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slega [8]
3 years ago
10

Which coefficients correctly balance the formula equation CaCo3(s) CaCo3+CO2(g)? a. 1, 1, 1 c. 2, 6, 3 b. 1, 3, 1 d. 3, 1, 2

Chemistry
1 answer:
qaws [65]3 years ago
8 0
Answer: option a: 1,1,1

Explanation:

1) The chemical equation given is wrong.  This is the right chemical equation

<span>CaCO3(s) → CaO(s) + CO2(g)?

2) To balance you start by counting the number of atoms of each element on both sides

element                              number of atoms                   conclusion
                                  reactant side         product side

Ca                                      1                        1                    balanced

C                                        1                        1                    balanced

O                                        3                       1+2 = 3           balanced

3) Therefore all the elements are balanced and the coefficients are 1, 1, and 1.
</span>
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Fill in the left side of this equilibrium constant equation for the reaction of benzoic acid (HC_6H_5CO_2) with water.
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Answer:

[C₆H₅COO⁻][H₃O⁺]/[C₆H₅COOH] = Ka

Explanation:

The reaction of dissociation of the benzoic acid in water is given by the following equation:

C₆H₅-COOH + H₂O  ⇄  C₆H₅-COO⁻  +  H₃O⁺    (1)

The dissociation constant of an acid is the measure of the strength of an acid:

HA ⇄ A⁻ + H⁺        (2)

K_{a} = \frac{[A^{-}][H^{+}]}{[HA]}      (3)

<em>Where the dissociation constant of the acid (Ka) is equal to the ratio of the concentration of the dissociated forms of the acid, [A⁻][H⁺], and the concentration of the acid, [HA].     </em>

So, starting from the equations (2) and (3), the constant equation for the dissociation reaction of benzoic acid in water, of the equation (1), is:

K_{a} = \frac{[C_{6}H_{5}COO^{-}][H_{3}O^{+}]}{[C_{6}H_{5}COOH]}

I hope it helps you!          

6 0
3 years ago
Which of the following is a weak base ?
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3 years ago
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A chemistry student weighs out of sulfurous acid , a diprotic acid, into a volumetric flask and dilutes to the mark with distill
nadezda [96]

The question is incomplete, here is the complete question:

A chemistry student weighs out 0.104 g of sulfurous acid, a diprotic acid, into a 250.0 mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.0700 M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the final equivalence point. Be sure your answer has the correct number of significant digits.

<u>Answer:</u> The volume of NaOH needed is 36.2 mL

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

Given mass of sulfurous acid = 0.104 g

Molar mass of sulfurous acid = 82 g/mol

Volume of solution = 250 mL

Putting values in above equation, we get:

\text{Molarity of sulfurous acid}=\frac{0.104\times 1000}{82\times 250}\\\\\text{Molarity of sulfurous acid}=5.07\times 10^{-3}M

To calculate the volume of base, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_3

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=5.07\times 10^{-3}M\\V_1=250mL\\n_2=1\\M_2=0.0700M\\V_2=?mL

Putting values in above equation, we get:

2\times 5.07\times 10^{-3}\times 250.0=1\times 0.0700\times V_2\\\\V_2=\frac{2\times 5.07\times 10^{-3}\times 250}{1\times 0.0700}=36.2mL

Hence, the volume of NaOH needed is 36.2 mL

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nexus9112 [7]

Answer:

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Explanation:

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