Write a balanced equation for the double-replacement precipitation reaction described, using the smallest possible integer coefficients. A precipitate forms when aqueous solutions of lead(II) nitrate and sodium hydroxide are combined.

Answer:

- carbon (C)

- hydrogen (H)

- oxygen (O)

- nitrogen (N)

- sulfur (S)

Explanation:

Amino acids are organic molecules which base chemical structure is composed by:

- a carbon atom (C)

- a carboxyl group (-COOH)

- an amino group (-NH₂)

- a side chain- called R group- which varies in different amino acids. It can be composed by N, H, O, C and sometimes S.

According to this, the five chemical elements that are present in amino acids are:

- carbon (C)

- hydrogen (H)

- oxygen (O)

- nitrogen (N)

- sulfur (S)

Answer:

92.75%

Explanation:

The overall chemical equation for the reaction in the preparation of alum from the aluminium can be expressed as:

[tex]\mathtt{2Al + 2KOH + 4H_2SO_4 +2H_2O \to 2KAl(SO_4)_2 2H_2O +3H_2}[/tex]

From above; we will see that 2 moles of Aluminium react with sulphuric acid and water to produce 2 moles o aluminium alum.

Therefore, the theoretical yield can be determined as:

[tex]=0.4395 \ g Al \times \dfrac{1 \ mol \ Al}{27 \ g Al }\times \dfrac{2 \ mol \ KAl(SO_4)_2}{2 \ mol \ Al}\times \dfrac{294.23 \ KAl(SO_4)_2}{1 \ mol \ KAl(SO_4)_2}[/tex]

= 4.789g of [tex]KAl(SO_4)_2[/tex]

To find the percent yield, we need to divide the actual yield by the theoretical yield and then multiply it with 100.

∴

percent yield = ( mass of alum(g)/theoretical yield(g) ) × 100

percent yield = ( 4.789g / 5.1629g ) × 100%

percent yield = 0.9275 × 100%

percent yield = 92.75%

Thus, the percent yield of the experiment 92.75%

pH of the solution = 4.8

NH₄Cl is a salt of a weak base and a strong acid so that it will be partially hydrolyzed

Reaction:

NH₄Cl -----> NH₄⁺ + Cl⁻

NH₄ will be hydrolyzed in water

NH₄⁺ + H₂O ------> NH₃ + H₃O⁺ (H⁺)

This solution releases H⁺ ions so they are acidic

Formula

[tex]\tt [H^+]=\sqrt{\dfrac{Kw}{Kb}.M }[/tex]

Kw = The water equilibrium constant  = 10⁻¹⁴

Kb = the ionization constant of a base = 1.8.10⁻⁵

M = 0.46

[tex]\tt [H^+]=\sqrt{\dfrac{10^{-14}}{1.8.10^{-5}}\times 0.46 }[/tex]

[H+]=√2.55 x 10⁻¹⁰ = 1.599 x 10⁻⁵

pH = - log [H+]

pH = 5-log 1.599 = 4.796≈ 4.8

Answer:

the second answer its science behind it

Answer:

b

Explanation:

Answer:

Changes in environment, food source changes and disease?

Explanation:

I dont know what the answer choices are

Answer:

changes in the environment

Explanation:

like digging big holes.

Hope this helps!

Answer:

The ball's kinetic energy is transforming to potential energy.

Explanation:

the ball is stopping resulting in potential energy

Answer:

CHO

Explanation:

Carbon = 41%,  Hydrogen = 4.58%, oxygen = 54.6%

Step 1:

Divide through by their respective relative atomic masses

41/ 12,         4.58/1,         54.6/16

3.41              4.58            3.41

Step 2:

Divide by the lowest ratio:

3.41/3.41,      4.58/3.41,     3.41/3.41

1,                    1,                  1

Hence the empirical formula is CHO

Answer:

The empirical formula of X is C3H4O3.

