How many elements make up most of the ocean?

Answers

Answer 1

Answer:

Hydrogen and oxygen

Explanation:

Answer 2

Answer:

eight

Explanation:

the elements of course are the following

hydrogen

oxygen

calcium

sodium

sulfur

magnesium

chlorine

potassium

i hope this helps you


Related Questions

Why does a chemical change occur when copper is heated?

Answers

Answer:

When copper is heated, it decomposes to form copper oxide and carbon dioxide. It is an endothermic reaction, which means that it absorbs heat. When heated, copper is easily bent or molded into shapes.

Explanation:

When comparing the two chair conformations for a monosubstituted cyclohexane ring, which type of substituent shows the greatest preference for occupying an equatorial position rather than an axial position

Answers

Answer:

See the explanation

Explanation:

In this case, we have to keep in mind that in the monosubstituted product we only have to replace 1 hydrogen with another group. In this case, we are going to use the methyl group [tex]CH_3[/tex].

In the axial position, we have a more steric hindrance because we have two hydrogens near to the [tex]CH_3[/tex] group. If we have more steric hindrance the molecule would be more unstable. In the equatorial positions, we don't any interactions because the [tex]CH_3[/tex] group is pointing out. If we don't have any steric hindrance the molecule will be more stable, that's why the molecule will the equatorial position.

See figure 1

I hope it helps!

Describe the buffer capacity of the acetic acid buffer solution in relation to the addition of both concentrated and dilute acids and bases.

Answers

Answer:

The answer is in the explanation

Explanation:

Acetic acid, CH₃COOH, is a weak acid that will produce a buffer when its conjugate base, CH₃COO⁻, acetate ion, is added to the solution.

That is because a buffer is the mixture of a weak acid and its conjugate base or vice versa.

When an acid (HX) is added to the solution, the acetate ion will react producing acetic acid, thus:

CH₃COO⁻ + HX → CH₃COOH + X⁻

For this reason, the pH doesn't change abruptly because H⁺ ions are not produced.

Now, if a  base (BOH) is added to the buffer, CH₃COOH will react producing acetate ion and water, thus:

CH₃COOH + BOH → CH₃COO⁻ + H₂O + B⁺.

In the same way, there are not produced free OH⁻ and the pH doesn't change significantly.

Air is compressed from an inlet condition of 100 kPa, 300 K to an exit pressure of 1000 kPa by an internally reversible compressor. Determine the compressor power per unit mass flow rate if the device is (a) isentropic, (b) polytropic with n =1.3, (c) isothermal

Answers

Answer:

(a) [tex]W_{isoentropic}=8.125\frac{kJ}{mol}[/tex]

(b) [tex]W_{polytropic}=7.579\frac{kJ}{mol}[/tex]

(c) [tex]W_{isothermal}=5.743\frac{kJ}{mol}[/tex]

Explanation:

Hello,

(a) In this case, since entropy remains unchanged, the constant [tex]k[/tex] should be computed for air as an ideal gas by:

[tex]\frac{R}{Cp_{air}}=1-\frac{1}{k} \\\\\frac{8.314}{29.11} =1-\frac{1}{k}\\[/tex]

[tex]0.2856=1-\frac{1}{k}\\\\k=1.4[/tex]

Next, we compute the final temperature:

[tex]T_2=T_1(\frac{p_2}{p_1} )^{1-1/k}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.4}=579.21K[/tex]

Thus, the work is computed by:

[tex]W_{isoentropic}=\frac{kR(T_2-T_1)}{k-1} =\frac{1.4*8.314\frac{J}{mol*K}(579.21K-300K)}{1.4-1}\\\\W_{isoentropic}=8.125\frac{kJ}{mol}[/tex]

(b) In this case, since [tex]n[/tex] is given, we compute the final temperature as well:

[tex]T_2=T_1(\frac{p_2}{p_1} )^{1-1/n}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.3}=510.38K[/tex]

And the isentropic work:

[tex]W_{polytropic}=\frac{nR(T_2-T_1)}{n-1} =\frac{1.3*8.314\frac{J}{mol*K}(510.38-300K)}{1.3-1}\\\\W_{polytropic}=7.579\frac{kJ}{mol}[/tex]

(c) Finally, for isothermal, final temperature is not required as it could be computed as:

[tex]W_{isothermal}=RTln(\frac{p_2}{p_1} )=8.314\frac{J}{mol*K}*300K*ln(\frac{1000kPa}{100kPa} ) \\\\W_{isothermal}=5.743\frac{kJ}{mol}[/tex]

Regards.

