Which accurately represents these building blocks of matter from the smallest to the largest?
atom -- molecule or compound
O molecule -- atom - element
compound - molecule -- element
molecule atom or element

Answer:

pressure

Explanation:

pressure is the amount of force exerted on an area. when you blow up the balloon you're filling it with gas particles. the gas particles move freely within the balloon and may collide with one another exerting pressure on the inside of the balloon.

The pressure of the gas is the amount of force that is exerted on a balloon by the gas inside the balloon. Therefore, option B is correct.

Pressure can be described as the force applied perpendicular to the surface of a body per unit area. Pressure can be defined as a standard mechanical quantity and is derived from a unit of force divided by a unit of area.

The SI unit of measurement of pressure, the pascal (Pa) or Newton per square meter (N/m²). Pressure can be defined as the amount of force exerted perpendicular to the surface per unit area.

Mathematically, the pressure exerted by force can be calculated as:

[tex]{\displaystyle p={\frac {F}{A}}}[/tex]

where, p is the pressure, F is the magnitude of the normal force, and A is the area of the surface.

Therefore, the amount of force that is exerted on the balloon by the gas inside the balloon is equal to pressure.

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Answer:

Phagocytosis

Explanation:

Answer: 2681.81

Explanation:

Answer:

6.25e-6 is the value of the equilibrium constant

Explanation:

we have this equation

[tex]PbBr(s) ----- Pb^{2+}(aq) + 2Br(aq)[/tex]

When at a state of equilibrium,

we have the concentration of Pb^2+ to be 0.01

we have the concentration of Br^- to be 0.025

the equilibrium constant concentration of both pure solids and liquid s are said to be equal to 1

[PbBR2] = 1

such tht

Keq = [Pb^2+] x [Br-]^2

we already know the values of these from the above.

0.01x0.025^2

= 0.01 x 0.000625

= 0.00000625

= 6.25 x 10^-6

= 6.25e^-6

Answer:

C3Br6

Explanation:

C= (1 X 12.011) = 12.011

Br= (2 X 79.904)= 159.808

159.808+12.011 = 171.819

515.46 divided by 171.819 = 3.00

so you mulitpy CBr2 by 3 which gives you C3Br6

Answer:

I'm on the exact same queston

Answer:

The test variable (independent variable) controls the outcome variable (dependent variable)

Explanation:

its right on study island

Explanation:

U need to draw the graph first and make a line at 17 pennies, where the line of 17 pennies and your graph meet is the mass of it(at y axis)

Answer:

See explanation

Explanation:

A popular experiment that describes the effect of heat on the position of equilibrum is the change of colour when copper II chloride is heated.

As the solution is heated, it's colour changes from blue to green, this implies the the colour change (blue to green) is an endothermic process (equilibrum position shifts to the right with increase in temperature)

The equilibrum is represented by the equation;

[Cu(H2O)6]^2+(aq) + 4Cl^-(aq)<------>[CuCl4]^2-(aq) + 6H2O(l) ∆H=positive

The equilibrium mixture undergoing cooling or heating have colour changes. The temperature affects the colour of the products formed and the forward reaction is endothermic.

In the reaction copper (II) chloride or [tex]\rm CuCl_{4}[/tex] is the main species. The heat or the temperature affects the colour formation of copper (II) chloride as the equilibrium change affects the colouration of the product.

The heating of the solution affects the colour change from blue to the green of the reactant to products and the forward reaction shifts the equilibrium towards the right when the temperature is increased and is an endothermic reaction.

The reaction at the equilibrium can be shown as,

[tex]\rm [Cu(H_{2}O)_{6}]^{2+} (aq) + 4Cl^{-} (aq) \Leftrightarrow [CuCl_{4}]^{2-}(aq) + 6H_{2}O(l), \Delta H=positive[/tex]

Therefore, temperature changes the colouration and the forward reaction is an endothermic reaction.

