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3 years ago

Significant Figures: How many significant figures are in 865,010?

Chemistry

1 answer:

Harrizon [31]3 years ago

7 0

Answer: There are five significant figures in 865,010.

Explanation:

When a degree of accuracy is stated by each digit present in a mathematical figure then it is called a significant figure.

Rules for counting significant figures is as follows.

Any non-zero digits and zeros present between a non-zero figure are counted. For example, 3580009 has seven significant figures.

Trailing zeros are counted in a non-zero figure. For example, 0.00250 has three significant figures.

Leading zeros are not counted. For example, 0.0025 has two significant figures.

So, in the given figure 865010 has five significant figures and the trailing zero will not be counted.

Thus, we can conclude that there are five significant figures in 865,010.

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Una muestra de 2,5000 (g) de piedra caliza, se disuelve y se precipita el Calcio como oxalato de Calcio (CaC2O4), este precipita

sergij07 [2.7K]

Answer:

$\%Ca=19.378\%$

Explanation:

¡Hola!

En este caso, basado en la información dada, es posible conocer la masa de calcio en carbonato de calcio, que de hecho es la misma en la muestra, al utilizar el siguiente esquema de cálculo:

$m_{Ca}=1,2093gCaCO_3*\frac{1molCaCO_3}{100.1gCaCO_3}*\frac{1molCa}{1molCaCO_3}*\frac{40.1gCa}{1molCa}\\\\m_{Ca}=0.48444gCa$

Finalmente, calculamos el porcentaje requerido como sigue:

$\% Ca=\frac{0.48444g}{2.5000g} *100\%\\\\\%Ca=19.378\%$

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3 years ago

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The number of protons in an atom’s nucleus is equal to its atomic number?

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3 years ago

[TG.05]Which of these statements best explains why P waves arrive earlier than S waves at the seismic station?

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The sugar deoxyribose is an important component of DNA. Deoxyribose is 44.8% 7.5% H, and 47.7% O by mass. What is the empirical

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Answer:

C₅H₁₀O₄

Explanation:

The empirical formula is the simplest whole-number ratio of atoms in a compound.

The ratio of atoms is the same as the ratio of moles.

So, our job is to calculate the molar ratio of C:H:O.

Assume 100 g of deoxyribose.

1. Calculate the mass of each element.

Then we have 44.8 g C, 7.5 g H, and 47.7 g O.

2. Calculate the moles of each element

$\text{Moles of C} = \text{44.8 g C} \times \dfrac{\text{1 mol C}}{\text{12.01 g C}} = \text{3.730 mol C}\\\\\text{Moles of H} = \text{7.5 g H} \times \dfrac{\text{1 mol H}}{\text{1.008 g H }} = \text{7.44 mol H}\\\\\text{Moles of O} = \text{47.7 g O} \times \dfrac{\text{1 mol O}}{\text{16.00 g O }} = \text{2.981 mol O}$

3. Calculate the molar ratio of the elements

Divide each number by the smallest number of moles

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4. Write the empirical formula

EF = C₅H₁₀O₄

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