What are 3 characteristics of chemical reactions

Answers

Answer:

Evolution of gas.

Formation of a precipitate.

Change in color.

Explanation:

How many grams of POCl3 are produced when 225.0 grams of P4O10 and 675.0 grams of PCl5 react? This is the balance equation P4O10 + 6PCl5 → 10POCl3

Answers

Answer:

900g of POCl₃

Explanation:

Hello,

To solve this question, we'll require the equation of reaction.

P₄O₁₀ + 6PCl₅ → 10POCl₃

Molar mass of P₄O₁₀ = 283.886 g/mol

Molar mass of PCl₅ = 208.24 g/mol

Molar mass of POCl₃ = 153.33 g/mol

But Number of moles = mass / molar mass

Mass = molar mass × number of moles

Mass of POCl₃ = 153.33 × 10 = 1533.3g

Mass of PCl₅ = 208.24 × 6 = 1249.44g

Mass of P₄O₁₀ = 283.886 × 1 = 283.886g

From the equation of reaction,

283.886g of P₄O₁₀ + 1249.44g of PCl₅ produces 1533.33g of POCl₃

I.e 1533.33g of reactants produces 1533.33g of product (law of conservation of mass)

Therefore, (225g of P₄O₁₀ + 675g of PCl₅) = 900g will give x g of POCl₃.

1533.33g of reactants = 1533.33g of products

900g of reactants = x g of products

x = (900 × 1533.33) / 1533.33

x = 900g of POCl₃

An experimenter studying the oxidation of fatty acids in extracts of liver found that when palmitate (16:0) was provided as substrate, it was completely oxidized to CO2. However, when undecanoic acid (11:0) was added as substrate, incomplete oxidation occurred unless he bubbled CO2 through the reaction mixture. The addition of the protein avidin, which binds tightly to biotin, prevented the complete oxidation of undecanoic acid even in the presence of CO2, although it had no effect on palmitate oxidation. Explain these observations in light of what you know of fatty acid oxidation reactions.

Answers

Answer:

Even-number fatty acids such as palmitate undergoes complete β-oxidation in the liver motochondria to CO₂ because the product, acetyl-CoA can enter the TCA cycle.

Oxidation of odd-number fatty acids such as undecanoic acid yields acetyl-CoA + propionyl-CoA in their last pass. Propionyl-CoA requires additional reactions including carboxylation in order to be able to enter the TCA cycle.

The reaction CO2 + propionyl-CoA ----> methylmalonyl-CoA is catalyzed by propionyl-CoA carboxylase, a biotin-containing enzyme, which is inhibited by avidin.  Palmitate oxidation however, does not involve carboxylation.

Explanation:

Even-number fatty acids such as palmitate undergoes complete β-oxidation in the liver motochondria to CO₂ because their oxidation product, acetyl-CoA, can enter the TCA cycle where it is oxidized to CO₂.

Undecanoic acid is an odd-number fatty acid having 11 carbon atoms. Oxidation of odd-number fatty acids such as undecanoic acid yields a five -carbon fatty acyl substrate for their last pass through β-oxidation which is oxidized and cleaved into acetyl-CoA + propionyl-CoA. Propionyl-CoA requires additional reactions including carboxylation in order to be able to enter the TCA cycle. Since oxidation is occuring in a liver extract, CO₂ has to be externally sourced in order for the carboxylation of propionyl-CoA to proceed and thus resulting in comlete oxidation of undecanoic acid.

The reaction CO2 + propionyl-CoA ----> methylmalonyl-CoA is catalyzed by propionyl-CoA carboxylase, a biotin-containing enzyme.  The role of biotin is to activate the CO₂ before its tranfer to the propionate moiety. The addition of the protein avidin prevents the complete oxidation of undecanoic acid by  binding tightly to biotin, hence inhibiting the activation and transfer of CO₂ to propionate.

Palmitate oxidation however, does not involve carboxylation, hence addition of avidin has no effect on its oxidation.