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Answer:

8.937g/cm³

Explanation:

To answer this question we need to know that, in 1 unit FCC cell you have:

Edge length = √8 * R

Volume = 8√8 * R³

And there are 4 atoms per unit cell

Mass of 4 atoms in g:

4 atom * (1mol / 6.022x10²³atom) * (63.55g / mol) = 4.221x10⁻²²g

Volume in cm³:

0.1278nm * (1x10⁻⁷cm / 1nm) = 1.278x10⁻⁸cm

Volume = 8√8 * (1.278x10⁻⁸cm)³

Volume = 4.723x10⁻²³cm³

And density is:

4.221x10⁻²²g / 4.723x10⁻²³cm³ =

Answer:

Eukaryotic (a plant cell)

Explanation:

The presence of a chloroplast indicates that the cell has membrane-bound organelles. This is not a feature of prokaryotic cells - only eukaryotic cells. Therefore, the cell is eukaryotic. The presence of a large vacuole, chloroplast, and rigid cell wall suggests its a plant cell as plant cells are the only eukaryotes with these features.

Answer:

C6H1206

Explanation:

C6H12O6 is a monomer of carbohydrates also known as glucose, so it is not an electrolyte at all.

Answer: The percent yield of this combination reaction is 41.3 %

Explanation : Given,

Mass of [tex]Na[/tex] = 5.0 g

Mass of [tex]Cl_2[/tex] = 10.0 g

Molar mass of [tex]Na[/tex] = 23 g/mol

Molar mass of [tex]Cl_2[/tex] = 71 g/mol

First we have to calculate the moles of [tex]Na[/tex] and [tex]Cl_2[/tex].

[tex]\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}[/tex]

[tex]\text{Moles of }Na=\frac{5.0g}{23g/mol}=0.217mol[/tex]

and,

[tex]\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}[/tex]

[tex]\text{Moles of }Cl_2=\frac{10.0g}{71g/mol}=0.141mol[/tex]

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation will be:

[tex]2Na+Cl_2\rightarrow 2NaCl[/tex]

From the balanced reaction we conclude that

As, 2 mole of [tex]Na[/tex] react with 1 mole of [tex]Cl_2[/tex]

So, 0.217 moles of [tex]Na[/tex] react with [tex]\frac{0.217}{2}=0.108[/tex] moles of [tex]Cl_2[/tex]

From this we conclude that, [tex]Cl_2[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]Na[/tex] is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of [tex]NaCl[/tex]

From the reaction, we conclude that

As, 2 mole of [tex]Na[/tex] react to give 2 mole of [tex]NaCl[/tex]

So, 0.217 mole of [tex]HCl[/tex] react to give 0.217 mole of [tex]NaCl[/tex]

Now we have to calculate the mass of [tex]NaCl[/tex]

[tex]\text{ Mass of }NaCl=\text{ Moles of }NaCl\times \text{ Molar mass of }NaCl[/tex]

Molar mass of [tex]NaCl[/tex] = 58.5 g/mole

[tex]\text{ Mass of }NaCl=(0.217moles)\times (58.5g/mole)=12.7g[/tex]

Now we have to calculate the percent yield of this reaction.

Percent yield = [tex]\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100[/tex]

Actual yield = 5.24 g

Theoretical yield = 12.7 g

Percent yield = [tex]\frac{5.24g}{12.7g}\times 100[/tex]

Percent yield = 41.3 %

Therefore, the percent yield of this combination reaction is 41.3 %

Answer:

Explanation:

Its a elemental potassium is soft ,white  in colour  and has one more electron than argon,an element that we know is extremely stable ... Potassium extra electron is easily lost to form the much more stable cation, K+

Answer:

k Nishant

Explanation:

i don't know sorry but u can search in google

Answer:

Explanation:

Caffeine is a weak base with pKb = 10.4

Kb = 10⁻¹⁰°⁴ = 3.98 x 10⁻¹¹

molecular weight of caffeine = 194.2

455 x 10⁻³ g / L = 455 x 10⁻³ / 194.2 moles / L

concentration of given solution a = 2.343 x 10⁻³ M

Let the caffeine be represented by B .

