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3 years ago

Dry ice goes directly from the solid phase to the gas phase. What phase

Chemistry

2 answers:

Marianna [84]3 years ago

8 0

Answer:

Sublimation is being represented.

Sublimation is the transition from solid to gas state

Hope this helps! Have a good day!

GalinKa [24]3 years ago

7 0

Answer:

Sublimation

Explanation:

Sublimation is when a solid turns into a gas

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State the functional groups that undergo cracking and substitution reaction.

omeli [17]

Unsaturated organic compounds with a carbon-to-carbon double bond and alkynes with a carbon-to-carbon triple bond, as well as aldehydes and ketones with a carbon-to-oxygen double bond, undergo addition reactions.

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2 years ago

Which equation is used to help form the combined gas law? eP, V, P, V, т.

alexdok [17]

Answer:

The combined gas law is formulated from PV/T =K.

Explanation:

The combined gas law comprises of Boyle's law, Charles's law and Gay lusaac's law. This laws were not discovered but simply put together considering other cases of ideal gas law. It states that if the amount of gas is left unchanged, the ratio between the pressure, volume, and temperature is constant.

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3 years ago

Calculate the empirical formula for each stimulant based on its elemental mass percent composition. a. nicotine (found in tobacc

Ira Lisetskai [31]

This an incomplete question, here is a complete question.

Calculate the empirical formula for each stimulant based on its elemental mass percent composition.

a. nicotine (found in tobacco leaves): C 74.03%, H 8.70%, N 17.27%

b. caffeine (found in coffee beans): C 49.48%, H 5.19 %, N 28.85% and O 16.48%

Answer:

(a) The empirical formula for the given compound is $C_5H_7N$

(b) The empirical formula for the given compound is $C_4H_5N_2O$

Explanation:

Part A: nicotine

We are given:

Percentage of C = 74.03 %

Percentage of H = 8.70 %

Percentage of N = 17.27 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 74.03 g

Mass of H = 8.70 g

Mass of N = 17.27 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{74.03g}{12g/mole}=6.17moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{8.70g}{1g/mole}=8.70moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{17.27g}{14g/mole}=1.23moles$

Step 2: 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 1.23 moles.

For Carbon = $\frac{6.17}{1.23}=5.01\approx 5$

For Hydrogen = $\frac{8.70}{1.23}=7.07\approx 7$

For Nitrogen = $\frac{1.23}{1.23}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 5 : 7 : 1

The empirical formula for the given compound is $C_5H_7N_1=C_5H_7N$

Part B: caffeine

We are given:

Percentage of C = 49.48 %

Percentage of H = 5.19 %

Percentage of N = 28.85 %

Percentage of O = 16.48 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 49.48 g

Mass of H = 5.19 g

Mass of N = 28.85 g

Mass of O = 16.48 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{49.48g}{12g/mole}=4.12moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{5.19g}{1g/mole}=5.19moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{28.85g}{14g/mole}=2.06moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{16.48g}{16g/mole}=1.03moles$

Step 2: 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 1.03 moles.

For Carbon = $\frac{4.12}{1.03}=4$

For Hydrogen = $\frac{5.19}{1.03}=5.03\approx 5$

For Nitrogen = $\frac{2.06}{1.03}=2$

For Nitrogen = $\frac{1.03}{1.03}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N : O = 4 : 5 : 2 : 1

The empirical formula for the given compound is $C_4H_5N_2O_1=C_4H_5N_2O$

6 0

3 years ago

Why is water considered the greatest solvent on earth

kotykmax [81]

1) H2O is able to dissolve both polar molecules and non polar ones
2) due to its extreme polarity it can even dissolve some I onic compounds
3 the h2o molecule itself is small in size

6 0

3 years ago

The mineral enargite is 48.41% cu, 19.02% as, and 32.57% s by mass. what is the empirical formula of enargite?

LuckyWell [14K]

Empirical formula is the simplest ratio of whole numbers of components in a compound.
Assuming for 100 g of the compound
Cu As S
mass 48.41 g 19.02 g 32.57 g
number of moles 48.41 / 63.5 g/mol 19.02 / 75 g/mol 32.57 / 32 g/mol
= 0.762 mol = 0.2536 mol = 1.018 mol
divide by the least number of moles
0.762 / 0.2536 0.2536 / 0.2536 1.018 / 0.2536
= 3.00 = 1.00 = 4.01
once they are rounded off
Cu - 3
As - 1
S - 4
therefore empirical formula is Cu₃AsS₄

8 0

3 years ago

Read 2 more answers