Harvey kept a balloon with a volume of 348 milliliters at 25.0˚C inside a freezer for a night. When he took it out, its new volume was 322 milliliters, but its pressure was the same. If the final temperature of the balloon is the same as the freezer’s, what is the temperature of the freezer?

Answers

Answer:

[tex]T2=276K[/tex]

Explanation:

Given:

Initial volume of the balloon V1 = 348 mL

Initial temperature of the balloon T1 = 255C

Final volume of the balloon V2 = 322 mL

Final temperature of the balloon T2 =

To calculate T1 in kelvin

T1= 25+273=298K

Based on Charles law, which states that the volume of a given mass of a ideal gas is directly proportional to the temperature provided that the pressure is constant. It can be applied using the below formula

[tex](V1/T1)=(V2/T2)[/tex]

T2=( V2*T1)/V1

T2=(322*298)/348

[tex]T2=276K[/tex]

Hence, the temperature of the freezer is 276 K

Answer: 276 kelvins

Explanation:

Be sure to answer all parts. Three 8−L flasks, fixed with pressure gauges and small valves, each contain 4 g of gas at 276 K. Flask A contains He, flask B contains CH4, and flask C contains H2. Rank the flask contents in terms of:

Answers

Here is the complete question.

Be sure to answer all parts. Three 8−L flasks, fixed with pressure gauges and small valves, each contain 4 g of gas at 276 K. Flask A contains He, flask B contains CH4, and flask C contains H2. Rank the flask contents in terms of:  the following properties. (Use the notation >, <, or =, for example B=C>A.)

(a) pressure

(b) average molecular kinetic energy

(c) diffusion rate after the valve is opened

(d) total kinetic energy of the molecules

Answer:

Explanation:

Given that:

Three flask A,B, C:

contains a volume of 8-L

mass m = 4g    &;

Temperature = 276 K

Flask A = He

Flask B = H₂

Flask C = CH₄

a) From the ideal gas equation:

PV = nRT

where;

n = number of moles = mass (m)/molar mass (mm)

Then:

PV = m/mm RT

If  T ,m and V are constant for the three flasks ; then

P ∝ 1/mm

As such ; the smaller the molar mass the larger the pressure.

Now; since the molecular weight of CH₄ is greater than He and H₂ and also between He and H₂,  He has an higher molecular weight .

Then the order of pressure in the flask is :

[tex]\mathbf{P_B >P_A>P_C}[/tex]

where :

[tex]P_A[/tex] = pressure in the flask A

[tex]P_B[/tex] = pressure in the flask B

[tex]P_C[/tex]= Pressure in the flask C

b)

average molecular kinetic energy

We all know that  the average molecular kinetic energy  varies directly proportional to the temperature.

Thus; the given temperature = 276 K

∴ The order of the average molecular kinetic energy is [tex]\mathbf{K.E_A =K.E_B =K.E_C}[/tex]

c)

The rate of diffusion of gas is inversely proportional to the square root of it density . Here the density is given in relation to their molar mass.

So;

rate of diffusion ∝ [tex]\dfrac{1}{\sqrt{mm} }[/tex]

where;

[tex]D_A[/tex] = rate of diffusion in flask A

[tex]D_B[/tex] = rate of diffusion in flask B

[tex]D_C[/tex] = rate of diffusion in flask C

Thus; the order of the rate of diffusion = [tex]D_B[/tex]  > [tex]D_A[/tex] > [tex]D_C[/tex]

d)  total kinetic energy of the molecules .

The kinetic energy deals with how the speed of particles of a  substance determines how fast the substances will diffuse in a given set of condition.

The the order of the total kinetic energy depends on the molecular speed

Thus; the order of the total kinetic energy  for the three flask is as follows:

[tex]\mathbf{ K.E_B>K.E_A>K.E_C}[/tex]

The acetic acid/acetate buffer system is a common buffer used in the laboratory. Write the equilibrium equation for the acetic acid/acetate buffer system. The formula of acetic acid is CH3CO2H .

Answers

Answer:

CH₃CO₂H + H₂O ⇄ CH₃CO₂⁻ + H₃O⁺

Explanation:

A buffer is defined as the mixture of a weak acid and its conjugate base or vice versa.

For the acetic acid buffer, CH₃CO₂H is the weak acid and its conjugate base is the ion without H⁺, that is CH₃CO₂⁻. The equilibrium equation in water knowing this is:

CH₃CO₂H + H₂O ⇄ CH₃CO₂⁻ + H₃O⁺

In the equilibrium, the acid is dissociated in the conjugate base and the hydronium ion.

