The diagram shows the different models of the atom that eventually led to the modern atomic theory.

Which model represents the current atomic theory based on our knowledge of the nucleus and electron cloud?

1) Model A

2) Model B

3) Model C

4) Model D

Answer:

Explanation:

The volume of the gas can be found by using the formula for Boyle's law

[tex]P_1V_1 = P_2V_2[/tex]

where

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume

Since we are finding the new volume

[tex]V_2 = \frac{P_1V_1}{P_2} \\[/tex]

We have

[tex]V_2 = \frac{10 \times 1}{5} = \frac{10}{5} \\ [/tex]

We have the final answer as

Hope this helps you

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:

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

Answer:

The charges are typically used to determine the mole ratio and composition of the individual atoms of elements in the compound

Explanation:

Take for instance the following compounds:

[tex]NaOH[/tex] and [tex]H_{2} O[/tex]

I hope this helps.

Answer:

Equilibrium

Explanation:

Correct on edge 2020

Answer:

A - Equilibrium

Explanation:

For edge:)

Answer: Entropy is increasing

Explanation:

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:

2,000,000

Explanation:

Answer:

yeah 2,000,000

Explanation:

Answer:

0.0003125 g

Explanation:

Given data:

Half life of iodine-125 = 60 days

Mass of iodine pellet  = 0.020 g

Mass of iodine remain after one year = ?

Solution:

one year = 365 days

Number of half lives = Time elapsed/ half life

Number of half lives = 365 days/ 60 days

Number of half lives = 6

At time zero = 0.020 g

At 1st half life = 0.020 g/2 = 0.01 g

At 2nd half life = 0.01 g / 2 = 0.005 g

At 3rd half life = 0.005 g/2 = 0.0025 g

At 4th half life = 0.0025 g / 2 = 0.00125 g

At 5tht half life = 0.00125 g / 2 =  0.000625 g

At 6th half life = 0.000625 g/ 2 = 0.0003125 g

Answer:

d. It contains one valence electron.

Since, potassium has one valence electron so it can easily donate one electron during chemical reaction.

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.

Answer:

m1v1=m2v2 change 35 ml to l = 0.035L

12*0.035 = M2*1.20

Solve for M2 = 0.35M

Explanation:

Answer:

177 grams

Explanation:

moles of CaF₂ = [tex]\frac{mass}{molar mass}[/tex] = [tex]\frac{365}{78}[/tex] = 4.67949

moles of F-atom = 2×moles of CaF₂ = 2×4.67949 = 9.35898

(mol)×(molar mass)= 9.35898×19 = 177.82062

Therefore, 177 grams of F are in the mixture.

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:

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:

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:

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: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:

They are in the same group on the periodic table indicating that they have the same number of valence electrons.

Explanation:

Answer:

Jupiter

Explanation:

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:

layers of epithelial tissue

Answer:

It's B

Explanation:

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.

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:

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