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There are 5.5 L of a gas present at -38.0 C. What is the temperature if the volume of the gas has changed to 1.30 L?

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

Answer:

The structure of the atom

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 number of moles of precipitate, [tex]PbBr_2[/tex] formed will be 0.01 moles.

Explanation : Given,

Concentration of NaBr = 0.200 M

Volume of solution = 100.0 mL = 0.1 L      (1 L = 1000 mL)

First we have to calculate the moles of NaBr.

[tex]\text{Moles of NaBr}=\text{Concentration of NaBr}\times \text{Volume of solution in L}[/tex]

[tex]\text{Moles of NaBr}=0.200M\times 0.1L=0.02mol[/tex]

Now we have to calculate the moles of precipitate, [tex]PbBr_2[/tex] formed.

The balanced chemical reaction is:

[tex]2NaBr(aq)+Pb(NO_3)_2(aq)\rightarrow PbBr_2(s)+2NaNO_3(aq)[/tex]

From the balanced chemical reaction we conclude that:

As, 2 moles of NaBr react to give 1 mole of [tex]PbBr_2[/tex]

So, 0.02 moles of NaBr react to give [tex]\frac{0.02}{2}=0.01[/tex] mole of [tex]PbBr_2[/tex]

Therefore, the number of moles of precipitate, [tex]PbBr_2[/tex] formed will be 0.01 moles.

The number of mole of the precipitate (i.e PbBr₂) formed when 100 mL of 0.2 M NaBr react with excess Pb(NO₃)₂ is 0.01 mole

Volume = 100 mL = 100 / 1000 = 0.1 L

Molarity of NaBr = 0.2 M

Mole = Molarity x Volume

Mole of NaBr = 0.2 × 0.1

2NaBr(aq) + Pb(NO₃)₂(aq) → PbBr₂(s) + 2NaNO₃ (aq)

From the balanced equation above,

2 moles of NaBr reacted to produce 1 mole of PbBr₂.

Therefore,

0.02 mole of NaBr will react to produce = [tex]\frac{0.02}{2} \\\\[/tex] = 0.01 mole of PbBr₂.

Thus, the number of mole of the precipitate (i.e PbBr₂) produced from the reaction is 0.01 mole

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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 =

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:

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: 2681.81

Explanation:

Answer:

A. exist in fixed amounts or available in limited supplies

Explanation:

Answer:

A

Explanation:

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:

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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#SPJ2

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:

= .7567105263

Explanation:

1 atm = 760 mmHg

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

Answer:

k Nishant

Explanation:

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

Answer:

6.7

⋅

10

23

atoms of H

Explanation:

Answer:(trick question) once the salt has dissolve  in the water it is no longer visible

Thank you! have an amazing day.

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:

The correct answer is b.

Explanation:

hope this helps u

Answer:

Bryna is correct it is B

Explanation:

Answer:

[tex]\%\ Error = 21.5\%[/tex]

Explanation:

Given

[tex]Measured = 24.5[/tex]

[tex]Actual = 31.2[/tex]

Required

Determine the percentage error

First, we need to determine the difference in the measurement

[tex]Difference = |Actual - Measured|[/tex]

[tex]Difference = |31.2 - 24.5|[/tex]

[tex]Difference = |6.7|[/tex]

[tex]Difference = 6.7[/tex]

The percentage error is calculated as thus:

[tex]\%\ Error = \frac{Difference * 100\%}{Actual}[/tex]

[tex]\%\ Error = \frac{6.7 * 100\%}{31.2}[/tex]

[tex]\%\ Error = \frac{670\%}{31.2}[/tex]

[tex]\%\ Error = 21.4743589744\%[/tex]

[tex]\%\ Error = 21.5\%[/tex] approximated

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:

Phagocytosis

Explanation:

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:

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