The acetic acid/acetate buffer system is a common buffer used in the laboratory, the equilibrium equation for the acetic acid/acetate buffer system. The formula of acetic acid is CH3CO2H -

CH₃CO₂H + H₂O ⇄ CH₃CO₂⁻ + H₃O⁺

An acid buffer is a solution that contains roughly the same concentrations of a weak acid and its conjugate base.

an acetate buffer contains roughly equal concentrations of acetic acid and acetate ion.Both are in chemical equilibrium with each other.

The equation is:

CH₃CO₂H + H₂O ⇄ CH₃CO₂⁻ + H₃O⁺

where CH₃CO₂H - acetic acid

and, CH₃CO₂⁻ acetate ion

Thus, CH₃CO₂H + H₂O ⇄ CH₃CO₂⁻ + H₃O⁺ is the equilibrium equation for the acetic acid/acetate buffer system.

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g The atomic mass of an element is equal to ________. The atomic mass of an element is equal to ________. its mass number one-twelfth of the mass of a carbon-12 atom a weighted average mass of all of the naturally occurring isotopes of the element its atomic number the average mass of all of the naturally occurring isotopes of the element

Answers

Answer:

Total numbe of protons and neutrons in a single atom of that element

Explanation:

Hello,

I'll answer the question by filling in the blank spaces

"The atomic mass of an element is equal to the total number of proton and neutron in a particular atom of the element. The atomic mass of an element is equal to the atomic weight. Its mass number one-twelfth of the mass of carbon-12 atom a weighted mass of all naturally occurring isotopes of the elements. Its atomic mass is the average mass of all the naturally occurring isotopes of the element."

The atomic mass of an element is the total number of protons and neutrons in a single atom of that element.

The atomic mass of an element is equal to a weighted average mass of all of the naturally occurring isotopes of the element. The correct answer is option 2.

Isotopes are elements with the same number of protons (atomic number) but differing numbers of neutrons (mass number).

Most elements exist in nature as a mixture of isotopes, each with a different mass number and abundance. The atomic mass of an element is computed by adding the masses of all isotopes, multiplying by their relative abundance, and dividing by the total abundance of all isotopes.

This gives a weighted average mass that corresponds to the normal mass of an element's atom in nature.

Therefore, the correct answer is option 2. to a weighted average mass of all of the naturally occurring isotopes of the element.

Learn more about isotopes here:

https://brainly.com/question/27475737

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A solution is prepared by mixing 5.00 mL of 0.100 M HCl and 2.00 mL of 0.200 M NaCl. What is the molarity of chloride ion in this solution?

Answers

Answer:

0.129 M

Explanation:

0.100 M HCl = 0.100 mol/L solution HCl

5.00 mL = 0.00500 L solution HCl

0.100 mol/L HCl * 0.00500 L = 0.000500 mol HCl

                             HCl ------> H+ + Cl-

                           1 mol                   1 mol

                    0.000500 mol           0.000500 mol

0.200 M NaCl = 0.200 mol/L solution NaCl

2.00 mL = 0.00200 L solution NaCl

0.200 mol/L NaCl*0.00200 L = 0.000400 mol NaCl

                              NaCl ------> Na+ + Cl-

                            1 mol                        1 mol

                     0.000400 mol               0.000400 mol

Chloride ion altogether (0.000500 mol + 0.000400 mol) =0.000900 mol

Solution altogether (0.00500 L+0.00200 L) = 0.00700L

Molarity (Cl-)= solute/solution = 0.000900 mol/0.00700L = 0.129 mol/L=

= 0.129 M

Carbon dioxide and water vapor are variable gases because _____.

