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Which method can be used to find the range

lidiya [134]

Answer:

To find the range, first order the data from least to greatest. Then subtract the smallest value from the largest

3 0

3 years ago

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What process occurs as pieces of dry ice get smaller

fenix001 [56]

Sublimation. It's basically, in simple terms, when a solid changes to a gas without going into liquid form.

8 0

3 years ago

A 17.11 gram sample of an organic compound containing only C, H, and O is analyzed by combustion analysis and 21.71 g CO2 and 5.

Andru [333]

Answer: The empirical formula and the molecular formula of the organic compound is $CHO$ and $C_4H_4O_4$ respectively.

Explanation:

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 $CO_2$ = 21.71 g

Mass of $H_2O$ = 5.926 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

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

So, in 21.71 g of carbon dioxide, = $\frac{12}{44}\times 21.71=5.921g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 5.926 g of water, = $\frac{2}{18}\times 5.926=0.658g$ of hydrogen will be contained.

Mass of oxygen in the compound = (17.11) - (5.921+0.658) = 10.53 g

Mass of C = 5.921 g

Mass of H = 0.658 g

Mass of O = 10.53 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{5.921g}{12g/mole}=0.493moles$

Moles of H= $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.658g}{1g/mole}=0.658moles$

Mass of O= $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{10.53g}{16g/mole}=0.658moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{0.493}{0.493}=1$

For H = $\frac{0.658}{0.493}=1$

For O= $\frac{0.658}{0.493}=1$

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

Hence the empirical formula is $CHO$ .

empirical mass of CHO = 12(1) + 1(1) + 1 (16) = 29

Molecular mass = 104.1 g/mol

$n=\frac{\text {Molecular mass}}{\text {Equivalent mass}}=\frac{104.1}{29}=4$

Thus molecular formula = $n\times {\text {Empirical formula}}=4\times CHO=C_4H_4O_4$

6 0

3 years ago

On a hot summer day you and some friends decide you want to cool down your pool. Determine the mass of ice you would need to add

sergeinik [125]

Answer: The mass of ice you would need to add to bring the equilibrium temperature of the system to 300 K is $16.14 \times 10^{4}$ kg.

Explanation:

We know that relation between heat energy and specific heat is as follows.

q = $m \times S \times \Delta T$

As density of water is 1 kg/L and volume is given as 400,000 L. Therefore, mass of water is as follows.

Mass of water = Volume × Density

= $400,000 L \times 1 kg/L$

= 400,000 kg

or, = $400,000 \times 10^{3}$ g (as 1 kg = 1000 g)

Specific heat of water is 4.2 J/gm K. Therefore, change in temperature is as follows.

$\Delta T$ = 305 K - 273 K

= 32 K

Now, putting the given values into the above formula and calculate the heat energy as follows.

q = $m \times S \times \Delta T$

= $400,000 \times 10^{3} \times 4.2 \times 32 K$

= $5376 \times 10^{7}$ J

or, = $5376 \times 10^{4}$ kJ

According to the enthalpy of melting of ice 333 kJ/Kg of energy absorbed by by 1 kg of ice. Hence, mass required to absorb energy of $5376 \times 10^{4}$ kJ is calculated as follows.

Mass = $\frac{5376 \times 10^{4} kJ}{333 kJ/Kg} \times 1 kg$

= $16.14 \times 10^{4}$ kg

Thus, we can conclude that the mass of ice you would need to add to bring the equilibrium temperature of the system to 300 K is $16.14 \times 10^{4}$ kg.

3 0

3 years ago

The energy required to remove the third electron from a dipositive ion is

jolli1 [7]

The energy required is Ionization energy

5 0

2 years ago

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