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2 years ago

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Which of theCardiorespiratory fitness is most likely to improve due to which of the following types of exercise? A. bicycling B.

weight lifting C. push-ups D. sit-ups Please select the best answer from the choices provided. A B C D

1 answer:

gizmo_the_mogwai [7]2 years ago

7 0

Answer: A (Bicycling)

Explanation:

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The characteristic odor of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen. Combustion of

gavmur [86]

<u>Answer:</u> The empirical formula for the given compound is $C_3H_6O$

<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:

Conversion factor: 1 g = 1000 mg

Mass of $CO_2=6.32mg=0.00632g$

Mass of $H_2O=2.58g=0.00258g$

Mass of compound = 2.78 mg = 0.00278 g

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

So, in 0.00632 g of carbon dioxide, $\frac{12}{44}\times 0.00632=0.00172g$ of carbon will be contained.

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

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

So, in 0.00258 g of water, $\frac{2}{18}\times 0.00258=0.000286g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.00278) - (0.00172 + 0.000286) = 0.000774 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 = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.00172g}{12g/mole}=1.43\times 10^{-4}moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.000286g}{1g/mole}=2.86\times 10^{-4}moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.000774g}{16g/mole}=4.83\times 10^{-5}moles$

<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 $4.83\times 10^{-5}mol$

For Carbon = $\frac{1.43\times 10^{-4}}{4.83\times 10^{-5}}=2.96\approx 3$

For Hydrogen = $\frac{2.86\times 10^{-4}}{4.83\times 10^{-5}}=5.92\approx 6$

For Oxygen = $\frac{4.83\times 10^{-5}}{4.83\times 10^{-5}}=1$

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

The ratio of C : H : O = 3 : 6 : 1

Hence, the empirical formula for the given compound is $C_3H_{6}O_1=C_3H_6O$

3 0

3 years ago

Aqueous sodium hydroxide forms a light blue precipitate. What is the formula of the blue precipitate?

lisabon 2012 [21]

Answer:

Cu2+(aq) + 2OH-(aq) => Cu(OH)2(s)

Explanation:

Use of aqueous sodium hydroxide is a precipitation reaction to test for anions or cations. Aqueous sodium hydroxide in a precipitate test forms a insoluble precipitates along with some colors characteristics.

Aqueous sodium hydroxide (NaOH) when mixed with copper(II) (Cu2+) forms a blue precipitate. The formula is as follows:

Cu2+(aq) + 2OH-(aq) => Cu(OH)2(s)

8 0

3 years ago

Approximately how many particles are in 2 moles?

schepotkina [342]

Answer:

$1.2\times10^2^4$ particles in 2 moles.

Explanation:

The number of particles that are contained in one mole, the international unit of amount of substance: by definition, exactly 6.022×10²³, and it is dimensionless. It is named after the scientist Amedeo Avogadro.

It is also known as Avogadro's constant.

∴ Number of particles in one mole = $6.022\times10^2^3$

∴ Number of particles in 2 mole = 2 times Number of particles in one mole

∴ Number of particles in 2 mole= $2\times6.022\times10^2^3=12.044\times10^2^3\approx1.2\times10^2^4$

Hence there are $1.2\times10^2^4$ particles in 2 moles.

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3 years ago

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How many significant figures are in the mass 3.0047 g?

Rudik [331]

Answer:

5 significant figures

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3 years ago

Which reaction leads to the formation of a precipitate?

Brrunno [24]

2Ca(OH)2(aq) + 2FeCl3(aq) on the dead locs

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3 years ago

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