Answers

Answer: their amounts vary throughout the atmosphere

Explanation:

There is very little that travels over the atmosphere

Vary=very little

Hope that helps

How many moles of CO2 can be produced by the complete reaction of 1.0 g of lithium carbonate with excess hydrochloric acid (balanced chemical reaction is given below)? Li2CO3(s) + 2HCl(aq) --> 2LiCl(aq) + H2O(l) + CO2(g) Question 1 options: 1.7 g 1.1 g 0.60 040 g

Answers

Answer:Mass of CO2 = 0.60g

Explanation:

Given the chemical rection

Li2CO3(s) + 2HCl(aq) --> 2LiCl(aq) + H2O(l) + CO2(g

No of moles = mass / molar mass

molar mass Li2CO3 = Molecular mass  calculation: 6.941 x 2 + 12.0107 + 15.9994 x 3 =  

= 73.8909 g/mol

therefore Number of moles Li2CO3 = 1.0g / 73.89 g/mol

= 0.0135 moles Li2CO3

From our given Balanced equation,  shows that  

Li2CO3(s) + 2HCl(aq) --> 2LiCl(aq) + H2O(l) + CO2(g

1 mole Li2CO3 produces 1 mole CO2

therefore 0.0135 mol Li2CO3 will produce  0.0135 moles of CO2

Also

No of moles = mass / molar mass

Mass = No of moles x molar mass

molar mass of CO2=12.0107 + 15.9994 x 2=44.0095 g/mol

Mass of CO2= 0.0135 X 44.0095 g/mol =0.594≈0.60g

what’s the SI unit of time ?

Answers

A first option. ......

Answer:

The answer is A

Explanation:

When ethanol, C2H5OH (a component in some gasoline mixtures) is burned in air, one molecule of ethanol combines with three oxygen molecules to form two CO2 molecules and three H2O molecules.


A) Write the balanced chemical equation for the reaction described.


B) How many molecules of CO2 and H2O would be produced when 2 molecules ethanol are consumed? Equation?


C) How many H2O molecules are formed, then 9 O2 molecules are consumed? What conversion factor did you use? Explain!


D) If 15 ethanol molecules react, how many molecules O2 must also react? What conversion factor did you use? Explain!

Answers

Answer:

1) C2H5OH(l)+3O2(g)⟶2CO2(g)+3H2O(l)

2) four molecules of CO2 will be produced and six molecules of water

3)9 molecules of water are formed when 9 molecules of oxygen are consumed.

4) 45 molecules of oxygen

Explanation:

The balanced chemical reaction equation is shown here and must guide our work. When ethanol is burned in air, it reacts as shown;

C2H5OH(l)+3O2(g)⟶2CO2(g)+3H2O(l)

Hence, if we use 2 molecules of ethanol, the balanced reaction equation will look like this;

2C2H5OH(l)+6O2(g)⟶4CO2(g)+6H2O(l)

Hence four molecules of CO2 are formed and six molecules of water are formed

From the balanced stoichiometric equation;

3 molecules of oxygen yields 3 molecules of water

Therefore, 9 molecules of oxygen will yield 9 × 3/3 = 9 molecules of water

Therefore, 9 molecules of water are formed when 9 molecules of oxygen are consumed.

From the reaction equation;

1 molecule of ethanol reacts with 3 molecules of oxygen

Therefore 15 molecules of ethanol will react with 15 × 3/1 = 45 molecules of oxygen

State the effect of anion hydrolysis on the pH of water​

Answers

Answer:

Depending on the anions and cations present within a hydrolysis reaction, the solution can be more... ... This lesson will explain how this occurs. ... that could react with water and create products that affect the characteristics of the solution.

Answer:

Salts of weak bases and strong acids do hydrolyze, which gives it a pH less than 7. This is due to the fact that the anion will become a spectator ion and fail to attract the H+, while the cation from the weak base will donate a proton to the water forming a hydronium ion.

Explanation:

I hope this is the answer your looking for

Give the characteristic of a zero order reaction having only one reactant. a. The rate of the reaction is not proportional to the concentration of the reactant. b. The rate of the reaction is proportional to the square of the concentration of the reactant. c. The rate of the reaction is proportional to the square root of the concentration of the reactant. d. The rate of the reaction is proportional to the natural logarithm of t

Answers

Answer:

a. The rate of the reaction is not proportional to the concentration of the reactant.

Explanation:

The rate expression for a zero order reaction is given as;

A → Product

Rate = k[A]⁰

[A]⁰ = 1

Rate = K

GGoing through the options;

a) This is correct because in the final form of the rate expression, the rate is independent of the concentration.

b) This option is wrong

c) This option is also wrong

d) Like options b and c this is also wrong becaus ethere is no relationship between either the concentration or t.

