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2 years ago

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Many diseases can be prevented with a proper diet, exercise, and abstinence from smoking, alcohol, drugs, and sexual activity. T

rue or false

1 answer:

tiny-mole [99]2 years ago

4 0

Answer:

True

Explanation:

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Why is there DNA in our food?

sukhopar [10]

Answer:

probably because most of our food comes from animals-

Explanation:

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2 years ago

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Mercury(II) oxide (HgO) decomposes to form mercury (Hg) and oxygen (O2). The balanced chemical equation is shown below.

zavuch27 [327]

15.63 mol. You need 15.63 mol HgO to produce 250.0 g O_2.

<em>Step 1</em>. Convert <em>grams of O_2 to moles of O_2</em>

Moles of O_2 = 250.0 g O_2 × (1 mol O_2/32.00 g O_2) = 7.8125 mol O_2

<em>Step 2</em>. Use the molar ratio of HgO:O_2 to convert <em>moles of O_2 to moles of HgO </em>

Moles of HgO = 0.8885 mol O_2 × (2 mol HgO/1 mol O_2) = <em>15.63 mol HgO</em>

7 0

3 years ago

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A tank of gas has partial pressures of nitrogen and oxygen equal to 1.61 Ã— 104 kPa and 4.34 Ã— 103 kPa , respectively.What is t

Alchen [17]

Answer:

$20.44\times 10^3 kPa$ is the total pressure of the tank.

Explanation:

Partial pressures of nitrogen = $p_{N_2}=1.61\times 10^4 kPa$

Partial pressure of oxygen = $p_{O_2}=4.34\times 10^3 kPa$

Total pressure of gases in the tank = P

Applying Dalton's law of partial pressures :

$P=p_{N_2}+p_{O_2}=1.61\times 10^4 kPa+4.34\times 10^3 kPa$

$P=20.44\times 10^3 kPa$

$20.44\times 10^3 kPa$ is the total pressure of the tank.

7 0

2 years ago

PLEASE ANSWER ASAP (WILL GIVE BRAINLIEST)

Flauer [41]

Answer:

I think it is either A. or B.

Explanation:

(I think!)

5 0

2 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