How many atoms are in 10 g of He

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:

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

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:

C6H1206

Explanation:

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

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:

seafloor spreading

Explanation:

i took the test

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:

48.4 g

Explanation:

Answer:

[tex]Kc=\frac{[SF_6]^8}{[F_2]^2^4}[/tex]

Explanation:

Hello.

In this case, for the undergoing chemical reaction:

[tex]S_8(s) + 24F_2(g) \rightleftharpoons 8SF_6(g)[/tex]

We consider the law of mass action in order to write the equilibrium expression yet we do not include S8 as it is solid and make sure we power each gaseous species to its corresponding stoichiometric coeffient (24 for F2 and 8 for SF6), thus we obtain:

[tex]Kc=\frac{[SF_6]^8}{[F_2]^2^4}[/tex]

Best regards!

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:

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:

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:

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:

A sample of an ideal gas has a volume of 2.21 L at 279 K and 1.01 atm. Calculate the pressure when the volume is 1.23 L and the temperature is 299 K.

You need to apply the ideal gas law PV=nRT

You have the pressure, P=1.01 atm

you have the volume, V = 2.21 L

The ideal gas constant R= 0.08205 L. atm/ mole.K at  273 K

find n = PV/RT = (1.01 atm x 2.21 L / 0.08205 L.atm/ mole.K x 273 K)

n= 0.1 mole, Now find the pressure for n=0.1 mole, T= 299K and

L=1.23 L

P=nRT/V= 0.1mole x 0.08205 (L.atm/ mole.K x 299 k)/ 1.23 L

= 1.994 atm

Explanation:

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: 2681.81

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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Answer:

0.328 mol/dm³

Explanation:

We have

I started this calculation from Rate's law.

Remember equilibrium constant has been given to be 60%

Our interest is Kc, that is the equilibrium constant.

Ca = 4(1-0.6)/1+(-0.5*0.6)

= 4-2.4/1-0.3

= 1.6/0.7

= 2.2857

Cb = 4x0.6/2(1+(-0.5*0.6))

= 2.4/2(0.7)

= 2.4/1.4

= 1.7143

Kc = Cb/Ca²

= 1.7143/2.2857²

= 1.7143/5.2244

= 0.328 mol/dm³

I have added an attachment showing earlier stages to the final answer

Answer:

Phagocytosis

Explanation:

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:

768g

Explanation:

We can use to formula [tex]N(A) = N_0(\frac{1}{2})^\frac{t}{t_{1/2}}[/tex] . Here, N(A) is the final amount. N0 is the initial amount. t is the time elapsed, and [tex]t_{1/2}[/tex] is the half life. Plugging in, we get the answer above.

The half life of I -137 is 8.07 days. if 24 grams are left after 40.35 days, 800 gram  were in the original sample.

The half-life (symbol t12) is the amount of time it takes for a volume (of material) to be reduced to half of its original value. In nuclear physics, the phrase is typically used to indicate how rapidly unstable atoms experience radioactive decay or even how long stable nuclei survive.

The phrase is sometimes used more broadly to describe any form of exponential (or, in rare cases, non-exponential) decay. The biological ½ of medications and other compounds in the human body, for example, is referred to in the medical sciences. In exponential growth, the inverse of half-life is doubling time.

ln P = kt + C

P = amount of I-137 at time t

C = constant

k = 1/time

t = time

1st condition:

P = Po, t = 0 days

2nd condition: (half-life)

P = 0.5Po, t = 8.07 days

3rd condition:

P = 25 grams, t = 40.35 days

Po = 800 grams

mass of I-137 = 800 gram

Therefore, 800 gram were in the original sample.

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