The cytochromes are heme‑containing proteins that function as electron carriers in the mitochondria. Calculate the difference in the reduction potential (ΔE∘′) and the change in the standard free energy (ΔG∘′) when the electron flow is from the carrier with the lower reduction potential to the higher. cytochrome c1 (Fe3+)+e−↽−−⇀cytochrome c1 (Fe2+)E∘′=0.22 V cytochrome c (Fe3+)+e−↽−−⇀cytochrome c (Fe2+)E∘′=0.254 V Calculate ΔE∘′ and ΔG∘′ .

Answers

Complete Question

The complete question is shown on the first uploaded image

Answer:

The change in reduction potential is  [tex]\Delta E^o=E^o_{cell} = 0.034 V[/tex]

The change in standard free energy is  [tex]\Delta G^o = -3.2805 \ KJ/mol[/tex]

Explanation:

From the question we are told that

At the anode

      [tex]cytochrome \ c_1 \ (Fe^{3+}) + e^-[/tex]⇔[tex]cytochrome \ c_1 \ (Fe^{2+}) \ \ E^o = 0.22 \ V[/tex]

At the cathode

      [tex]cytochrome \ c \ (Fe^{3+}) + e^-[/tex]⇔[tex]cytochrome \ c \ (Fe^{2+}) \ \ E^o = 0.254 \ V[/tex]

The difference in the reduction potential is mathematically represented as

     [tex]\Delta E^o = E^o_{cathode} - E^o_{anode}[/tex]

substituting values

      [tex]\Delta E^o = 0.254 - 0.220[/tex]

     [tex]\Delta E^o=E^o_{cell} = 0.034 V[/tex]

The change in the standard free energy is mathematically represented as

      [tex]\Delta G^o = -n * F * E^o_{cell}[/tex]

Where  F is the Faraday constant with value  F = 96485 C

and  n i the number of the number of electron = 1

   So

       [tex]\Delta G^o = -(1) * 96485 * 0.034[/tex]

       [tex]\Delta G^o = -3.2805 \ KJ/mol[/tex]

Consider the insoluble compound zinc carbonate , ZnCO3 . The zinc ion also forms a complex with hydroxide ions . Write a balanced net ionic equation to show why the solubility of ZnCO3 (s) increases in the presence of hydroxide ions and calculate the equilibrium constant for this reaction. For Zn(OH)42- , Kf = 2.9×1015 . Use the pull-down boxes to specify states such as (aq) or (s).

Answers

Answer:

The net ionic equation is  [tex]ZnCO_3 _{(s)} + 4 OH^{-}_{(aq)} \to [Zn(OH)_4]^{2-} _{(aq)} + CO_3^{2-} _{(aq)}[/tex]

The equilibrium constant is  [tex]K = 4.06 *10^{4}[/tex]

Explanation:

From the question we are that

      The  [tex]K_f = 2.9 *10^{15 }[/tex]

The ionic equation is chemical represented as

    Step 1

         [tex]ZnCO_3 _{(s)}[/tex]  ⇔   [tex]Zn^{2+} _{aq} + CO_3^{2-} _{aq}[/tex]   The  solubility product constant for stage is     [tex]K_{sp} = 1.4*10^{-11}[/tex]

 Step 2

        [tex]Zn^{2+} _{(aq)} + 4 0H^{-} _{(aq)}[/tex]    ⇔  [tex][Zn(OH_4)]^{2-} _{(aq)}[/tex]  The formation constant for this step is given as [tex]K_f = 2.9 *10^{15 }[/tex]

 The net reaction is  

           [tex]ZnCO_3 _{(s)} + 4 OH^{-}_{(aq)} \to [Zn(OH)_4]^{2-} _{(aq)} + CO_3^{2-} _{(aq)}[/tex]

The equilibrium constant is mathematically evaluated as

         [tex]K = K_{sp} * K_f[/tex]

substituting values

         [tex]K = 1.4*10^{-11} * 2.9 *10^{15}[/tex]

        [tex]K = 4.06 *10^{4}[/tex]

Draw structural formulas for all the alkene(s) formed by treatment of each haloalkane or halocycloalkane with sodium ethoxide in ethanol. Assume that elimination occurs by an E2 mechanism.

Answers

Answer:

Explanation:

Kindly note that I have attached the complete question as an attachment.

Here, we are told that elimination occurs by an E2 mechanism. What this means is that the hydrogen and the halogen must be above and below for the reaction to proceed.

