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zimovet [89]
2 years ago
14

An experiment requires 50.0 mL of 0.040 M NaOH for the titration of 1.00 mmol of acid. Mass analysis of the acid shows 2.24% hyd

rogen, 26.7% carbon, and 71.1% oxygen. Draw the Lewis structure of the acid.
Chemistry
1 answer:
tino4ka555 [31]2 years ago
8 0

The   molecular  formula    is  calculated  as  follows

that  is   for  carbon = 26.7/12=  2.23  moles

                  hydrogen=  2.24/1=2.24  moles

                oxygen=  71.1/16=4.44  moles

<h3> What is molecular  formula?</h3>

A chemical formula is a way to describe the chemical ratios of the atoms that make up a specific chemical compound or molecule in chemistry. Chemical element symbols, numbers, and occasionally other symbols, such as parentheses, dashes, brackets, commas, and plus (+) and minus () signs, are used to represent the chemical elements. These can only include one typographic line of symbols, which may also include subscripts and superscripts. A chemical formula has no words and is not the same as the chemical name. A chemical formula does not equal a complete chemical structural formula, despite the fact that it may suggest some basic chemical structures. Chemical formulae are often less powerful than chemical names and structural formulae, and they can only fully describe the structure of the simplest molecules and chemical compounds.

To learn more about molecular  formula from the given link:

brainly.com/question/14425592

#SPJ4

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Answer:

Moderate reaction means those reactions which proceed with a measurable rates at the normal room temperature.

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Compute the equilibrium constant at 25 ∘C for the reaction between Sn2+(aq) and Cd(s), which form Sn(s) and Cd2+(aq). Express yo
Radda [10]

Answer:

6.1×10^8

Explanation:

The reaction is;

Sn^2+(aq) + Cd(s) -----> Sn(s) + Cd^2+(aq)

E°cell = E°cathode - E°anode

E°cathode= -0.14 V

E°anode= -0.40 V

E°cell = -0.14-(-0.40)

E°cell= -0.14+0.40

E°cell= 0.26 V

But

E°cell= 0.0592/n log K

E°cell= 0.0592/2 log K

0.26= 0.0296log K

log K = 0.26/0.0296

log K= 8.7838

K= Antilog (8.7838)

K= 6.1×10^8

8 0
3 years ago
A gas at 750 mmhg and with a volume of 2. 00 l is allowed to change its volume at constant temperature until the pressure is 600
Gemiola [76]

Answer:

The new volume of a gas at 750 mmhg and with a volume of 2. 00 l when allowed to change its volume at constant temperature until the pressure is 600 mmhg is 2.5 Liters.

Explanation:

Boyle's law states that the pressure of a given amount of gas is inversely proportional to it's volume at constant temperature. It is written as;

P ∝ V

P V = K

P1 V1 = P2 V2

Parameters :

P1 = Initial pressure of the gas = 750 mmHg

V1 = Initial pressure of the gas = 2. 00 Liters

P2 = Final pressure of the gas = 600 mmHg

V2 = Fimal volume of the gas = ? Liters

Calculations :

V2 = P1 V1 ÷ P2

V2= 750 × 2. 00 ÷ 600

V2 = 1500 ÷ 600

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Read 2 more answers
Calculate the initial rate for the formation of C at 25 ∘C, if [A]=0.50M and [B]=0.075M.Express your answer to two significant f
N76 [4]

The question is incomplete, here is the complete question:

Calculate the initial rate for the formation of C at 25°C, if [A]=0.50 M and [B]=0.075 M. Express your answer to two significant figures and include the appropriate units.Consider the reaction

A + 2B ⇔ C

whose rate at 25°C was measured using three different sets of initial concentrations as listed in the following table:

The table is attached below as an image.

<u>Answer:</u> The initial rate for the formation of C at 25°C is 2.25\times 10^{-2}Ms^{-1}

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

A+2B\rightleftharpoons C

Rate law expression for the reaction:

\text{Rate}=k[A]^a[B]^b

where,

a = order with respect to A

b = order with respect to B

  • Expression for rate law for first trial:

5.4\times 10^{-3}=k(0.30)^a(0.050)^b ....(1)

  • Expression for rate law for second trial:

1.1\times 10^{-2}=k(0.30)^a(0.100)^b ....(2)

  • Expression for rate law for third trial:

2.2\times 10^{-2}=k(0.50)^a(0.050)^b ....(3)

Dividing 2 by 1, we get:

\frac{1.1\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.30)^a(1.00)^b}{(0.30)^a(0.050)^b}\\\\2=2^b\\b=1

Dividing 3 by 1, we get:

\frac{2.2\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.50)^a(0.050)^b}{(0.30)^a(0.050)^b}\\\\4.07=2^a\\a=2

Thus, the rate law becomes:

\text{Rate}=k[A]^2[B]^1       ......(4)

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

5.4\times 10^{-3}=k[0.30]^2[0.050]^1\\\\k=1.2M^{-2}s^{-1}

Calculating the initial rate of formation of C by using equation 4, we get:

k=1.2M^{-2}s^{-1}

[A] = 0.50 M

[B] = 0.075 M

Putting values in equation 4, we get:

\text{Rate}=1.2\times (0.50)^2\times (0.075)^1\\\\\text{Rate}=2.25\times 10^{-2}Ms^{-1}

Hence, the initial rate for the formation of C at 25°C is 2.25\times 10^{-2}Ms^{-1}

8 0
3 years ago
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