Are the cells in your body right now the same exact cells you was born with true or false

Answer:

[tex]\%\ ionization=6.45\%[/tex]

Explanation:

Hello.

In this case, since the ionization of a hypothetical base MOH is:

[tex]MOH(aq)\rightleftharpoons M^+(aq)+OH^-(aq)[/tex]

The equilibrium expression is:

[tex]Kb=\frac{[M^+][OH^-]}{[MOH]}[/tex]

And the percent ionization is:

[tex]\%\ ionization=\frac{[OH^-]}{[MOH]} *100\%[/tex]

For the equilibrium concentration of hydroxide, it turns out:

[tex]\%\ ionization=\frac{0.050M}{0.775M}*100\%\\\\\%\ ionization=6.45\%[/tex]

Best regards!

Answer:

8.78 L

Explanation:

The balanced equation for the combustion of butane (C₄H₁₀) in presence of oxygen (O₂) to produce carbon dioxide (CO₂) and water (H₂O) is the following:

2C₄H₁₀ + 13O₂ ⇒ 8CO₂ + 10H₂O

First, we calculate the moles there are in 5.69 g of butane, by dividing the mass into the molecular weight (MM butane= 58 g/mol):

moles C₄H₁₀= mass/MM= 5.69 g/(58 g/mol)= 0.098 mol

According to the equation, 2 moles of C₄H₁₀ produce 8 moles of CO₂. Thus, if we have 0.098 moles C₄H₁₀:

moles CO₂= 0.098 mol C₄H₁₀ x 8 mol CO₂/(2 mol C₄H₁₀)= 0.392 mol CO₂

Finally, we know that 1 mole of any gas occupies a volume of 22.4 L measured at STP, so we calculate the volume of CO₂ as follows:

volume CO₂= 0.392 mol CO₂ x 22.4 L/1 mol CO₂= 8.78 L

Answer:Approximate density of the high-leaded brass(alloy) =8.306g/cm³

Explanation:

The density of an  alloy is its mass (100g) divided by its volume

Therefore we have that the alloy (high-leaded brass) has a composition of

60.5 wt% Cu with density  8.94g/cm3

34.5 wt% Zn, wth densty  7.13g/cm3

5.0 wt% Pb with densty g/cm3

The total volume of the alloy will be the mass / density of ts composition given as :

60.5gCu/DCu + 34.5gZn/DZn + 5.0gPb/Dpb

= 60.5/8.94 + 34.5/7.13+ 5.0/11.35

= 6.76 cm³ + 4.838 cm³+0.4405 cm³ = 12.0385cm³

Approximate density of the high-leaded brass(alloy) = 100g/  12.0385cm³ =8.306g/cm³

Answer:

m1v1=m2v2 change 35 ml to l = 0.035L

12*0.035 = M2*1.20

Solve for M2 = 0.35M

Explanation:

Answer:

Explanation:

Enantiomeric excess of this compound X is given by the formula

EE= Observed specific rotation*100 / maximum specific rotation

EE= 14*100/24 = 58.33%

Percentage enantiomeric excess

%(-) = EE/2 +50

       = 58.33/2 + 50

        = 79.12% OF (-)X and 20.88% of (+) X

Answer:

Jupiter

Explanation:

Answer:

Molecules from food and molecules of oxygen move from the mouth and the nose to cells of the body through a series of blood vessels, including veins, arteries, and microscopically small blood vessels (capillaries), that extend throughout the body.Explanation:

Answer:

Molecules from food and air enter your body through your nose and mouth. These particles go down your throat and your lungs, and to your stomach. The blood also distributes these molecules.

Hope this helps!

Answer:

pH = 9.30

Explanation:

We can obtain the pH of a buffer using H-H equation:

pH = pKa + log [A⁻] / [HA]

Where pH is the pH of the solution,

pKa the pKa of the buffer (9.24 for boric acid)

[A⁻] the molarity of the conjugate base, NaH2BO3

[HA] molarity of the weak acid, H3BO3

Replacing:

pH = 9.24 + log [0.36M] / [0.31M]

Answer:

40.92%

Explanation:

CaCo₃(s)-->CaO(s)+CO₂(g)

mass of Co2(g)= 1.33grams

mass of CaO(s)=(3.25-1.33)=1.92grams

mass % of CaCO₃ = [tex](\frac{1.33}{3.25} )[/tex]×100%

40.92%

Answer:

I would say all of the above

Explanation:

I say this because we use material and energy, we grow and develop, we are made up of cells, and we respond to the environment.