Explanation:

Answer:

D

Explanation:

Identifying unique features of evidence

A laboratory is a place where scientific investigations and experiments are conducted. Observing a phenomenon in the lab includes identifying unique features of evidence. Thus, option D is correct.

An observation is said to be the deduction that is seen or heard and is due to the actions of the process or event that are gained from the information from a conducted experiment.

It is an important factor in an experiment and research and is used in the lab to falsify or prove the hypothesis. It is made based on the experimental setup and involves various steps.

The observations are used to draw conclusions and inferences based on the detailed identification of the characteristics of the evidence. The evidence supports the observation.

Therefore, option D. the features of evidence are the correct option.

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Answer:

1st one

Explanation:

The question is incomplete, the complete question is;

A 8.00 L tank at 2.64 °C is filled with 9.82 g of chlorine pentafluoride gas and 10.1 g of dinitrogen difluoride gas. You can assume both gases behave as ideal gases under these conditions. Calculate the mole fraction and partial pressure of each gas, and the total pressure in the tank. Be sure your answers have the correct number of significant digits.

Answer:

See explanation for details

Explanation:

Number of moles of N2F2 = mass/ molar mass

Molar mass of N2F2 = 66 g/mol

Number of moles = 10.1 g/66 = 0.15 moles

Number of moles of ClF5 = 9.82 g/130 g/mol= 0.08 moles

Molar mass of ClF5= 130g/mol

Total number of moles = 0.15 moles + 0.08 moles = 0.23 moles

Given that;

T= 2.64 °C + 273 = 275.64 K

n= 0.23 moles

R= 0.082 Latmmol-1K-1

V= 8.00 L

P= ??

From;

PV =nRT

P= nRT/V

P= 0.23 ×0.082 × 275.64/8.00

P= 0.65 atm

Mole fraction of N2F2= 0.15/0.23 = 0.65

Partial pressure = mole fraction × total pressure = 0.65 × 0.65 = 0.42 atm

Mole fraction of ClF5 = 0.08/0.23 = 0.35

Partial pressure of ClF5 = mole fraction × total pressure = 0.35 × 0.65 = 0.22 atm

Answer:

2 should be potential energy and 3 should be A

Explanation:

Answer:

Yah

Explanation:

Answer:

false

Explanation:

Elements cannot be broken down into a simpler substance.

Answer:

-241 kJ/mol

Explanation:

Let's consider the reaction of hydrogen with excess oxygen to form water.

2 H₂ + O₂ ⟶ 2 H₂O

When 2.16g of hydrogen reacts with excess oxygen, 258 kJ of heat are released, that is, Q = -258 kJ. Considering that the molar mass of hydrogen is 2.02 g/mol, the change of enthalpy associated with the reaction of 1.00 mol of hydrogen gas is:

ΔH° = -258 kJ/2.16 g × (2.02 g/1.00 mol) = -241 kJ/mol

Answer:

Explanation:

Answer: Matter can be broken down into two categories: pure substances and mixtures. Pure substances are further broken down into elements and compounds. Mixtures are physically combined structures that can be separated into their original components. A chemical substance is composed of one type of atom or molecule :)!

Answer:

metamorphic rock from heat and pressure

Answer:

The concentration is  [tex]   [Cu^{2+}]_a  = 10^{-10.269} [/tex]  

Explanation:

From the question we are told that

    The voltage of the cell is  [tex]E =  0.22 \  V[/tex]

Generally the reaction at the cathode is  

  [tex]Cu^{2+} _{(aq)} + 2e^{-} \to  Cu_{s}[/tex] the half cell voltage is  V_c =  0.337 V

Generally the reaction at the anode is    

   [tex]Cu _{(s)} \to Cu^{2+} _{(aq)} +  2e^{-}[/tex]  the half cell voltage is  V_a = -0.337 V

Gnerally the reaction of the cell is  

    [tex]Cu_{(s)} + Cu^{2+} _{(aq)} \to Cu^{2+}_{(aq)} +  Cu_{(s)}[/tex]

