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2 years ago

What is the solution to the problem expressed to he correct number of significant figures? 12.0/7.11

Chemistry

1 answer:

Law Incorporation [45]2 years ago

6 0

Answer: The solution to the problem expressed to he correct number of significant figures is 1.69

Explanation:

Significant figures : The figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

The rule applied for multiplication and division is :

The least precise number determines the number of significant figures in the answer.

As 12.0 has 3 significant digits and 7.11 also has 3 significant digits, the answer would also contain 3 significant digits.

For $\frac{12.0}{7.11}=1.69$

Thus the solution to the problem expressed to he correct number of significant figures is 1.69

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A reaction was performed in which 3.6 g 3.6 g of benzoic acid was reacted with excess methanol to make 1.3 g 1.3 g of methyl ben

Anit [1.1K]

Answer: The percent yield of the reaction is 32.34 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For benzoic acid:

Given mass of benzoic acid = 3.6 g

Molar mass of benzoic acid = 122.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of benzoic acid}=\frac{3.6g}{122.12g/mol}=0.0295mol$

The chemical equation for the reaction of benzoic acid and methanol is:

$\text{Benzoic acid + methanol}\rightarrow \text{methyl benzoate}$

By Stoichiometry of the reaction

1 mole of benzoic acid produces 1 mole of methyl benzoate

So, 0.0295 moles of benzoic acid will produce = $\frac{1}{1}\times 0.0295=0.108$ moles of methyl benzoate

Now, calculating the mass of methyl benzoate from equation 1, we get:

Molar mass of methyl benzoate = 136.15 g/mol

Moles of methyl benzoate = 0.0295 moles

Putting values in equation 1, we get:

$0.0295mol=\frac{\text{Mass of methyl benzoate}}{136.15g/mol}\\\\\text{Mass of methyl benzoate}=(0.0295mol\times 136.15g/mol)=4.02g$

To calculate the percentage yield of methyl benzoate, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of methyl benzoate = 1.3 g

Theoretical yield of methyl benzoate = 4.02 g

Putting values in above equation, we get:

$\%\text{ yield of methyl benzoate}=\frac{1.3g}{4.02g}\times 100\\\\\% \text{yield of methyl benzoate}=32.34\%$

Hence, the percent yield of the reaction is 32.34 %

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Identifica el elemento según su configuración electrónica, 1s2 2s2 2p6 3s2 3p6 4s2 3d8 *

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Answer:

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2 years ago

which of the following pairs of elements could react to form an ionic compound? A. potassium and calcium B. sulfur and oxygen C.

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The correct option is FLUORINE AND COPPER.
An ionic compound is usually formed by the combination of a metal and a non metal, the metal usually act as an electron donor while the non metal act as an electron acceptor. Thus, in ionic compounds, there is total transfer of electrons from the metal to the non metal. In the question given here, copper is the metal while the fluorine is the non metal.

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Is metal rusting a chemical reaction? Yes or no

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Explanation:

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3 years ago

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Oxygen gas, generated by the reaction 2KClO3(s)---2KCl(s)+3O2(g), is collected over water at 27•C in 3.72L vassel at a total pre

Julli [10]

Answer:

moles = 0.093 moles

Explanation:

In this case, we know that this reaction is taking plave in a vessel that has a 730 torr of total pressure.

The total pressure is a value obtained by:

Pt = Pwater + PO2

We need to know the pressure of O2, because then, with stoichiometry, we can calculate the moles of KClO3

The pressure of oxygen is:

PO2 = 730 - 26 = 704 Torr

Now, this pressure is in Torr, and we need to convert it to Atm, so:

704 Torr / 760 Torr = 0.9263 atm

Now, let's use the ideal gas equation:

PV = nRT

With this expression, we will calculate the moles of O2, and then, the moles of KClO3:

n = PV/RT

R = 0.082 L atm /K mol

P = 0.9263 atm

V = 3.72 L

T = 27 + 273 = 300 K

Replacing the data:

n = 0.9263 * 3.72 / 300 * 0.082

n = 0.14 moles

Finally, by stoichiometry, we know that 2 moles of KClO3 produces 3 moles of O2, so:

moles of KClO3 = 0.14 * 2/3 = 0.093 moles of KClO3

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