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3 years ago

10. Which statement best describes evidence that a chemical reaction occurs as a cake bakes?

Chemistry

2 answers:

BigorU [14]3 years ago

7 0

B the cake rises as gas bubbles from in the baking cake

nydimaria [60]3 years ago

5 0

Answer:

B.The cake rises as gas bubbles form in the baking cake preferably carbon dioxide

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A 0.08024 M solution of NaOH was used to titrate a solution of unknown concentration of HCl. A 32.08 mL sample of the HCl soluti

iris [78.8K]

Answer:

0.08024

Explanation:

4 0

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:

<u>Part A: nicotine </u>

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 $C_5H_7N_1=C_5H_7N$

<u>Part B: caffeine</u>

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

A 2.3 kg object has 15 ) of kinetic energy. Calculate its speed.

chubhunter [2.5K]

Answer:

<h3>The answer is option D</h3>

Explanation:

To find the speed given the kinetic energy and mass we use the formula

$v = \sqrt{ \frac{2KE}{m} } \\$

where

m is the mass

v is the speed

From the question

KE = 15 J

m = 2.3 kg

We have

$v = \sqrt{ \frac{2 \times 15}{2.3} } = \sqrt{ \frac{30}{2.3} } \\ = 3.61157559...$

We have the final answer as

Hope this helps you

7 0

3 years ago

Choose the correct total number of electron domains (bonding and nonbonding) about a central atom if the angle(s) between the el

slamgirl [31]

Answer:

The central atom has 3 electron domains.

Explanation:

According to the Valence Shell electron pair repulsion theory (VSEPR) put forward by Gillespie and Nyholm in 1957, the shape of a molecule is determined by repulsion between all the electron pairs (electron domains) present in the valence shell.

The electron pairs or electron domains are known to position themselves as far apart in space as possible in order to minimize repulsions.

Hence, when the central atom of a molecule contains three electron domains, they are positioned at an angle of 120° from each other to minimize repulsions. Hence the answer.

5 0

3 years ago

Which of the following elements is not diatomic?

KATRIN_1 [288]

Answer:

Neon

Explanation:

Elements from group 8A stay alone

8 0

3 years ago