B    +   H₂O =  BH + OH⁻

a - x                   x        x  

x² / ( a - x ) = Kb

x² / ( a - x ) = 3.98 x 10⁻¹¹

x is far less than a so a -x is almost equal to a

x² = 3.98 x 10⁻¹¹ x 2.343 x 10⁻³ = 9.32  x 10⁻¹⁴

x = 3.05 x 10⁻⁷

[ OH⁻ ] = 3.05 x 10⁻⁷

pOH = - log ( 3.05 x 10⁻⁷ )

= 7 - log 3.05

= 7 - 0.484 = 6.5

pH = 14 - 6.5 = 7.5  

The pH of 455 mg/L of caffeine is 7.5

Using the formula;

Mass concentration = molar concentration × molar mass

Molar mass of caffeine = 194 g/mol

Mass concentration of caffeine = 455 mg/L

Molar concentration = Mass concentration/molar mass

Molar concentration = 455 × 10^-3g/L/194 g/mol

= 0.00235 M

Let Caffeine by depicted by the general formula BH

We can now set up the ICE table as follows;

           :B + H2O   ⇄ BH       +       OH^-

I      0.00235              0                      0

C           - x                  +x                    +x

E    0.00235  - x           x                      x

Note that water is present in large excess

Again; pKb of caffeine =10.4

Kb = Antilog[-pKb]

Kb = Antilog [-10.4]

Kb = 3.98 × 10^-11

Kb = [BH] [OH^-]/[:B]

3.98 × 10^-11 = [x] [x]/[ 0.00235  - x  ]

3.98 × 10^-11 [ 0.00235  - x  ] = [x] [x]

9.4 × 10^-14 - 3.98 × 10^-11x = x^2

x^2 + 3.98 × 10^-11x  - 9.4 × 10^-14 = 0

x = 3.1 × 10^-7 M

Recall  [BH] = [OH^-] = 3.1 × 10^-7 M

Now;

pOH = - log [OH^-]

pOH = log [3.1 × 10^-7 M]

pOH = 6.5

But;

pH + pOH = 14

pH = 14 - pOH

pH = 14 - 6.5

pH = 7.5

The pH of 455 mg/L of caffeine is 7.5

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Missing parts

Caffeine (C8H10N4O2) is a weak base with a pKb of 10.4.  Calculate the pH of a solution containing a caffeine concentration  of 455 mg/L.

Answer:

Ca(ClO₃)₂(s)      →      CaCl₂(s) + 3O₂(g)

Explanation:

Chemical equation:

Ca(ClO₃)₂(s)      →      CaCl₂(s) + O₂(g)

Balance chemical equation:

Ca(ClO₃)₂(s)      →      CaCl₂(s) + 3O₂(g)

Step 1:

Ca(ClO₃)₂(s)      →      CaCl₂(s) + O₂(g)

Left hand side                     Right hand side

Ca = 1                                     Ca = 1

Cl = 2                                     Cl = 2

O =  6                                     O = 2

Step 2:

 Ca(ClO₃)₂(s)      →      CaCl₂(s) + 3O₂(g)

Left hand side                     Right hand side

Ca = 1                                     Ca = 1

Cl = 2                                     Cl = 2

O =  6                                     O = 6

Answer:

ΔU = −55.45 kJ

Explanation:

From first law of thermodynamics in chemistry, we have;

ΔU = Q + W

where;

ΔU is change in internal energy

Q is the net heat transfer

W is the net work done

We are given;

Q = 74.6 kJ

But Q will be negative since heat is released

Thus;

ΔU = -74.6 kJ + W

We are given;

Constant pressure; P = 35 atm = 35 × 101325 = 3546375 N/m²

Volume before reaction; Vi = 8.2 L = 0.0082 m³

Volume after reaction; V_f = 2.8 L = 0.0028 m³

Now,

W = -P(V_f - V_i)

