2. When do chemical reactions happen?

Some athletes have used steroids, often with negative consequences to their health. Steroids belong to which class of biochemica

pochemuha

Steriods belong to the class of lipids which contains four carbon rings that are fused. Examples of these are estradiol and testosterone. They are important in cell membranes since they are responsible in altering membrane fluidity and also they are molecules used in signaling.

6 0

3 years ago

An ion is an atom with a net electrical charge due to ______.

Serga [27]

<span>the loss of one or more electrons & the addition of one of more electrons</span>

7 0

3 years ago

Read 2 more answers

A 1.50 g sample of solid NH₄NO₃ was added to 35.0 mL of water in a styrofoam cup (insulated from the environment) and stirred un

GaryK [48]

Answer : The heat of the reaction is, 1.27 kJ/mole

Explanation :

First we have to calculate the heat released.

Formula used :

$Q=m\times c\times \Delta T$

or,

$Q=m\times c\times (T_2-T_1)$

where,

Q = heat = ?

m = mass of sample = 1.50 g

c = specific heat of water = $4.81J/g^oC$

$T_1$ = initial temperature = $22.7^oC$

$T_2$ = final temperature = $19.4^oC$

Now put all the given value in the above formula, we get:

$Q=1.50g\times 4.81J/g^oC\times (19.4-22.7)^oC$

$Q=-23.8095J=-0.0238kJ$

Now we have to calculate the heat of the reaction in kJ/mol.

$\Delta H=\frac{Q}{n}$

where,

$\Delta H$ = enthalpy change = ?

Q = heat released = 0.0238 kJ

n = number of moles NH₄NO₃ = $\frac{\text{Mass of }NH_4NO_3}{\text{Molar mass of }NH_4NO_3}=\frac{1.50g}{80g/mol}=0.01875mole$

$\Delta H=\frac{0.0238kJ}{0.01875mole}=1.27kJ/mole$

Therefore, the heat of the reaction is, 1.27 kJ/mole

8 0

3 years ago

Enter your answer in the provided box. Find the pH of a buffer that consists of 0.34 M HBrO and 0.89 M KBrO (PK, of HBrO = 8.64)

11111nata11111 [884]

Answer : The pH of buffer is 9.06.

Explanation : Given,

$pK_a=8.64$

Concentration of HBrO = 0.34 M

Concentration of KBrO = 0.89 M

Now we have to calculate the pH of buffer.

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[KBrO]}{[HBrO]}$

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

$pH=8.64+\log (\frac{0.89}{0.34})$

$pH=9.06$

Therefore, the pH of buffer is 9.06.

6 0

4 years ago

A nitrogen oxide, containing 53.85% N, acts as a vasodilator, lowering blood pressure in the human body. What is its empirical f

jenyasd209 [6]

The empirical formula of the nitrogen oxide that acts as a vasodilator and lowers human blood pressure is $N_4O_3$ . Empirical formula is the smallest whole number ratios of elements in a compound.

FURTHER EXPLANATION

To get the empirical formula from mass percent data, the following steps are followed:

1. Get the mass percent data for all the elements that make up the compound.

2. Assume that there is 100 grams of sample compound. Determine the equivalent masses of each element using the mass percent given.

3. Convert the mass of each element to number of moles.

4. Divide each calculated mole by the smallest mole value.

5. The quotient will be the subscript of the element in the compound. If a decimal is obtained, round off the answers to the nearest whole number. Some exceptions to this are the following decimals: x.25, x.33, x.50, and x.75. When these decimals are obtained, all the subscripts are multiplied by a number that will result in a whole number.

Applying the steps to the problem,

STEP 1: Get the mass percent.

Mass % of N = 53.85

Mass % of O = (100 - 53.85) = 46.15

STEP 2: Assume that 100 g sample is used and convert the mass % to mass in grams.

mass of N = 53.85% x 100 g = 53.85 g

mass of O= 46.15% x 100 g = 46.15 g

STEP 3: Convert mass to moles by dividing the given mass by the formula mass.

$moles \ of \ N \ = 53.85 \ g (\frac{1 \ mol}{14 \ g}) = 3.85 \\\\moles \ of \ O \ = 46.26 \ g \ (\frac{1 \ mol}{16 \ g}) \ = 2.89$

The moles may be written as the temporary subscripts of the elements in the compound as follows:

$N_{3.85}O_{2.89}$

STEP 4: Divide the moles by the smallest mole value.

$N_{3.85}O_{2.89}\\N_{ \frac{3.85}{2.89}} \ O _{ \frac{2.89}{2.89}} \\ N_{1.33}O_{1}\\$

<u>STEP 5:</u> Multiply the subscripts by a number that would give the smallest whole number ratio.

Since the decimal is 1.33 it cannot be rounded off to 1. It should be multiplied by 3 to get the smallest whole number.

NOTE: All subscripts must be multiplied by the same factor, 3.

$N_{1.33}O_{1}\\3(N_{1.33}O_{1})\\\boxed {N_{4}O_{3}}$

To check if the empirical formula is correct, calculate the mass % of each element based on the formula and compare with the given in the problem.

$mass \ \%\ = (\frac{mass \ of \ element}{mass \ of \ compound}) x 100\\\\mass \ \% \ of \ N = \ \frac{(4)(14)}{(4)(14) \ + \ (3)(16)}$

$mass \ \% \ of \ N \ = 53.85%\\mass \ \% \ of \ O \ = (\frac{(3)(16)}{(4)(14)+(3)(16)}) \ 100\\mass \ \% \ of \ O \ = 46.15%$

Since the values are the same from the given, the answer is correct.

LEARN MORE

Writing Chemical Formula brainly.com/question/4697698
Naming Compounds brainly.com/question/8968140
Mole Conversion brainly.com/question/12979299

Keywords: empirical formula, compounds

6 0

3 years ago

Read 2 more answers