The number of moles of H₂ formed when 2.97 g sample of Mg reacts with excess HCl is 0.122 moles

To determine the number of moles of H₂ that can be formed if 2.97 g sample of Mg reacts with excess HCl,

First, we will write a balanced chemical equation for the reaction

Mg + 2HCl → MgCl₂ + H₂

From the reaction above,

It means that 1 mole of Mg will react with 2 moles of HCl to produce 1 mole of MgCl₂ and 1 mole of H₂

Now, we will determine the number of moles of Mg present in the given sample.

From the question,

Mass of Mg = 2.97 g

Using the formula

[tex]Number \ of \ moles = \frac{Mass}{Molar \ mass}[/tex]

Molar mass of Mg = 24.305 g/mol

∴ [tex]Number \ moles \ of \ Mg = \frac{2.97}{24.305}[/tex]

Number of moles of Mg in the sample = 0.122 moles

Since, 1 mole of Mg will react with 2 moles of HCl to produce 1 mole of MgCl₂ and 1 mole of H₂

∴ 0.122 moles of Mg will react with excess HCl to produce 0.122 moles of  H₂

Hence, the number of moles of H₂ formed when 2.97 g sample of Mg reacts with excess HCl is 0.122 moles

Learn more here: https://brainly.com/question/18516616

i think science is important bcoz it has a choice whether to save or not

like now scientist are finding a cure for corona

Answer:

31.94 g

Explanation:

Water

ΔT = 35 C - 25 C = 10

m = ?

c = 4.182 J/gC

Heat = mc ΔT

Iron

ΔT = 85 C - 50 C = 35 (Negative sign is not relevant)

m = 85

c = 0.449 J/gC

Heat = mc ΔT= 85 * 0.449 *  35 = 1335.76 J

Substituting the value of heat in the equation for water;

Heat = m * 4.182 * 10

1335.76 = m * 41.82

m = 31.94 g

Answer:

2(NH4)3PO4 + 3CaBr2 → Ca3(PO4)2 + 6NH4Br

Explanation:

Tribasic Ammonium Phosphate + Calcium Bromide = Tricalcium Phosphate + Ammonium Bromide

Reaction Type

Double Displacement (Metathesis)

Reactants

Tribasic Ammonium Phosphate - (NH4)3PO4

Ammonium Phosphate

Molar Mass of H12N3O4P Oxidation State of H12N3O4P

Calcium Bromide - CaBr2

Kalziumbromid CaBr2 Calcium Bromide Hydrate

Molar Mass of Br2Ca Oxidation State of Br2Ca

Products

Tricalcium Phosphate - Ca3(PO4)2

Bone Phosphate Of Lime BCP TCP Tricalcium Orthophosphate Calcium Phosphate Tertiary Calcium Phosphate Calcium Orthophosphate Ca3(PO4)2

Molar Mass of Ca3O8P2 Oxidation State of Ca3O8P2

Ammonium Bromide - NH4Br

NH4Br

Molar Mass of BrH4N Oxidation State of BrH4N

Answer:

2,000,000

Explanation:

Answer:

yeah 2,000,000

Explanation:

Answer:

CrCl₃.

Explanation:

From the question given above, the following data were obtained:

Mass of chromium (Cr) = 22.60 g.

Mass of metal chloride = 68.83 g.

Empirical formula =.?

Next, we shall determine the mass of chlorine (Cl) in the metal chloride. This can be obtained as follow:

Mass of chromium (Cr) = 22.60 g.

Mass of metal chloride = 68.83 g.

Mass of chlorine (Cl) =.?

Mass of chlorine (Cl) = (Mass of metal chloride) – (Mass of chromium)

Mass of chlorine (Cl) = 68.83 – 22.60

Mass of chlorine (Cl) = 46.23 g

Finally, we shall determine the empirical formula for the metal chloride as follow:

Cr = 22.60 g.

Cl = 46.23 g

Divide by their molar mass

Cr = 22.60 / 52 = 0.435

Cl = 46.23 / 35.5 = 1.302

Divide by the smallest

Cr = 0.435 /0.435 = 1

Cl = 1.302 /0.435 = 3

The empirical formula for the metal chloride is CrCl₃

Answer:

See explanation

Explanation:

Evaporation is an endothermic process. In an endothermic process, heat is absorbed from the surrounding. This causes the system to feel cool.

Since evaporation is endothermic and heat is absorbed from the surrounding, the system will feel cool and the temperature of the system decreases accordingly.

Hence, during evaporation, the temperature of the system decreases.

Answer:

Dehydrogenation of alkanes is endothermic

Explanation:

The dehydrogenation of alkanes is an endothermic reaction. The enthalpy change for the reaction is highly positive.

Recall that for an endothermic reaction, increase in temperature increases the rate of forward reaction.

As a result of that, the dehydrogenation of alkanes proceeds in the foward direction at elevated temperatures.