At initial the voltage is  V  =  0 V

Generally the voltage of the cell at 25°C is  

       [tex]E =  V  - \frac{0.0591}{n} log \frac{[Cu^{2+}] _a}{[Cu^{2+}]_c}[/tex]

Here n is number of  of electron and it is 2

So from the question we are told that one cell has a concentration 1.5 x 10-3 M

Let assume it is  [tex][Cu^{2+}]_c[/tex]

So

      [tex]0.22=  0  - \frac{0.0591}{2} log \frac{[Cu^{2+}] _a}{  1.5 * 10^{-3} }[/tex]

=>    [tex]-7.445 =     log \frac{[Cu^{2+}] _a}{  1.5 * 10^{-3} }[/tex]

=>  [tex]-7.445 =     log [Cu^{2+}_a] - log [1.5*10^{-3}][/tex]

=>    [tex]-7.445  + log [1.5*10^{-3} =     log [Cu^{2+}_a] [/tex]

=>    [tex]-7.445  - 2.824 =     log [Cu^{2+}_a] [/tex]

Taking the antilog

=>    [tex]   [Cu^{2+}]_a  = 10^{-10.269} [/tex]    

=>     [tex]   [Cu^{2+}]_a  = 5.38 *10^{-11} \  M  [/tex]  

Answer:

Mass = 23.232 g

Explanation:

Given data:

Mass of C₈H₁₈ = 7.58 g

Mass of CO₂ produced = ?

Solution:

Chemical equation:

2C₈H₁₈ + 25O₂      →     16CO₂ + 18H₂O

Number of moles of octane:

Number of  moles =  mass/molar mass

Number of moles = 7.58 g/ 114.23 g/mol

Number of moles = 0.066 mol

Now we will compare the moles of CO₂ with octane from balance chemical equation.

                  C₈H₁₈            :           CO₂

                       2              :              16

                 0.066             :           16/2×0.066 = 0.528

Mass of CO₂ produced:

Mass = number of moles × molar mass

Mass = 0.528 mol × 44 g/mol

Mass = 23.232 g

Answer:

mixture

Explanation:

Kool-Aid  is a mixture of sugar, water, dye, and flavoring.

Answer:

ΔH°rxn = -827.5 kJ

Explanation:

Let's consider the following balanced equation.

2 PbS(s) + 3 O₂(g) → 2 PbO(s) + 2 SO₂(g)

We can calculate the standard enthalpy of reaction (ΔH°rxn) from the standard enthalpies of formation (ΔH°f) using the following expression.

ΔH°rxn = [2 mol × ΔH°f(PbO(s)) + 2 mol × ΔH°f(SO₂(g) )] - [2 mol × ΔH°f(PbS(s)) + 3 mol × ΔH°f(O₂(g) )]

ΔH°rxn = [2 mol × ΔH°f(PbO(s)) + 2 mol × ΔH°f(SO₂(g) )] - [2 mol × ΔH°f(PbS(s)) + 3 mol × ΔH°f(O₂(g) )]

ΔH°rxn = [2 mol × (-217.32 kJ/mol) + 2 mol × (-296.83)] - [2 mol × (-100.4) + 3 mol × 0 kJ/mol]

ΔH°rxn = -827.5 kJ

The standard enthalpy of the reaction is -827.5 kJ/mol

The standard enthalpy of reaction [tex]\mathbf{\Delta H^0_{rxn}}[/tex] is the enthalpy change that happens in a system whenever one mole of the matter is converted through a chemical process under normal conditions.

