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myrzilka [38]
3 years ago
7

How many significant figures are there in the following numbers: 10.78, 6.78, 0.78? If these were pH values, to how many signifi

cant figures can you express the [H]? Explain any discrepancies between your answers to the two questions.
Chemistry
1 answer:
Ksivusya [100]3 years ago
3 0

Answer:

10.78 → 4 significant figures, pH = 10.78 → [H⁺] = 1.66ₓ10⁻¹¹ M

6.78 → 3 significant figures, pH = 6.78 → [H⁺] = 1.66ₓ10⁻⁷ M

0.78 → 2 significant figures, pH = 0.78 → [H⁺] = 0.166 M

pH always can be expressed by at least 4 significant figures. The [H⁺], can be expressed by, at least 3 significant figures

Explanation:

Significant figures are the numbers of a measurement that have certainty plus a doubtful number (it is associated with the uncertainty in the measurement). For example, if we measure a paper with a ruler and the ruler measures up to centimeters we can say that the paper is 7.5 cm long, with which we know that the paper is 7 cm + 0.5 cm which we associate with uncertainty. In this case we talk about two significant figures. If the sheet measured 7.57 cm we would already be talking about a more precise measurement and in this case with 3 significant figures.

10.78 → 4 significant figures

6.78 → 3 significant figures

0.78 → 2 significant figures

To determine [H⁺], we apply 10^-pH

10⁻¹⁰°⁷⁸ = 1.66ₓ10⁻¹¹ M

10⁻⁶°⁷⁸ = 1.66ₓ10⁻⁷ M

10⁻⁰°⁷⁸ = 0.166 M

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Lesechka [4]

Answer:

the final volume of the gas is V_2 = 1311.5 mL

Explanation:

Given that:

a sample gas has an initial volume of 61.5 mL

The workdone = 130.1 J

Pressure = 783 torr

The objective is to determine the final volume of the gas.

Since the process does 130.1 J of work on its surroundings at a constant pressure of 783 Torr. Then, the pressure is external.

Converting the external pressure to atm ; we have

External Pressure P_{ext}:

P_{ext} = 783 \ torr \times \dfrac{1 \ atm}{760 \ torr}

P_{ext} = 1.03 \ atm

The workdone W = P_{ext}V

The change in volume ΔV= \dfrac{W}{P_{ext}}

ΔV = \dfrac{130.1 \ J  \times \dfrac{1 \ L  \ atm}{ 101.325 \ J}  }{1.03 \ atm }

ΔV = \dfrac{1.28398717 }{1.03  }

ΔV = 1.25 L

ΔV = 1250 mL

Recall that the initial  volume = 61.5 mL

The change in volume V is \Delta V = V_2 -V_1

-  V_2= -  \Delta V  -V_1

multiply through by (-), we have:

V_2=   \Delta V+V_1

V_2 =  1250 mL + 61.5 mL

V_2 = 1311.5 mL

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Answer: The coefficient in front of AgCl when the equation is properly balanced is 2.

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

Decomposition is a type of chemical reaction in which one reactant gives two or more than two products.

Decomposition of silver chloride is represented as:

2AgCl(s)\rightarrow 2Ag(s)+Cl_2(g)

Thus the coefficient in front of AgCl when the equation is properly balanced is 2.

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Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produ
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<u>Answer:</u> The molecular formula for the given organic compound is C_{18}H_{20}O_2

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=39.61g

Mass of H_2O=9.01g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 39.61 g of carbon dioxide, \frac{12}{44}\times 39.61=10.80g of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18 g of water, 2 g of hydrogen is contained.

So, in 9.01 g of water, \frac{2}{18}\times 9.01=1.00g of hydrogen will be contained.

Mass of oxygen in the compound = (13.42) - (10.80 + 1.00) = 1.62 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = \frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{10.80g}{12g/mole}=0.9moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1g}{1g/mole}=1moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.62g}{16g/mole}=0.10moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.10 moles.

For Carbon = \frac{0.9}{0.10}=9

For Hydrogen = \frac{1}{0.10}=10

For Oxygen = \frac{0.10}{0.10}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 9 : 10 : 1

Hence, the empirical formula for the given compound is C_9H_{10}O

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

n=\frac{\text{Molecular mass}}{\text{Empirical mass}}

We are given:

Mass of molecular formula = 268.34 g/mol

Mass of empirical formula = 134 g/mol

Putting values in above equation, we get:

n=\frac{268.34g/mol}{134g/mol}=2

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(9\times 2)}H_{(10\times 2)}O_{(1\times 2)}=C_{18}H_{20}O_2

Thus, the molecular formula for the given organic compound is C_{18}H_{20}O_2.

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