W = - 3546375(0.0028 - 0.0082)

W = 19.15 KJ

Thus;

ΔU = Q + W

ΔU = -74.6 kJ + 19.15 KJ =

ΔU = −55.45 kJ

Answer:

= .7567105263

Explanation:

1 atm = 760 mmHg

575.1 mmHg (1 atm/760mmHg) = .7567105263 atm

Answer:

Explanation:

The percentage error of a certain measurement can be found by using the formula

[tex]P(\%) = \frac{error}{actual \: \: number} \times 100\% \\ [/tex]

From the question

actual density = 2.49g/mL

error = 2.49 - 1.47 = 1.02

We have

[tex]p(\%) = \frac{1.02}{2.49} \times 100 \\ = 40.96385542...[/tex]

We have the final answer as

Hope this helps you

Answer:

1.3x10⁻⁸ mol/L

Explanation:

0.0013μmol, Calculate concentration in mol/L

To obtain concentration in mol/L we need to convert the μmoles to moles and mL to liters:

Moles silver(II) oxide:

0.0013μmol × (1mol / 1x10⁶μmol) = 1.3x10⁻⁹ moles

Liters solution:

100mL * (1L / 1000mL) = 0.1L

That means concentration in mol/L is:

1.3x10⁻⁹ moles / 0.1L =

Answer:

We are given:

mass of the block = 3500 grams

volume of the block = 1000 mL

Finding the density:

Density = mass of the object (in grams) / volume of the object (in mL)

Density = 3500 / 1000

Density = 3.5 g / mL

Answer:

The structure of the atom

Answer:

false you can not get a exact location of electrons from just modern science

Complete Question

The pKb values for the dibasic base B are pKb1=2.10 and pKb2=7.54. Calculate the pH at each of the points in the titration of 50.0 mL of a 0.60 M B(aq) solution with 0.60 M HCl(aq).

(a) before addition of any HCl (b) after addition of 25.0 mL of HCl

Answer:

a The value  is  [tex]pH =12.81[/tex]

b [tex]pH  = 11.9[/tex]

Explanation:

From the question we are told that

  The first pKb value  for B is [tex]pK_b_1  =  2.10[/tex]

   The second pKb value  for B is [tex]pK_b_2  =  7.54[/tex]

     The volume is  [tex]V =   50.0 mL   =[/tex]

     The  concentration  of  B is  [tex][B]  =  0.60 M[/tex]

     The concentration of [tex]C_A =  0.60 M[/tex]

Generally the reaction equation showing the first dissociation of B is  

[tex]\ce{B_{(aq) } + H_2O _{(l)} <=> BH^+ _{(aq)}  +  OH^- _{(aq)} }[/tex]

Here the ionic  constant for B is mathematically represented as

      [tex]K_i  =  \frac{[BH^+] [OH^-]}{[B]}[/tex]

Let denot the concentration of  [BH^+]  as  z  and  since [tex][BH^+] =  [OH^-][/tex] then [tex][OH^-][/tex] is also  z

So  [B] =  0.60  -  z  

Here [tex]K_i[/tex] is ionic constant for the first reaction of a dibasic base B and the value is

   [tex]K_i  =  7.94 *10^{-3}[/tex]

So

      [tex] 7.94 *10^{-3}=  \frac{z^2}{ 0.60 - z}[/tex]

=>   [tex]z^ 2 + 0.00794 z - 0.00476[/tex]

using quadratic formula to solve this equation

     [tex]z = 0.0651[/tex]

Hence the concentration of  [tex]OH^{-}[/tex] is   [tex][OH^-] =0.0651[/tex]

Generally  [tex]pOH =  -log [OH^-][/tex]

=>    [tex]pOH =  -log (0.065)[/tex]

=>    [tex]pOH = 1.187 [/tex]

