6.0x10⁻⁵

The empirical formula is the molecule formula with the small subscripts numbers, which represent the number of moles of the atoms in the molecule. So, let's divide all the number of moles for the small on 6.0x10⁻⁵.

nC = (6.4x10⁻⁴)/(6.0x10⁻⁵) = 10

nH = (1.28x10⁻³)/(6.0x10⁻⁵) = 20

nO = (6.0x10⁻⁵)/(6.0x10⁻⁵) = 1

So, the empirical formula of methol is C₁₀H₂₀O.

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3 years ago
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The following reaction was performed in a sealed vessel at 760 ∘C : H2(g)+I2(g)⇌2HI(g) Initially, only H2 and I2 were present at
zubka84 [21]

Answer:

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Explanation:

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ICE table can be written as -

                             H₂(g)    +      I₂(g)    ⇄     2HI(g)

initial moles         3.85             2.35                -

at equilibrium      3.85 - x       2.35 - x           2x

From question , at equilibrium the concentration of I₂ = 0.0500 M

The concentration of I₂ ( ICE table ) = concentration of I₂ (given in question )

2.35 - x = 0.0500

x = 2.3

Putting the value of x in ICE table , to obtain the concentration  terms as-

[H₂] = 3.85 - x

[H₂] = 3.85 - 2.3

[H₂] = 1.55 M

[HI] = 2x

[HI] = 2* 2.3

[HI] = 4.6 M

[I₂] = 0.0500M       (Given)

Equilibrium Constant ( Kc )

The value of equilibrium constant is written as the concentration of the products each raised to the power of their respective stoichiometry by the concentration of reactants each raised to the power of their respective stoichiometry.

Kc = [HI]² / [H₂][I₂]

Kc = ( 4.6 )² / (1.55)*(0.0500)

Kc = 273.0322 .

6 0
4 years ago
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