The question is incomplete; the complete question is;

For the following reaction:

H2CO3 + S2- ---------> HCO3- + HS-

The equilibrium constant is greater than one. The strongest acid in this system is ___________ and the weakest base is ____________.

A) H2CO3 and HS-                  

B) S2- and HS-                        

C) HCO3- and S2-                    

D) HS- and S2-

E) H2CO3 and HCO3-

Answer:

H2CO3 and HS-  

Explanation:

According to the Brownstead- Lowry theory of acids and bases, an acid is a proton donor while a base is specie that accepts protons.

A strong acid has a greater tendency to donate protons. If we look at the reaction, H2CO3 has the greatest ability to donate protons hence it is the strongest acid.

On the other hand, HS- has the least ability to accept protons so it is the weakest base.

Answer:

Two major categories of mutations are germline mutations and somatic mutations. Germline mutations occur in gametes. These mutations are especially significant because they can be transmitted to offspring and every cell in the offspring will have the mutation. Somatic mutations occur in other cells of the body

Explanation:

g o o g l e

Two major categories of mutations are germline mutations and somatic mutations. Germline mutations occur in gametes.

A mutation is a change to an organism's DNA sequence. Errors in DNA replication during cell division, exposure to mutagens, or viral infection can all cause mutations. Somatic mutations (which happen in body cells) cannot be passed on to offspring, whereas germline mutations (which happen in eggs and sperm) can.

Our cells are constantly undergoing mutations, yet very few of these have any negative effects on our health. Compared to what we typically see in science fiction movies, this is completely different.

A mutation is never so advantageous in real life that it makes someone a superhero or does anything strange like give them wings.

Therefore, Two major categories of mutations are germline mutations and somatic mutations. Germline mutations occur in gametes.

To learn more about gametes, refer to the link:

https://brainly.com/question/2569962

#SPJ2

Answer:

B = tissue paper will stick to the balloon because it is positively charged.

Explanation:

The option B is correct. Balloon is negatively charged while it is stated in question and can also see from the picture tissue paper is positively charged.

We know that opposite charges attract each other. Thus when balloon and tissue paper come close to each other the stick together.

All other options are incorrect because,

Option A:

In option it is stated that tissue paper is negatively charged. Although negative charges are present but there is more positive charge in balloon we can say that more number of protons are present thus balloon have positive charge.

Option C:

Tissue paper is negatively charged and will move away.

It is wrong. Tissue paper is positively charged.

Option D:

Tissue paper will move away because it is  positively charged. It is also wrong statement because tissue paper will not move away it will stick to the balloon because of positive charge.

Answer:

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

[tex]n = \frac{N}{L} \\ [/tex]

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

[tex]n = \frac{3.01 \times {10}^{23} }{6.02 \times {10}^{23} } = \frac{3.01}{6.02} \\ [/tex]

We have the final answer as

Hope this helps you

The model depicted that has been resembling the modern atomic theory has been Model D. Thus option 4 is correct.

The atom has been the smallest unit of the matter, that has been consisted of the electrons, protons, and neutrons as the subatomic particles.

The modern atomic theory has stated that the structure of the atom has a positively charged centrally located nucleus, with the negatively charged body revolving around it.

The central nucleus has consisted of the neutral neutrons and positive protons, while the revolving bodies have been the negative electrons. The model depicted that has been resembling the modern atomic theory has been Model D. Thus option 4 is correct.

For more information about the modern atomic theory, refer to the link:

https://brainly.com/question/7910610

Answer: Model D; answers 4

Explanation: Is the most modern and recent atom model based on our knowledge.

Answer:

[tex]U=-6852\frac{kJ}{mol}[/tex]

Explanation:

Hello there!

In this case, since the heat capacity of the bomb calorimeter, we can use this found value on ethernet: 5.86 kJ/°C, thus, we can compute the heat gained by the calorimeter:

[tex]Q_{calorimeter}=C\Delta T=5.86kJ/\°C*(26.6-21.9)\°C=27.54kJ[/tex]

Thus, since the heat gained by the calorimeter is actually the heat released due to the combustion of byphenyl, we compute it as the negative of that of the calorimeter:

[tex]Q_{combustion}=-Q_{calorimeter}=-27.54kJ[/tex]

Finally the energy of reaction, U, in kJ/mol for the combustion of byphenyl (molar mass = 154 g/mol) is:

[tex]U=\frac{-27.54kJ}{0.627g*\frac{1mol}{156g} } \\\\U=-6852\frac{kJ}{mol}[/tex]

Best regards!