Generally the pH is mathematically represented as

    [tex]pH = 14 - 1.187[/tex]

      [tex]pH =12.81[/tex]

Generally the volume of [tex]HCl[/tex] at the second dissociation of the base B is   [tex] 50 mL [/tex]

The volume of the [tex]HCl[/tex] half way to the first dissociation of the base is 25mL

Now the pOH at half way to the first dissociation of the base is  

     [tex]pOH  =  -log(K_i)[/tex]

=>   [tex]pOH  =  -log(0.00794)[/tex]

=>   [tex]pOH  =  2.100[/tex]

Generally the pH after addition of 25.0 mL of HCl is  

    [tex]pH  =  14 -  2.100[/tex]\

=>   [tex]pH  = 11.9[/tex]

The first dissociation's equation is as follows:

[tex]B(aq) + H_2O(l) \leftrightharpoons BH^{+} (aq) + OH^{-}(aq) \\\\[/tex]

Constant of base ionization

[tex]\to K_{bl}=\frac{[BH^{+}][OH^{-}]}{[B]}\\\\ \to 7.94\times 10^{-3} = \frac{x\times x}{(0.95- x)} \\\\\to 7.94\times 10^{-3} = \frac{x^2}{(0.95- x)} \\\\\to x^2=7.94\times 10^{-3} (0.95-x) \\\\\to x^2=7.543\times 10^{-3} - 7.94\times 10^{-3} x) \\\\\to x^2=7.543\times 10^{-3} - 7.94\times 10^{-3} x) \\\\\to x = 0.0830\ M\\\\[/tex]

So,

[tex]\to [OH^{-}] = 0.0830\ M\\\\[/tex]

The second dissociation of the base equation is

[tex]BH^{+}\ (aq) + H_20\ (l) \leftrightharpoons BH_2^{2+}\ (aq) + OH^{-}\ (aq) \\\\[/tex]

Constant of base ionization

[tex]\to K_{bl}=\frac{[BH^{+}][OH^{-}]}{[B]}\\\\ \to 3.2 \times 10^{-8} =\frac{y \times (0.0830+y)}{(0.0830- y)}\\\\[/tex]

[tex]\to y= \frac{ 3.2 \times 10^{-8} \times (0.0830- y)}{ (0.0830+y)} \\\\ \to y= \frac{ 3.2 \times 10^{-8} \times (0.0830- y)}{ (0.0830+y)} \\\\ \to y = 3.2\times 10^{-8}[/tex]

So,

[tex]\to [OH^{-}] = 0.0830\ M \\\\\to pOH = 1.08 \\\\\to pH = 14.00 - pOH = 12.92\\\\[/tex]

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Answer:

6.7

⋅

10

23

atoms of H

Explanation:

The heat of vaporization of benzene is required.

The heat of vaporization of benzene is 33009 J/kg.

[tex]T_0[/tex] = Normal boiling point = 80.1+273.15 K

[tex]T_B[/tex] = Boiling point at given pressure = 26.1+273.15 K

[tex]R[/tex] = Gas constant = 8.314 J/mol K

[tex]P[/tex] = Pressure at given [tex]T_B[/tex] = 100 torr

[tex]\Delta H[/tex] = Heat of vaporization

From the Clausius–Clapeyron equation

[tex]\dfrac{1}{T_B}=\dfrac{1}{T_0}-\dfrac{R\ln(\dfrac{P}{P_0})}{\Delta H}\\\Rightarrow \Delta H=\dfrac{R\ln\dfrac{P}{P_0}}{\dfrac{1}{T_0}-\dfrac{1}{T_B}}\\\Rightarrow \Delta H=\dfrac{8.314\times \ln\left(\frac{100}{760}\right)}{\frac{1}{80.1+273.15}-\frac{1}{26.1+273.15}}\\\Rightarrow \Delta H=33008.99\ \text{J/kg}[/tex]

The heat of vaporization of benzene is 33009 J/kg.

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