The possible products are as follows;

Please check attachment for complete equations and diagrams of compounds too.

s the following nuclear equation balanced? yes no

Answers

Answer:

Yes.

Explanation:

The nuclear equation {226/88 Ra → 222/26 Rn + 4/2 He} is balanced. As we know that an alpha particle is identical to a helium atom. This implies that if an alpha particle is eliminated from an atom's nucleus, an atomic number of 2 and a mass number of 4 is lost.

Therefore, the equation will be reduced to:

226 - 4 = 222

88 - 2 = 86

Hence, the equation is balanced.

If one contraction cycle in muscle requires 55 kJ55 kJ , and the energy from the combustion of glucose is converted with an efficiency of 35%35% to contraction, how many contraction cycles could theoretically be fueled by the complete combustion of one mole of glucose? Round your answer to the nearest whole number.

Answers

Answer:

18 moles

Explanation:

Here the combustion of one mole of glucose ----> carbon dioxide + water, releases 2870 kilojoules / moles.

_______________________________________________________

With one contraction cycle requiring 55 kilojoules,

2870 / 55 ≈ 52.18

And with the efficiency being 35 percent,

52.1818..... * 0.35 = ( About ) 18 moles

Hope that helps!

The atomic mass of gallium is 69.72 . The density of iron is 7.87 . The atomic mass of iron is 55.847 . Calculate the number of gallium atoms in one ton (2000 pounds) of gallium. (Enter your answer to three significant figures.)

Answers

Answer:

the atomic mass of any elemet contains avogardo numberof atoms

In case of Gallium,

69.72 gram is atomic mass and it cotnains around 6.023*10^23 atoms of Gallium

but, 2000 punds = 907184.7 grams

907184.7 gram of gallium contains= 6.023*10^23* 907184/69.72

                                                          = 79 *10^26 atoms

Explanation:

Enter your answer in the provided box. On a cool, rainy day, the barometric pressure is 739 mmHg. Calculate the barometric pressure in centimeters of water (cmH2O) (d of Hg = 13.5 g/mL; d of H2O = 1.00 g/mL).

Answers

Answer:

997.65cmH2O

Explanation:

Barometric pressure = 739 mmHg

density of Hg = 13.5 g/ml

density of water (H2O) = 1.00 g/ml

Calculate Barometric pressure in centimetres of water ( cmH20)

equate the barometric pressure of Hg and water

739 * 13.5 * 9.8 = x * 1 * 9.81

x ( barometric pressure of water in mmH2O ) = 739 *13.5 / 1 = 9976.5mmH2O

in cmH2O = 997.65cmH2O

Enter an equation for the formation of C2H5OH(l) from its elements in their standard states. Enter any reference to carbon as C(s). Express your answer as a chemical equation. Identify all of the phases in your answer.

Answers

Answer:

C(s) + 3 H₂(g) + 1/2 O₂(g) ⇒C₂H₅OH(l)

Explanation:

Ethanol (C₂H₅OH) is an alcohol and it is formed by carbon (C), H (hydrogen) and O (oxygen) atoms. These elements in their standard states are:

C: C(s), it is solid, could be graphite, diamond, among others.

H: H₂(g), it is a diatomic gas.

O: O₂(g), it is a diatomic gas.

So, we can write the equation for the formation of  C₂H₅OH from C(s), H₂ and O₂ as follows:

C(s) + H₂(g) + O₂(g) ⇒C₂H₅OH(l)

Finally, we have to balance the equation by adding the estequiometrical coefficients:

C(s) + 3 H₂(g) + 1/2 O₂(g) ⇒C₂H₅OH(l)

2C(s)+3[tex]H_{2} [/tex](g)+[tex]\frac{1}{2} [/tex][tex]O_{2} [/tex](g)→[tex]C_{2} [/tex][tex]H_{5} [/tex]OH(l)

Explanation:

Standard state of carbon: C(s)

Standard state of oxygen: [tex]O_{2} [/tex](g)

Standard State of Hydrogen: [tex]H_{2} [/tex](g)

Then balance the equation C2H5OH(l) to get 2C(s)+3[tex]H_{2} [/tex](g)+[tex]\frac{1}{2} [/tex][tex]O_{2} [/tex](g)→[tex]C_{2} [/tex][tex]H_{5} [/tex]OH(l).

