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bazaltina [42]
3 years ago
7

What is a weak acid?

Chemistry
1 answer:
Helen [10]3 years ago
4 0

Answer:

B

Explanation:

strong acids ionize completely in water

You might be interested in
What is the mass in grams, if you have 8.42 x 10^18 atoms of Bromine
Dahasolnce [82]

Answer:

84.2 grams

Explanation:

you have to multiply 8.42x10^18 to get the answer above

8 0
3 years ago
In a particular experiment, 2.50-g samples of each reagent are reacted. The theoretical yield of lithium nitride is ________ g.
Neporo4naja [7]

Answer:

4.18 g

Explanation:

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Given: For Li

Given mass = 2.50 g

Molar mass of Li  = 6.94 g/mol

<u>Moles of Li  = 2.50 g / 6.94 g/mol = 0.3602 moles</u>

Given: For N_2

Given mass = 2.50 g

Molar mass of N_2 = 28.02 g/mol

<u>Moles of N_2 = 2.50 g / 28.02 g/mol = 0.08924 moles</u>

According to the given reaction:

6Li+N_2\rightarrow 2Li_3N

6 moles of Li react with 1 mole of N_2

1 mole of Li react with 1/6 mole of N_2

0.3602 mole of Li react with \frac {1}{6}\times 0.3602 mole of N_2

Moles of N_2 that will react = 0.06 moles

Available moles of N_2 = 0.08924 moles

N_2 is in large excess. (0.08924 > 0.06)

Limiting reagent is the one which is present in small amount. Thus,

Li is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

6 moles of Li gives 2 mole of Li_3N

1 mole of Li gives 2/6 mole of Li_3N

0.3602 mole of Li react with \frac {2}{6}\times 0.3602 mole of Li_3N

Moles of Li_3N = 0.12

Molar mass of Li_3N = 34.83 g/mol

Mass of Li_3N = Moles × Molar mass = 0.12 × 34.83 g = 4.18 g

<u>Theoretical yield = 4.18 g</u>

5 0
3 years ago
A chemist studies the reaction below. 2NO(g) + Cl2(g) 2NOCl(g) He performs three experiments using different concentrations and
Dmitry_Shevchenko [17]

Answer:

1. Rate =k [NO]^{2}[Cl_{2}]

2. k= 0.42 \frac{L^{2}}{mol^{2}*s}

Explanation:

Rate =k [NO]^{m}[Cl_{2}]^{n}

Rate1 = k[0.4]^{m}[0.3]^{n}=0.02\\Rate 2=k [0.8]^{m}[0.3}]^{n}=0.08\\\\\frac{Rate1}{Rate2}=\frac{0.02}{0.08} =\frac{k[0.4]^{m}[0.3]^{n}}{k[0.8]^{m}[0.3]^n}} \\\\\frac{1}{4} =(\frac{1}{2} )^{m},\\m=2

Rate3 =k [0.8]^{m}[0.6]^{n}=0.16\\Rate 2= k[0.8]^{m}[0.3}]^{n}=0.08\\\\\frac{Rate3}{Rate2}=\frac{0.16}{0.08} =\frac{k[0.8]^{m}[0.6]^{n}}{k[0.8]^{m}[0.3]^n}} \\\\\frac{2}{1} =(\frac{2}{1} )^{n},\\n=1

Rate =k [NO]^{2}[Cl_{2}]^{1}

Rate =k [NO]^{2}[Cl_{2}]^{1}\\Rate 1=k [0.4]^{2}[0.3]^{1} =0.02\\k*0.16*0.3=0.02\\k=\frac{0.02}{0.16*0.3}=\frac{1}{8*(\frac{3}{10} )}=\frac{5}{12}  = 0.42 \frac{L^{2}}{mol^{2}*s}

6 0
3 years ago
If you use about 4.5 x 1023 atoms of carbon in your pencil as you complete this
raketka [301]

Answer:

0.74mol

Explanation:

4.5×10^23/6.02×10^23=0.74mol

3 0
3 years ago
The standard free energy of formation, ΔG∘f, of a substance is the free energy change for the formation of one mole of the subst
OLEGan [10]

Answer:

B. 2 Na(s) + O₂(g) → Na₂O₂(s); ΔG∘f=−451.0 kJ/mol

D. 2 SO(g) + O₂(g) → 2 SO₂(g); ΔG°f=−600.4 kJ/mol

Explanation:

The spontaneity of a reaction  is given by the value of the standard Gibbs free energy of the reaction (ΔG°rxn). The more negative is the ΔG°rxn, the more spontaneous is a reaction.

The ΔG°rxn can be calculated using the following expression:

ΔG°rxn = ∑np × ΔG°f(products) − ∑nr × ΔG°f(reactants)

By definition, the standard Gibbs free energy of formation of simple substances in their most stable state is zero. That is why, in the reaction of formation of a compound ΔG°rxn = ΔG°f(product).

<em>Based on the standard free energies of formation, which of the following reactions represent a feasible way to synthesize the product? </em>

<em>     A. N₂(g) + H₂(g) → N₂H₄(g); ΔG°f=159.3 kJ/mol. </em>

<em>     </em>Not feasible. ΔG°rxn = ΔG°f(product) > 0.

    <em>B. 2 Na(s) + O₂(g) → Na₂O₂(s); ΔG°f=−451.0 kJ/mol</em>

    Feasible. ΔG°rxn = ΔG°f(product) < 0.

    <em>C. 2 C(s) + 2 H₂(g) → C₂H₄(g); ΔG°f=68.20 kJ/mol</em>

    Not feasible. ΔG°rxn = ΔG°f(product) > 0.

    <em>D. 2 SO(g) + O₂(g) → 2 SO₂(g); ΔG°f=−600.4 kJ/mol</em>

    Feasible. ΔG°rxn = ΔG°f(product) < 0.

3 0
4 years ago
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