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solmaris [256]
3 years ago
13

The ability to do work or cause change is

Chemistry
1 answer:
Ilia_Sergeevich [38]3 years ago
4 0
The ability to do work or cause change is
Energy
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Yes, it would be considered as physical change.
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Which one IS an example of a homogenous mixture?
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Calculate the number of mg of Allura Red present in an 10 fluid ounce glass of the beverage you used in this lab. Assume that th
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Answer : The mass of Allura Red present in an 10 fluid ounce glass of the beverage is, 7.634 mg.

Explanation : Given,

Volume of solution = 10 fl oz = 295.735 mL

Molar mass of Allura Red = 496.42 g/mole

Concentration of Allura Red (Molarity) = 0.000052 M

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

\text{Molarity}=\frac{\text{Mass of Allura Red}\times 1000}{\text{Molar mass of Allura Red}\times \text{Volume of solution (in mL)}}

Now put all the given values in this formula, we get:

0.000052M=\frac{\text{Mass of Allura Red}\times 1000}{496.42g/mole\times 295.735mL}

\text{Mass of Allura Red}=0.007634g=7.634mg

Therefore, the mass of Allura Red present in an 10 fluid ounce glass of the beverage is, 7.634 mg.

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3 years ago
The name "penicillin" is used for several closely-related antibiotics. A 1.2177 g sample of one of these compounds is burned, pr
masya89 [10]

<u>Answer:</u> The empirical formula for the given organic compound is C_7H_{11}O_2SN

<u>Explanation:</u>

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

S_vN_wC_xH_yO_z+O_2\rightarrow CO_2+H_2O+SO_2+NO

where, 'v', 'w' 'x', 'y' and 'z' are the subscripts of sulfur, nitrogen, carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=0.2829g

Mass of H_2O=0.1159g

Mass of SO_2=0.4503g

Mass of NO = 0.2109 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

Molar mass of sulfur dioxide = 64 g/mol

Molar mass of nitrogen monoxide = 30 g/mol

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

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

So, in 2.1654 g of carbon dioxide, \frac{12}{44}\times 2.1654=0.590g 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 0.6965 g of water, \frac{2}{18}\times 0.6965=0.077g of hydrogen will be contained.

  • <u>For calculating the mass of sulfur:</u>

In 64 g of sulfur dioxide, 32 g of sulfur is contained.

So, in 0.4503 g of sulfur dioxide, \frac{32}{64}\times 0.4503=0.225g of sulfur will be contained.

  • <u>For calculating the mass of nitrogen:</u>

In 30 g of nitrogen monoxide, 14 g of nitrogen is contained.

So, in 0.2109 g of nitrogen monoxide, \frac{14}{30}\times 0.2109=0.098g of nitrogen will be contained.

  • Mass of oxygen in the compound = (1.2177) - (0.590 + 0.077 + 0.225 + 0.098) = 0.2277 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{0.590g}{12g/mole}=0.049moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.077g}{1g/mole}=0.077moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.2277g}{16g/mole}=0.0142moles

Moles of Sulfur = \frac{\text{Given mass of Sulfur}}{\text{Molar mass of sulfur}}=\frac{0.225g}{32g/mole}=0.007moles

Moles of Nitrogen = \frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{0.098g}{14g/mole}=0.007moles

  • <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 0.007 moles.

For Carbon = \frac{0.049}{0.007}=7

For Hydrogen  = \frac{0.077}{0.007}=11

For Oxygen  = \frac{0.0142}{0.007}=2.03\approx 2

For Sulfur  = \frac{0.007}{0.007}=1

For Nitrogen  = \frac{0.007}{0.007}=1

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

The ratio of C : H : O : S : N = 7 : 11 : 2 : 1 : 1

Hence, the empirical formula for the given compound is C_7H_{11}O_2S_1N_1=C_7H_{11}O_2SN

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