Answer:

The bond formation of C-Cl bond by applying the given data in  question 10 is 325.4 kJ

Explanation:

From the data in question 10:

The equation for the given reaction can be expressed as:

[tex]CH_{4(g)} + 2Cl_{(g)} \to CH_2Cl_2_{(g)}+H_2_{(g)}[/tex]

The enthalpy of formation of the compounds  are as follows:

Substance                       [tex]\Delta \ H^0_f \ (kJ/mol)[/tex]

[tex]CH_{4(g)}[/tex]                               - 74.8

[tex]Cl_{(g)}[/tex]                                   120.9

[tex]CH_2Cl_{2(g)}[/tex]                           -95.8

From the data obtained above, the enthalpy of the reaction [tex]\Delta \ H ^0_{rxn}[/tex] can be computed as follows:

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

[tex]\Delta H^0_{rxn } = [ 1 \ mol \times ( \Delta H^0_{CH_2Cl_{2(g)}}) + 1 \ mol \times ( \Delta H^0_{H_{2(g)}})] - [ 1 mol \times (\Delta H^0_{CH_4}_{(g)} +2 \ mol \times (\Delta \ H^0 _{Cl(g)})]}[/tex]

[tex]\Delta^0_{rxn } = [ 1 \ mol \times ( -95.8 \ kJ/mol) + 1 \ mol \times ( 0 \ kJ/mol] - [ 1 mol \times (-74.8 \ kJ/mol +2 \ mol \times (120.9 \ kJ/mol)]}[/tex]

[tex]\Delta H^0_{rxn } =-262.8 \ kJ[/tex]

However; since we knew [tex]\Delta H^0_{rxn }[/tex] to be  -262.8  kJ

At standard conditions

Bond     Bond Energy (kJ/mol)

C-H        410

H-H        432

The bond energy can be calculated by using the expression:

[tex]\Delta H^0_{rxn } = \sum \Delta H _{bond \ broken }- \sum \Delta H _{bond \ formed}[/tex]

[tex]-262. 8 KJ =[4 \times ( \Delta H_{CH})] - 1 [ (2 \times \Delta H_{CH} ) + ( 2 \times \Delta H_{C-Cl}) + (1 + \Delta H _{H-H})][/tex]

[tex]-262. 8 KJ =[4 \times ( 410 \ kJ/mol) ] - 1 [ (2 \times 410 \ kJ/mol ) + ( 2 \times \Delta H_{C-Cl}) + (1 +432 \ kJ/mol][/tex]

[tex]-262. 8 KJ = 388 \ kJ - ( 2 \times \Delta H_{C-Cl})[/tex]

[tex]( 2 \times \Delta H_{C-Cl}) = 388 \ kJ +262. 8 KJ[/tex]

[tex]( 2 \times \Delta H_{C-Cl}) = 650.8 \ KJ[/tex]

[tex]\Delta H_{C-Cl} = \dfrac{650.8 \ KJ}{2}[/tex]

[tex]\Delta H_{C-Cl} = 325.4 \ kJ[/tex]

Thus; the bond formation of C-Cl bond by applying the given data in  question 10 is 325.4 kJ

Answer:

KCH₃COO

K₂CrO₄

Explanation:

Write a formula for the compound that forms from potassium and acetate.

Potassium has a charge +1 (K⁺) and acetate has a charge -1 (CH₃COO⁻). Thus, the neutral compound is KCH₃COO (potassium acetate).

Write a formula for the compound that forms from potassium and chromate.

Potassium has a charge +1 (K⁺) and chromate has a charge -2 (CrO₄²⁻). Thus, the neutral compound is K₂CrO₄ (potassium chromate).

Answer:

[Hg₂²⁺] = 1.3 × 10⁻¹⁶ M

Explanation:

Step 1: Given data

[Cl⁻]: 0.10 M

Solubility product constant of Hg₂Cl₂ at 25 °C (Ksp): 1.3 × 10⁻¹⁸

Step 2: Write the reaction for the solution of mercury (I) chloride

Hg₂Cl₂(s) ⇄ Hg₂²⁺(aq) + 2 Cl⁻(aq)

Step 3: Use the solubility product constant to calculate [Hg₂²⁺]

Ksp = 1.3 × 10⁻¹⁸ = [Hg₂²⁺].[Cl⁻]² = [Hg₂²⁺].(0.10)²

[Hg₂²⁺] = 1.3 × 10⁻¹⁶ M

Answer:

C

Explanation:

The reference compound for naming D and L isomers of sugar is glyceraldehyde.

D and L stand for dexter (right) and laevus (left). The D and L system references molecules to that of glyceraldehyde which is chiral. By chirality, it means that glyceraldehyde molecules have the capacity to produce non-superposable mirror images. The images are labeled D and L isomers for 'right rotary' and 'left rotary' respectively.

This property allows some chemical manipulations to be performed on the molecules of glyceraldehyde without changing its configuration and as such, other compounds with similar ability are named by analogy to glyceraldehyde.

The correct option is C.