The given reaction can be expressed as:

2PbS(s) + 3O₂(g) → 2PbO(s) + 2SO₂(g)

The standard enthalpy can be represented by the equation:

[tex]\mathbf{\Delta H^0_{rxn} = \sum \Delta _f ^0(products) - \sum \Delta _f^0(reactants)}[/tex]

At standard conditions, the standard enthalpies of formation of the given species are:

∴

[tex]\mathbf{ \Delta H^0_{rxn} = \Big[2 mol \times \Delta H^0_f(PbO(s)) + 2 mol \times \Delta H^0_ f(SO_2(g) )\Big] - \Big[2 mol \times \Delta H^0_f (PbS(s))} + \mathbf{ 3 mol \times \Delta H^0_f(O_2(g) )\Big] }}[/tex]

[tex]\mathbf{\Delta H^0rxn = [2 mol \times (-217.32 kJ/mol) + 2 mol \times (-296.83)] - [2 mol \times (-100.4)} \\ \mathbf{+ 3 mol \times 0 kJ/mol]}}[/tex]

[tex]\mathbf{\Delta H^0rxn = -827.5 \ kJ/mol}}[/tex]

Therefore, we can conclude that the standard enthalpy of the reaction is -827.5 kJ/mol

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Answer:

Explanation:

pH = 10.3

[ H] = 10⁻¹⁰°³

= 5 x 10⁻¹¹ M

concentration of CsOH C = 3 M

pKa of carbonate = 6.35

Ka = 10⁻⁶°³⁵ = 4.46 x 10⁻⁷

Buffer capacity = 2.303 x C x Ka x [ H⁺] / ( Ka + [ H⁺]² )²

= 2.303 x 3 x 4.46 x 10⁻⁷  x 5 x 10⁻¹¹ / ( 4.46 x 10⁻⁷  + 25 x 10⁻²² )²

= 154 x 10⁻¹⁸ / 19.9 x 10⁻¹⁴

= 7.74 x 10⁻⁴ .

Answer:

Repeated experimentation introduced new discoveries.

Explanation:

Science is not a static body of  knowledge, scientific theories change as new evidence emerges from repeated experiments.

For many years, the wave theory of light was generally held by almost all scientists.

Further experiments such as photoelectric effect and Compton effect lent credence to the the particle theory of light, hence the answer

Answer:

D

Explanation:

Repeated experimentation introduced new discoveries.

Explanation:

The degree of unsaturation is given as;

Degree of unsaturation = (2C + 2 + N - H - X) / 2

(a) C5H6

Using the formular above;

Degree of unsaturation = [ 2(5) + 2 - 6] / 2

Degree of unsaturation = [ 12 - 6 ] / 2 = 3

(b) C10H6Cl2

Using the formular above;

Degree of unsaturation = [ 2(10) + 2 - 6 -2 ] / 2

Degree of unsaturation = [ 22 - 8 ] / 2 = 7

(c) C4H3NO

Using the formular above;

Degree of unsaturation = [ 2(4) + 2 - 3 -1 ] / 2

Degree of unsaturation = [ 10 - 4 ] / 2 = 3

Note: Oxygen is ignored because its presence has no effect on the degree of unsaturation.

Answer:

k = -0.09165 years^(-1)

Explanation:

The exponential decay model of a radioactive isotope is generally given as;

A(t) = A_o(e^(kt))

Where;

A_o is quantity of isotope before decay, k is decay constant and A(t) is quantity after t years

We are given;

A_o = 5 kg

A(10) = 2kg

t = 10 years

Thus;

A(10) = 2 = 5(e^(10k))

Thus;

2 = 5(e^(10k))

2/5 = (e^(10k))

0.4 = (e^(10k))

In 0.4 = 10k

-0.9164 = 10k

k = -0.9164/10

k = -0.09165 years^(-1)

Answer:

See explanation

Explanation:

Friedel-Crafts alkylation frequently undergo polysubstitution because the process  leads to the addition of an electron donating alkyl group, which activates the benzene ring to further alkylation. The acyl group rather deactivates the ring towards electrophilic substitution.

When benzene is reacted with t-butyl chloride, the chief disubstituted product is the para rather than ortho isomer because the increased bulk of the tert-butyl group hinders attack at the ortho-sites.