On a hot summer day, the density of air at atmospheric pressure at 35.5°C is 1.1970 kg/m3. (a) What is the number of moles contained in 1.00 m3 of an ideal gas at this temperature and

Answers

Complete question:

On a hot summer day, the density of air at atmospheric pressure at 35.5°C is 1.1970 kg/m3. (a) What is the number of moles contained in 1.00 m3 of an ideal gas at this temperature and pressure.

Answer:

The  number of moles contained by an ideal gas at this temperature and pressure is 41.32 moles.

Explanation:

Given;

density of dry air, ρ = 1.1970 kg/m³

temperature of the air, T = 35.5°C  = 273 + 35.5 = 308.5 K

air volume, V = 1 m³

Apply ideal gas law for dry to calculate the air pressure;

[tex]P = \rho R_dT[/tex]

where;

P is the air pressure

ρ is the air density

Rd is gas constant for dry air = 287 J/kg/K

P = 1.197 x 287 x 308.5 = 105,981.78 Pa

(a) Now, determine the number of moles contained by an ideal gas at this temperature and pressure, by applying ideal gas law;

PV = nRT

where;

P is the pressure of the gas (Pa)

V is the volume of the gas (m³)

n is number of gas moles

R is gas constant = 8.314 m³.Pa / mol.K

T is temperature (K)

n = (PV) / (RT)

n = (105,981.78 x 1) / (8.314 x 308.5)

n = 41.32 moles

Therefore, the  number of moles contained by an ideal gas at this temperature and pressure is 41.32 moles.

The number of moles of an ideal gas at this temperature and pressure is 41.5 moles.

Given that;

Density of dry air = 1.1970 kg/m3

Pressure of dry air = ?

Temperature of dry air = 35.5°C + 273 = 308.5 K

Hence;

P = Density × gas constant of dry air × Temperature

P = 1.1970 kg/m3 × 287.1 J/Kg/K × 308.5 K

P = 106019 Pa or 1.05 atm

Using the ideal gas equation;

PV = nRT

n = PV/RT

n = 1.05 atm × 1000 L/0.082 atmL/K.mol × 308.5 K

n = 41.5 moles

Learn more: https://brainly.com/question/11897796

A certain element consists of two stable isotopes. The first has a mass of 62.9 amu and a percent natural abundance of 69.1 %. The second has a mass of 64.9 amu and a percent natural abundance of 30.9 %. What is the atomic weight of the element?

Answers

Answer:

63.518

Explanation:

The following data were obtained from the question:

Mass of Isotope A = 62.9 amu

Abundance of isotope A (A%) = 69.1%

Mass of isotope B = 64.9 amu

Abundance of isotope B (B%) = 30.9%

Atomic weight of the element =..?

The atomic weight of the element can be obtained as follow:

Atomic weight = [(Mass of A x A%)/100] + [(Mass of B x B%) /100]

Atomic weight = [(62.9 x 69.1)/100] + [(64.9 x 30.9)/100]

Atomic weight = 43.4639 + 20.0541

Atomic weight = 63.518

Therefore, the atomic weight of the element is 63.518.

Duncan knows that it takes 36400 cal of energy to heat a pint of water from room temperature to boiling. However, Duncan has prepared ramen noodles so many times he does not need to measure the water carefully. If he happens to heat 0.800 pint of room-temperature water, how many kilojoules of heat energy will have been absorbed by the water at the moment it begins to boil?

Answers

Answer:

[tex]\large \boxed{\text{122 000 J}}[/tex]

Explanation:

1. Calculate the energy needed

[tex]\text{Energy} = \text{0.800 pt} \times \dfrac{\text{36 400 cal}}{\text{1 pt}} = \text{ 29 120 cal}[/tex]

2. Convert calories to joules

[tex]\text{Energy} = \text{29 120 cal} \times \dfrac{\text{4.184 J}}{\text{1 cal}} = \textbf{122 000 J}\\\\\text{The water will have absorbed $\large \boxed{\textbf{122 000 J}}$}[/tex]

Look at the picture and observations below.
Observations: The bee's wings are moving very fast.
The bee's wings are much smaller than its body.
what’s the answer ?

Answers

First of all what is it asking for and second of all there is no picture of the observations, sorry can’t help with this question.

Answer:

How are bees able to fly?

Explanation:

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